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Both but she is going to get you to pick just one so it was more economical. “and they [Yankees] are marked ... with such a perversity of character, as to constitute, from that circumstance, the natural division of our parties” Thomas Jefferson In the 1770s, the South had every reason to continue the relationship with England, one of its best customers. It was the manufacturing North that was getting the short end of that stick. Southerners joined the Revolutionary War out of patriotism, idealism, and enlightened political philosophy such as motivated Jefferson, not patriotism, philosophy, and economic betterment which inspired the North. In 1860, the shoe was on the other foot. Southern agrarians were at heel to the nation's bankers and industrialists. That just got worse with the election of the Republican Lincoln, bringing back into power the party favoring the wealthy supply side, as it still does. Then as now central to that, party's interest was keeping down the cost of manufacture. Today labor is the big cost, so today they move the plants offshore and leave US workers to their fate. Back before the US labor movement existed the big cost was raw materials, and the GOP was just as unprincipled toward its Southern suppliers as it is today toward labor. Thanks to modern graveyard science and surviving records, researchers know that in 1760, 100 years before the War Between the States, Charleston, South Carolina, had the largest population of slaves and we say proudly the second largest slave population was in New York City. One of the main quarrels was about taxes paid on goods brought into this country from foreign countries. This tax was called a tariff. Southerners felt these tariffs were unfair and aimed toward them because they imported a wider variety of goods than most Northern people. Taxes were also placed on many Southern goods that were shipped to foreign countries, an expense that was not always applied to Northern goods of equal value. An awkward economic structure allowed states and private transportation companies to do this, which also affected Southern banks that found themselves paying higher interest rates on loans made with banks in the North. As industry in the North expanded, it looked towards southern markets, rich with cash from the lucrative agricultural business, to buy the North's manufactured goods. The situation grew worse after several "panics", including one in 1857 that affected more Northern banks than Southern. Southern financiers found themselves burdened with high payments just to save Northern banks that had suffered financial losses through poor investment. However, it was often cheaper for the South to purchase the goods abroad. In order to "protect" the northern industries Jackson slapped a tariff on many of the imported goods that could be manufactured in the North. When South Carolina passed the Ordinance of Nullification in November 1832, refusing to collect the tariff and threatening to withdraw from the Union, Jackson ordered federal troops to Charleston. A secession crisis was averted when Congress revised the Tariff of Abominations in February 1833. The Panic of 1837 and the ensuing depression began to gnaw like a hungry animal on the flesh of the American system. The disparity between northern and southern economies was exacerbated. Before and after the depression the economy of the South prospered. Southern cotton sold abroad totaled 57% of all American exports before the war. The Panic of 1857 devastated the North and left the South virtually untouched. The clash of a wealthy, agricultural South and a poorer, industrial North was intensified by abolitionists who were not above using class struggle to further their cause. In the years before the Civil War the political power in the Federal government, centered in Washington, D.C., was changing. Northern and mid-western states were becoming more and more powerful as the populations increased. Southern states lost political power because the population did not increase as rapidly. As one portion of the nation grew larger than another, people began to talk of the nation as sections. This was called sectionalism. Just as the original thirteen colonies fought for their independence almost 100 years earlier, the Southern states felt a growing need for freedom from the central Federal authority in Washington. Southerners believed that state laws carried more weight than Federal laws, and they should abide by the state regulations first. This issue was called State's Rights and became a very warm topic in congress. God Bless You and The Southern People.
Josey Wales - the united kingdom political gadget is lots from failed as you proclaim. we've been a civilised united states of america for 2000 years, no longer 500 like u . s . of america. you have not got any historic previous, no traditions and no lifestyle and your gadget of Capitalism has infrequently been a roaring achievement because of the fact it became into u . s . of america and it relatively is greed that led to the international recession that we are all suffering. you have probably infrequently ever been out of your individual united states of america and perchance no longer even been to Britain so how could you already know what this united states of america is like? We have been an empire long in the previous your united states of america existed and ruled the worldwide with socialist recommendations and a feeling of justice which the full worldwide accompanied as a stable occasion. the adaptation between the Tea party and Mao's Cultural Revolution is easy - the Tea party contributors are individuals, on a similar time as the Maoists have been chinese language. different than for that they the two needed political substitute interior the rustic on opposite aspects of the political spectrum.
It was a bit of both really, but I think mostly economical. The South had an economy based on agriculture (think of the slaves picking cotton and tobacco), where the North had a more industrial economy with factories and such.
Like most things in life it was all about the money.
Http://en.wikipedia.org/wiki/Origins_of_...
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4b-3(b+1)-b=2(b-1)-2b 4b-3b-3-b=2b-2-2b 4b-4b-3=2b-2b-2 -3=-2 There is no possible value for b
This question is a bit wierd but this is how it works.you have to multiply everything in the brackets by the number infront -3(b+1)= -3 xb +-3 x1= -3b -3 4b-3(b+1)-b=2(b-1)-2b 4b -3b + -3 -b=2b -2 -2b -3 = -2 ?? this question is wrong because the sides arent equal. look y its wrong: i simplyfied it to this 4b -3b + -3 -b=2b -2 -2b now i take the bs away and nothing is left 4b-3b= b b-b=0 -3 left now the other side 2b-2b=0 -2 left -3 = -2 XwrongX this = means equal and -3 isnt the same as -2 i hope u know that. btw dont listen to bobby, he probably doesnt know a thing about maths. im best in my class and my maths teacher has a master degree. there is no c in this equation.
B=c (abc's simple)
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Recall P(R | S) = P(R intersection S) / P(S). So the left hand side can be written as: P((A intersection B) intersection (B union C)) / P(B union C) = P(A intersection B) / P(B union C) (*) Now the right hand side equals: P((A intersection B) intersection B) / P(B) times P(B intersection (B union C)) / P(B union C) = P(A intersection B) / P(B) times P(B) / P(B union C) = P(A intersection B) / P(B union C) (**). Clearly (*) and (**) are the same expression, hence the left hand side equals the right hand side.... You should go ahead and draw some Venn diagrams for checking why for instance (A intersection B) intersection (B union C) = (A intersection B), etc. The fact that none of the probabilities is zero insures we won't be dealing with fractions with zero denominators.
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B = -4(c - 2d) B = -4c + 8d B + 4c = 8d (B + 4c) / 8 = d
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B.C. stands for "Before Christ". But when the term was invented, they got the year wrong, so now we think that Christ was born in the year 4 B.C. 1450 BC to 1410 BC is 40 years 1050 BC to 465 BC is 585 years 1000 BC to 300 BC is 700 years.
B.C. -> Before Christ 1450 - 1410 B.C. -> 2860y 1050 - 465 B.C. -> 1515y 1000 - 300 B.C. -> 1300y
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Sec(b)^2 - tan(b)^2 = 1 sec(b)^2 = 1 + tan(b)^2 sec(b)^2 = 1 + 5.76 sec(b)^2 = 6.76 sec(b) = -2.6 , 2.6 Since we're in Q3, then sec(b) is negative sec(b) = -2.6 sec(b) = -13/5 cos(b) = -5/13 sin(b)^2 + cos(b)^2 = 1 sin(b)^2 + 25/169 = 19/169 sin(b)^2 = 144/169 sin(b) = -12/13 , 12/13 Once again, since we're in Q3, sin(b) < 0 sin(b) = -12/13
Tan B = 2.4 tan B = 24/10 =12//5 = -12/-5 but, tan θ = o/a thus, if we let B = θ, for 180˚ =< θ =< 270˚ o = -12 and a = -5 using the relationship o² + a² = h², find h h² = o² + a² h² = (-12)² + (-5)² h² = 144 + 25 h² = 169 h = 13 Using the relationship sin θ = o/h, then, sin B = -12/13 Using the relationship cos θ = a/h, then, cos B = -5/13
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Discrete Maths Showing all your working and giving a reason for each step, use the laws of set algebra (set identities) to simplify (A u B)' u (A u B') Where: u is union ' is complement ^ = Intersection {} = Non-empty set My working, not sure if I'm right: = A' ^ B' u (A u B') De Morgans = A' ^ B; u (B' u A) Commutive Law = A' ^ (B' u B') u A Associate Law = A' ^ B' u A Idempotent = A' ^ A u B Commutive Law = {} u B Domination = B Identity Law Need help urgently! :(
Hi = (A' ^ B') u (A u B') De Morgans = (A' ^ B') u (B' u A) Commutative Law = ((A' ^ B') u B') u ((A' ^ B') u A) Distributive Law = ((A' u B') ^ (B' u B')) u ((A' u A) ^ (B' u A)) Distributive Law = ((A' u B') ^ B') u ((A' u A) ^ (B' u A)) Idempoten = ((A' u B') ^ B') u (U ^ (B' u A)) Complement Law = ((A' u B') ^ B') u (B' u A) Identity Law = ((A' u B') u (A u B')) ^ (B' u (B' u A) Distributive Law = ((A' u B') u (A u B')) ^ ((B' u B') u A) Associative Law = ((A' u B') u (A u B')) ^ (B' u A) Idempotent = [ (A' u (A u B')) u (B' u (A u B'))] ^ (B' u A) Distributive Law = [ (A' u A) u B') u (B' u (A u B'))] ^ (B' u A) Associative Law = [ (U u B') u (B' u (A u B'))] ^ (B' u A) Complement Law = [ U u (B' u (A u B'))] ^ (B' u A) Absortion Law = U ^ (B' u A) Absortion Law = B' u A Identity Law = A u B' Commutative Law ********************************
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N(A U B U C ) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(A ∩ C) + n (A ∩ B ∩ C) 50 = 25+21+21 - 7 - 5 - 8 + n (A ∩ B ∩ C) n (A ∩ B ∩ C) = 3
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There are at least three ways to solve this. Let them equal each other. Subtraction/Addition. Or Matrix reduction. I will go over the first two. a + 2b = 4 2a - b = -2 Now, in order to solve this by Subtraction/Addition, you must multiply one equation so that a variable will cancel. We can do this by multiplying the second by 2 and eliminate the b's, or the top by -2 to eliminate the a's and still work with whole numbers. So: a + 2b = 4 2*( 2a - b = -2) => 5a - 0 = 0 So, a=0 From this we can then substiude 0 for a and get b = 2 Another way is to solve for one variable in terms of the other. The first will become a = 4 - 2b and the second will be a = (-2 + b)/2 Now, we can let the two a values equal each other and we will get: 4 -2b = (-2 + b)/2 => 8 -4b = -2 +b => 10 = 5b => 2 = b Then subsititude this value in to either original equation and we will find that a = 0
Using the 2nd equation.. 2a-b=-2 .... add 2 to both sides and add b 2a+2=b Now we have solved for b so we plug it in to the first equation a+2(2a+2)=4 a+4a+4=4 5a=o a=o Now we plug a into the second equation 0-b=-2 b=0+2 .... b=2 You can then proceed to check your answers
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A=1/2h(B+b) Assuming A=(1/2)h(B+b) and not A=1/(2h(B+b)) 2A = h(B + b) 2A = hB + hb 2A - hb = hB B = (2A - hb)/h or B = (2A/h) - b
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P(A|B) P(B) = P(A and B) => P(A|B) = P(A and B) / P(B) thus P(A|B ') = P(A and B ' ) / P(B ') P(B|A ') = P(B and A ' ) / P(A ') and P(A ') = 1 - P(A) = 0.6 and P(B ' ) = 1 - P(B) = 0.5 P(A) = P(A and B ') + P(A and B ) => P(A and B ' ) = P(A) - P(A and B) => P(B and A ' ) = P(B) - P(A and B) and P(A and B) = P(A) P(B) if the events are independant
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First, get everything out of the denominator: a(bx-1) + b(ax-1) = (a+b)(ax-1)(bx-1) 2abx - a - b = (a+b)(abx^2 -(a+b)x + 1) (a+b)ab x^2 - (a+b)^2 x + (a+b) = 2ab x - (a+b) (a+b)ab x^2 - ((a+b)^2 + 2ab) x + 2(a+b) = 0 You can now solve for x with the quadratic equation: x = [(a+b)^2 + 2ab) +/- √((a+b)^2 + 2ab)^2 - 4((a+b)ab)(2(a+b))] / 2(a+b)ab x = 1/2a + 1/2b + 1/(a+b) +/- [√((a+b)^4 + 4ab(a+b)^2 + 4a^2b^2) - 8ab(a+b)^2] / 2(a+b)ab x = 1/2a + 1/2b + 1/(a+b) +/- [√((a+b)^4 - 4ab(a+b)^2 + 4a^2b^2)] / 2(a+b)ab x = 1/2a + 1/2b + 1/(a+b) +/- [√((a+b)^2 - 2ab)^2] / 2(a+b)ab x = 1/2a + 1/2b + 1/(a+b) +/- ((a+b)^2 - 2ab) / 2(a+b)ab x = 1/2a + 1/2b + 1/(a+b) +/- [1/2a + 1/2b - 1/(a+b)] x = 1/a + 1/b OR 2/(a+b) We can now go back to the original equation, regroup and solve for b: a(bx-1) + b(ax-1) = (a+b)(ax-1)(bx-1) ax b - a + ax b - b = (ax-1) (x b^2 + ax b - b - a) (2ax-1) b - a = (ax^2 - x) b^2 + (ax-1)^2 b - a(ax-1) (ax^2 - x) b^2 + [(ax-1)^2 - (2ax-1)] b - a(ax-1) + a = 0 (ax^2 - x) b^2 + [(ax-1)^2 - (2ax-1)] b - a(ax-2) = 0 Again, solve with the quadratic equation. You'll get a similar equation for a - I'll leave these last two solutions to you.
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(t-b)e^-b = t e^-b - b e^-b Integrate t e^-b db = -t e^-b ----(1) Integrate b e^-b db by parts dv=e^-b v=-e^-b u=b du= db ∫ u dv = u v - ∫ v du ∫ b e^-b db = -be^-b + ∫e^-b db = -b e^-b + e^-b ------(2) (1)+(2) = -t e^-b -b e^-b + e^-b +C
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A. Let A = (a_ij) and B = (b_ij). Let A + B = (c_ij). Then c_ij = a_ij + b_ij. Of course, A and B need to be of the same dimensions (ex: Both are (n x n)). As A is diagonal, then a_ij = 0, ∀ i ≠ j. As B is diagonal, then b_ij = 0. ∀ i ≠ j. Then, a_ij + b_ij = 0, ∀ i ≠ j. Hence c_ij = 0, ∀ i ≠ j. And so, (A + B) is diagonal. B. If A and B are symmetric, prove that A + B is symmetric*. A and B are symmetric, so A = t(A) and B = t(B). (Definition of symmetry). But t(A + B) = t(A) + t(B) (Should be a stated somewhere in your textbook). Hence t(A + B) = A + B. Therefore A + B is symmetric.
Well A. let n be the dimension of the matrices (will not be used, nevertheless, whenever we say for any i and j, this means : i and j from 1 to n) and let be : A=(a_ij) and B = (b_ij) two diagonal matrices then : a matrix is diagonal iif for any indexes i ≠ j : a_ij = 0 now let be C = A + B = (c_ij) then for any i and j : c_ij = a_ij + b_ij and as A and B are diagonal, for any i and j so that : i ≠ j we have : a_ij = 0 and b_ij = 0 therefore : for any i and j so that : i ≠ j, c_ij = 0 finally : C = A + B is diagonal B. a matrix A = (a_ij) is symmetric iif a_ij = a_ji (can be summarized by : A^T = A wher A^T is the transposed matrix) so let A and B symmetric and C = A + B = (c_ij) means that for any i and j (from 1 to n) c_ij = a_ij + b_ij then : c_ji = a_ji + b_ji = a_ij + b_ij because A and B are symmetric = c_ij therefore : C=A+B is symmetric hope it' ll help !!
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F(b) is base 8 meaning it has an exponent of 8 to the power b. so f(b) = 8^b = 3804 then b = ln(3804) / ln(8) =3.964433844 b. f(10) = 8 ^10 = 1,073,741,824 / f(16) = 8^16 = 2.814749767 * 10^14 c. the integer values of b are not bounded, so any integer value is okay but the result as b gets larger (or approaches infinity) our answer will go to infinity. the only catch is if this function means it is in octal numbers then the answer would be totally different: octal means our way of counting is: 1, 2, 3, 4, 5, 6, 7, 11, 12, 13,14,15,16,17,21,.... then f(b)=3804 = 3*b^3 + 8*b^2 + 0*b^1 + 4*b^0 3*8^3 + 8*8^2 +0*8^1 + 4*8^0 = 2052
F(b) is supposed to be a function of b, but I don't see any b on the RHS. 3804base8 gives you a real value, so f(b) is always the same number. Perhaps you put 8 instead of b, I don't know, but you need a b somewhere on the RHS.
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Let d = (a, b) and e = (a, b - ta). (i) Since d = (a, b), we have d | a and d | b. Thus, d | (-ta) and so d | (b - ta). Since d | a and d|(b - ta), we have that e | d. (ii) Since e = (a, b - ta), we have e | a and e | (b - ta). So, e | ta ==> e | (ta + (b - ta)) = b. Since e | a and e | b, we have that d | e. By (i) and (ii), we conclude that d = e (since the gcd is positive). I hope this helps!
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I'm trying to solve this limit and I know that it is 0. My teacher rewrote the expression ln(b)-(1/2)*ln((b^2)+1) in a certain way so that when you put in infinity for b, the answer was zero, but i don't remember how he did it. So far I have ln(b)/(ln(b^2+1)^1/2) but I don't know if I'm doing it right. Thank you.
Lim(b→∞) [ln b - (1/2) ln(b^2 + 1)] = lim(b→∞) (1/2) [2 ln b - ln(b^2 + 1)] = lim(b→∞) (1/2) [ln(b^2) - ln(b^2 + 1)] = lim(b→∞) (1/2) ln(b^2/(b^2 + 1)) = (1/2) ln(1), since lim(b→∞) b^2/(b^2 + 1) = 1 by L'Hopital's Rule or otherwise = 0. I hope this helps!
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You can rearrange b^x as e^(x ln b) using the laws of logarithms ie. e^(x ln b) = e^(ln b^x) = b^x The general rule for differentiating e^(f(x)) is that it equals f '(x) e^ (f(x)) So (d/dx) (b^x) = (d/dx) [e^ (x ln b)] = ln b [e^ (x ln b)] = ln b [b^x] which is a multiple of b^x. So it is proportional. So answer C. :)
The derivative is (b^x)*log(b)...so id say c.
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If A, B are independent, then P(A and B) = P(A)P(B) = (0.625)(0.8) = 0.5
The probability of both A and B happening is them multiplied together. .5 is the answer
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X/a + y/b = 1 P = (a , 0) Q = (0, b) g = m = y2 - y1/x2 - x1 = - b/a PQ = √ (a^2+b^2) = 20 gradient = -b/a = - 1 a = b √ (a^2+b^2) = 20 a^2+b^2 = 400 (2a^2) = 400 a^2 = 200 a = √ (200) a = b = 10√2
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B^c means the complement of B. P(B^c) means the probability of B not happening. So P(A|B^c) is the conditional probability that A happens, given that we know B doesn't happen. P(B^c) = 1 - 0.6 = 0.4 P(A and B^c) = P(A|B^c) x P(B^c) = 0.1 x 0.4 = 0.04 P(A and B) = P(A|B) x P(B) = 0.3 x 0.6 = 0.18 P(A) = P(A and B^c) + P(A and B) = 0.04 + 0.18 = 0.22 P(B|A) = P(A and B) / P(A) = 0.18 / 0.22 = 9/11 or 0.818 to 3 d.p.
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Since a and b are real, we have limsup a_n <= limsup(a_n + b_n) + limsup(-b_n) = a + b - liminf b_n <= a + b - b = a Hence, a <= liminf a_n <= limsup a_n <= a, so that liminf a_n = limsup a_n = lim a_n = a. It's immediate this implies lim_n = b. For the second part, put a_n = n, b_n = (-1)^n, n = 1, 2, 3 .... Then, a = oo, b= -1 lim (a_n + b_n) = oo = a + b liminf a_n = a liminf b_n = b But (b_n) doesn't have a limit.
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This site--http://www.webref.org/anthropology/b/b_p... showed the letters "B.P." are used in radiocarbon dating --like with fossils, etc--and means "Before Present", so 1250 B.P. would be the year 759 A.D (or C.E.--current/Christian Era)..
Where did you find this date? Commonly, dates are AD and BC, or CE and BCE.
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(a + b - c)(a + b - c)(a + b - c) = (a(a + b - c) + b(a + b - c) - c(a + b - c))(a + b - c) = (a^2 + ab - ac + ab + b^2 - bc - ac - bc + c^2)(a + b - c) = (a^2 + b^2 + c^2 + 2ab - 2ac - 2bc)(a + b - c) = a^2(a + b - c) + b^2(a + b - c) + c^2(a + b - c) + 2ab(a + b - c) - 2ac(a + b - c) - 2bc(a + b - c) = a^3 + a^2 b - a^2 c + a b^2 + b^3 - b^2 c + a c^2 + b c^2 - c^3 + 2a^2 b + 2a b^2 - 2abc - 2a^2 c - 2abc + 2a c^2 - 2abc - 2b^2 c + 2b c^2 = a^3 + b^3 - c^3 + 3a^2 b - 3a^2 c + 3a b^2 - 3b^2 c + 3a c^2 + 3b c^2 - 6abc
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D First use the sum and difference formulas for Sine. sin(a+b) = sin(a)cos(b) + cos(a)sin(b) sin(a-b) = sin(a)cos(b) - cos(a)sin(b) Plugging these in we have: [sin(a)cos(b) + cos(a)sin(b)] / [sin(a)cos(b) - cos(a)sin(b)] Divide the numerator and denominator by sin(a)cos(b) and remember sin(a) / cos(a) = tan(a) cos(a) / sin(a) = cot(a) = 1 / tan(a) After dividing the top and bottom by sin(a)cos(b) we have: [1 + cot(a)tan(b)] / [1 - cot(a)tan(b)] using the identity cot(a) = 1 / tan(a) this becomes [1 + (tan(b) / tan(a))] / [1 - (tan(b) / tan(a))] Now we get a common denominator of tan(a) on the top and bottom. Separately the top and bottom look like this: 1 + [tan(b) / tan(a)] = [tan(a) + tan(b)] / tan(a) (Numerator) 1 - [tan(b) / tan(a)] = [tan(a) - tan(b)] / tan(a) (Denominator) Canceling the common tan(a) in the denominators of each fraction we end with [tan(a) + tan(b)] / [tan(a) - tan(b)]
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A²/(a+1) + b²/(b+1) = (a²−1)/(a+1) + (b²−1)/(b+1) + 1/(a+1) + 1/(b+1) = a−1 + b−1 + 1/(a+1) + 1/(b+1) = −1 + 1/(a+1) + 1/(b+1) … (i) By Cauchy’s inequality (1/(a+1)+1/(b+1))(a+1+b+1) ≥ (1+1)² → 1/(a+1)+1/(b+1) ≥ 4/3 ( this is valid provided a>−1, b>−1 so the √s are OK ) ∴ from (i) a²/(a+1) + b²/(b+1) ≥ −1+4/3 = ⅓ There is equality when (a+1) / 1/(a+1) = (b+1) / 1/(b+1) → a=b → a=b=½
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B^2+6b=0 b(b+6)=0 b=0 and -6
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(b, a) f(x) maps a -> b. f inverse maps b -> a
A. graph translated 7 contraptions to the impressive, so the factor (5,13) might desire to be on the graph. b. graph translated 6 contraptions down, so the factor (-a million,6) might desire to be on the graph. c. graph translated 2 contraptions to the left and 3 contraptions up, so the factor (3,9) might desire to be on the graph.
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(A-B)-C = (A-C) - (B-C) x is in (A-B)-C <=> x is in A-B and x is not in C <=> x is in A, x is not in B and x is not in C. <=> x is in A-C and x is not in B Suppose x is in B-C, then x is in B, which is a contradiction, so x is not in B-C. => x is in A-C and x is not in B-C <=> x is in (A-C) - (B-C) This shows that (A-B)-C is a subset of (A-B)-(B-C). To show that (A-B)-(B-C) is a subset of (A-B)-C, note that all the above steps are reversible except for one, so we have x is in (A-B)-(B-C) <=> x is in A-C and x is not in B-C <=> x is in A, x is not in C, and x is not in B-C. Suppose that x is in B. Since x is not in C, then x is in B-C, which is a contradiction. Therefore x is not in B => x is in A, x is not in C, and x is not in B. <=> x is in (A-B)-C. This shows that (A-B)-(B-C) is a subset of (A-B)-C. Since the LHS is a subset of the RHS and the RHS is a subset of the LHS, they're equal. Q.E.D.
Maybe we can try to solve this by taking an example: Let A = {1,2,3,4,5} B = {3, 6, 8} C = {1, 2, 3} Now, A-B = {1, 2, 4, 5} (A-B)-C = {4,5} = LHS A-C = {4, 5} B-C = {6, 8} RHS = (A-C)-(B-C) = {4 , 5} Therefore, LHS = RHS
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(b^2)^6=b12 (multiply the exponents 2 and 6). It's redundant to say b^-4, because all exponential numbers in the denominator are automatically negative. So, you have b^12/b^4. Divide the numerator and denominator by their Greatest Common Factor...in this case, 4. b^12 divided by 4 is b^3, while b^4 divided by 4 is b. You should have b^3/b. Now, divide numerator and denominator by b (so that you can get the simplest answer). b^3 divided by b is b^2, and b divided by b is one. You get b^2/1, or plain old b^2 as your final answer.
Again -- I am AMAZED at what they are teaching (or not!) in Algebra I. Here ya go -- Multiply the exponents in the numerator together: b^(2*-6) = b^(-12) Now subtract the exponents (top minus bottom): b^(-12 - (-4)) = b^(-8) Now change to a positive exponent by forming the reciprocal and changing the exponent to +8: 1/b^8 So the answer is C. All these steps are properties of exponents that can be found in your text. Good luck!
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Let a, b be odd with a > b so a = 2n + 1 b = 2m +1 with n > m Let c = gcd (a,b) = gcd (2n+1, 2m+1) then a = 2n +1 = p c for some p b = 2m+1 = q c for some q with p > q, p,q has no common factors, hence cannot be both even, p-q must be odd (a-b)/2 = (n - m) = (p -q)/2 now (a-b)/2 = n - m = (p - q) c/2 so gcd (a-b/2, b) = gcd ( (p-q) c , qc) = c since p, q has no common factors, (p-q)/2 and q has no common factors! gcd( a , b ) = gcd ( [a-b]/2 , b)
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Angle A + angle B + angle C = 180 => A + B + C = 180 => A = 180 - ( B + C ) =>applying cos on both sides cos A = cos ( 180 - ( B + C ) ) => cos A = - cos ( B + C ) .....................(1) => cos A + cos ( B + C ) = 0 cos A + cos ( B - C ) = - cos ( B + C ) + cos ( B - C ) => - cos B cos C + sin B sin C + cos B cos C + sin B sin C => 2 sin B sin C
This could have a given assertion as, A,B,C are the angles of a triangle. So, A+B+C=a hundred and eighty A=a hundred and eighty-(B+C) cos (-A+B+C) +cos(A-B+C) +cos(A+B-C) +cos(A+B+C) = cos(-A+a hundred and eighty-A)+cos(A-a hundred and eighty-A)+cos(a hundred and eighty-C-C)+... And carry on with alterations from now. stable success.
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By Fermat's little theorem a^p = a mod p and b^q = b mod q for all a and b. If a^p = b^q mod pq then a^p = b^q mod p ==> a = b^q mod p ==> b = b^q mod p (since a = b mod p by assumption), but this last statement is generally false. In fact I can select b such that b <> b^q mod p if q <>p. For example take p = 11, q = 5 and a = b = 2. Then 2^5 <> 2^11 mod 55 just compute 2^11 mod 55 = 13 <> 32 = 2^5 mod 55.
Take help of math teacher.
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For any function f, we automatically have f(A ∩ B) ⊂ f(A) ∩ f(B). [If y ∈ f(A ∩ B), then y = f(x) for some x ∈ A ∩ B. ==> y ∈ f(A) ∩ f(B).] If f(A ∩ B) and f(A) ∩ f(B) are not equal, then there exists some y ∈ f(A) ∩ f(B) which is not in f(A ∩ B). So, for some a ∈ A and b ∈ B with y = f(a) = f(b) and for all x ∈ A ∩ B, we have y ≠ f(x). Hence, a ≠ b and f is not an injection, which is a contradiction. -------------------- Now assume that f(A ∩ B) = f(A) ∩ f(B). To show that f is injective, suppose to the contrary that f is not injective. Then, there exist x ≠ y such that f(x) = f(y). Let A = {x} and B = {y}. ==> f(A) ∩ f(B) = {f(x)}, but f(A ∩ B) = ∅. This contradicts the hypothesis that f(A ∩ B) = f(A) ∩ f(B). -------------------- I hope this helps!
I'm undecided besides the incontrovertible fact that it variety of feels there could desire to be. Do re mi fa sol l. a. si do is a level of vocalizing an escalating scale. The letters a by way of g are notes of the main suitable scale. thinking the 1st and final vocalization is "do", and the 1st and final notice of a scale are the comparable, i could propose that perhaps this trend is utilized while vocalizing a scale.
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Since A is in Quadrant II, cos(A) < 0 and tan(A) < 0; since B is in Quadrant III, cos(B) < 0 and tan(B) > 0. By drawing 3-4-5 triangles such that sin(A) = 3/5 and sin(B) = -4/5, we see that: (a) For angle A: opp = 3, adj = 4, and hyp = 5 ==> cos(A) = -adj/hyp = -4/5 and tan(A) = -opp/adj = -3/4. (b) For angle B: opp = 4, adj = 3, and hyp = 5 ==> cos(B) = -adj/hyp = -3/5 and tan(B) = opp/adj = 4/3. Using these values and the sine, cosine, and tangent addition formulas, you can calculate sin(A + B), cos(A + B), and tan(A + B). Using the signs of these functions, you can determine the quadrant in which A + B terminates.
If triangle is given 2 lengths and a million perspective interior the format of SSA (geometry's area-area-perspective), then your answer is ambiguous because of the fact there could nicely be 2 techniques, particularly than purely one whilst utilising the regulation of Sines. remember, SSA spelled backwards isn't an excellent be conscious. subsequently, you will opt to apply the regulation of Cosine formulation which you wrote 2d. If the triangle is given 2 lengths and a million perspective interior the format of SAS, then you definately can nevertheless use the 1st formulation (regulation of Sines).
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We will use Extensionality. First, we must prove that every member of ( A - (B∩C) ) is a member of (A-B)∪(A-C) , and vice versa. Let D be an arbitrary object (→) Assume D∈ ( A - (B∩C) ) . Then d∈A and D not∈ (B∩C). So , ¬(D∈B&D∈C). By De Morgans law , it follows that D not∈ B or D not∈ C Case 1. D not∈ B. In this case, D∈A and D not∈ B, so D ∈ A-B. By disjunction introduction, ( D ∈ A-B∨D∈ A-C ). So, D ∈ (A-B)∪(A-C). Case 2. D not∈ C. Similar to case 1. Thus, D∈ (A-B)∪(A-C). (←) Assume D ∈ (A-B)∪(A-C). So, either D∈ (A-B) or D ∈ (A-C). Case 1. D∈ (A-B). Thus, D∈A and D not∈ B. Trivially, since D not∈ B. ¬(D∈B&D∈C). So, D not∈(B∩C). Since D∈A and D not∈(B∩C) We have that D∈( A- (B∩C) ) Case 2. D∈ (A-C). Similar to case 1. Thus, D∈ ( A- (B∩C) )
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1) 15b=10a, so a/b=15/10=3/2. 2) Divide by b then multiply by c: a/b=c/d. 3) I don't think you can find a closed ratio for this one.
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-(a+b)³=-(a+b)*(a+b)^2 =-(a+b)*(a^2+2ab+b^2)= -(a^3+2*a^2*b+a*b^2+b*a^2+2*a*b^2+b^3)... -(a^3+2*a^2*b+3*a*b^2+3*b*a^2+b^3) (a-b)³=(a-b)*(a-b)^2 =(a-b)*(a^2-2ab+b^2)= (a^3-2*a^2*b+a*b^2-b*a^2+2*a*b^2-b^3)= (a^3-3*a^2*b+3*a*b^2-b^3)
(a+b) (a+b) (a+b) i just did that in geometry class!!! :]
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B.ed through correspondence may not be considered as quality education in future..so its good to do b.ed in regular...all the best
You Can get the list of B.Ed colleges through correspondence from http://www.way2college.com/B-EdColleges.php
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A) if a<b, then a by definition will be some number less than b. Since a and b must be integers, the largest integer a can be would be b-1 (the integer immediately preceding b). So a can be b-1 or any integer less than that, so a<=b-1. B) If a<=b-1, then a<b. To contradict: assume a is not less than b. So we'll assume a=b. If a=b, then a<=b-1 is incorrect (because this implies a<=b-1<b, or a<=b-1<a, which is obviously a contradiction), so the assumption is false. If we assume a>b (the only other way to contradict a<b), then again a<=b-1 cannot be true (because this means a<=b-1<b<a, again a contradiction). So this assumption is false as well, and it follows that a<b.
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B=3a...............eq1 a+b=-1............eq2 a+3a=-1..........sub b=3a in eq2 4a=-1 a=-1/4.............sub a=-1/4 in eq1 or eq2 b=-3/4
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Assume by contradiction that (A - B) n (B - A) is nonempty. Thus, there exists an x in (A - B) n (B - A). Then x is in A - B and x is in B - A. Since x is in A - B, we see that x is in A and x is not in B. Since x is in B - A, we see that x is in B and x is not in A. Thus, x is in A and x is not in A, a contradiction. Therefore, (A - B) n (B - A) is the empty set. Thus, A - B and B - A are disjoint.
It is because A-B and B-A yield sets that do not overlap/intersect. Their union is simply A-B and B-A. See this website for the exact proof. http://cnx.org/content/m15198/latest/
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So ill split into sections 1st section (a+b)^2 (a+b)(a+b) a^2 + ab+ ab + b^2 a^2 +2ab + b^2 2nd section (25a +b - 5)^2 (25a + b - 5) (25a + b - 5) 625a^2 +25ab -125a +25ab+b^2 -5b -125a - 5b +25 625a^2 +50ab - 250a +b^2 -10b +25 3rd section (49a+b−10)^2 (49a+b−10)(49a+b−10) 2401a^2 + 49ab - 490a + 49ab + b^2 - 10b - 490a -10b +100 2401a^2 +98ab - 980a + b^2 - 20b +100 so adding section 1st and 2nd - this is 4th section a^2 +2ab + b^2 + 625a^2 +50ab - 250a +b^2 -10b +25 =626a^2 + 52ab - 250a +2b^2 -10b +25 so adding 4th and 3rd section 626a^2 + 52ab - 250a +2b^2 -10b +25 + 2401a^2 +98ab - 980a + b^2 - 20b +100 = 3027a^2 +150ab - 1230a +3b^2 - 30b +125 currently working on how to factor this.
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I am assuming you wish to prove that: sec(A - B) = cos(A + B)/[cos^2(A) - sin^2(B)]. By converting the LHS into the RHS LHS = sec(A - B) = 1/cos(A - B) = 1/[cos(A)cos(B) + sin(A)sin(B)], by the cosine subtraction formula. Then, multiplying the numerator and denominator by cos(A)cos(B) - sin(A)sin(B): 1/[cos(A)cos(B) + sin(A)sin(B)] = [cos(A)cos(B) - sin(A)sin(B)]/[cos^2(A)cos^2(B) - sin^2(A)sin^2(B)] = [cos(A)cos(B) - sin(A)sin(B)]/{cos^2(A)[1 - sin^2(B)] - sin^2(B)[1 - cos^2(A)]} = [cos(A)cos(B) - sin(A)sin(B)]/[cos^2(A) - cos^2(A)sin^2(B) - sin^2(B) + cos^2(A)sin^2(B)] = [cos(A)cos(B) - sin(A)sin(B)]/[cos^2(A) - sin^2(B)] = cos(A + B)/[cos^2(A) - sin^2(B)], by the cosine addition formula = RHS. I hope this helps!
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This is (a*(b+a^4) + b*(a+b^4))(1+a^2b^2) > (a+b^4)(b+a^4) or (a^5 + 2ab + b^5)(1+a^2b^2) > ab + b^5 + a^5 + a4b^4 or ab + a^2b^2 (a^5 + 2ab + b^5) > a^4b^4 or 1 + ab(a^5 + 2ab + b^5) > a^3b^3. If ab < 1 it is obvious. If ab> 1 it is enough to show (a^5 + 2ab + b^5) > a^2b^2. Since a^5 + b^5 >= 2 (a^5b^5)^(1/2) >= a^2b^2, we are done. This just looks a little too easy. Are you sure your text is correct?
Do you mean a/(a+b^4) + b/(b+a^4) > 1/(1+a^2 b^2) ? This is undefined if a=b=-1, so can you be a little clearer on the allowed values of a and b? --- ---- ---- Edit: if a=-10 and b=3 the statement is false. The left side is negative and the right side is positive.
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A intersection B means both A and B A union B means A or B or both A and B are sets or events. A=(1,2,3} B={1,5} A union B = {1,2,3,5} A intersection B = {1}
First lets talk about the difference between Union and Intersection. Union means the word "AND" Thus A Union B, means that we include everything from A and B. Intersection on the otherhand means "OR" So we include what is present is A OR B. They don't have to be in both, just at least one. Thus our proof would look like this... ABC = AB( C U B) = ABC U AB = ABC Note that AB is a subset of ABC. Thus if we apply a Union we already with ABC we are already including all of the set AB.
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A = {3,6,9,12,15,18,21,24,27,30} ---> P(A) = 10/30 B = {4,8,12,16,20,24,28} --------------> P(B) = 7/30 1. the two are not independent 2. P(A & B) = 2/30 3. P(A or B) = P(A) + P(B) - P(A & B) = (10+7-2)/30 = 15/30 4. P(A|B) = P(A & B) /P(B) = 2/7 5. P(B|A) = P(A & B) /P(A) = 2/10 = 1/5 5.
A and B = {12, 24} A or B = {3, 4, 6, 8, 9, 12, 15, 16, 18, 20, 21, 24, 27, 28} Does that help?
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The equation given can be written: a * (bc + k) -- b Now, let's look at each of the possible answers. (A) a [ (c + k) / b ] If you distribute a, you get: a(c +k) / b (ac + ak) / b But with the equation above, you get: (abc + ak) / b You're missing a b. This is probably the answer, but let's continue to be sure. (B) a [ c + (k/b) ] ac + [ (ak) / b ] Above, you get: (abc + ak) / b Which can be broken into two components: (abc / b) + (ak / b) The b in the first term cancels out, giving you: ac + (ak / b) Which is identical to what was given. (C) [a/b] (k + bc) This is the same as the initial equation; the only difference is the change in the placement of k and bc, which we know doesn't alter the way the equation turns out in the end. If the + was a - , that would be a different story, but in this case, it's okay. (D) ac + [ak / b] See part B (E) [abc + ak] / b See part A. The answer is A.
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E. (b-a) proof: If a = b, you can replace all 'b's in the equation, with 'a's. The equation becomes: 1/x + 1/a = 1/a Then you can subtract (1/a) from both sides, leaving you 1/x=0 Now, looking at the answers, anything that will give you a 0 on the bottom, will be undefined, therefore not zero. a. is really 1/a - 1/a, which equals 0. if you put this on the bottom (1/x=0, your answer) you will get undefined b. is really a - a, which also equals 0. c. is really 1/a^2. which, if put into the answer, will show a^2=0, which will be sometimes true, but not always. d. is really a - a / a^2, which equals 0/a^2, which seems right because you have a 0 in the numerator, but once you put it into your equation (1/x = 0) it switches, making it undefined, like the others. e. e is the correct answer, because it is the same as d, only switched. you get 0/a^2 = 0, which is true.
This is not a wordplay question. There is no real value of x for which that equation holds. (If a and b are the same, then 1/a and 1/b are the same, so 1/x would have to be 0.) If you meant to say "if a does not equal b and ...", then (e) is the correct answer.
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P(B|A) > P(B) => P(B|A)+P(B'|A) > P(B)+P(B'|A) so 1 > P(B)+P(B'|A) so P(B'|A) < 1-P(B)=P(B') The hint does make it easier, but you can use P(B|A)=P(B and A)/P(A).
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A U B is the union of both sets, that is, the total sum of both sets, which in this case A= {7,1,5} U B={8,9,3} = {1,3,5,7,8,9} (Note that order does not matter here) A ∩ B is the intersection of the sets, that is the elements, (integers here), that they both share in common. In this case, since A does not contain a number that is in B, and B does not contain an integer that is in A, then A ∩ B = {empty set}
A U B = (7, 1, 5, 8, 9, 3) or (1,3,5,7,8,9) A intersection B = ( empty set )
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Sin(b) = - 1/5 → where b is in quadrant III → cos(b) < 0 cos²(b) + sin²(b) = 1 cos²(b) = 1 - sin²(b) cos²(b) = 1 - (- 1/5)² cos²(b) = (25/25) - (1/25) cos²(b) = 24/25 cos²(b) = [± (√24)/5]² cos(b) = ± (√24)/5 → recall: cos(b) < 0 cos(b) = - (√24)/5 Do you know this identity? cos(a + b) = cos(a).cos(b) - sin(a).sin(b) → suppose that: a = b cos(b + b) = cos(b).cos(b) - sin(b).sin(b) cos(2b) = cos²(b) - sin²(b) cos(2b) = [1 - sin²(b)] - sin²(b) cos(2b) = 1 - sin²(b) - sin²(b) cos(2b) = 1 - 2.sin²(b) 2.sin²(b) = 1 - cos(2b) → you divide the angle b by 2 2.sin²(b/2) = 1 - cos(b) → recall: cos(b) = - (√24)/5 2.sin²(b/2) = 1 - [- (√24)/5] 2.sin²(b/2) = (5/5) + [(√24)/5] 2.sin²(b/2) = (5 + √24)/5 sin²(b/2) = (5 + √24)/10 sin²(b/2) = (5 + 2√6)/10 sin(b/2) = ± √[(5 + 2√6)/10] → recall: b is in quadrant III → (b/2) is in quadrant I → sin(b/2) > 0 sin(b/2) = √[(5 + 2√6)/10] sin(b/2) ≈ 0.995
Cos(B) = sqrt(24/25) = (2/5)*sqrt(6) sin(B/2) = sqrt[1/2 - 1/2 cos(B)] = sqrt[1/2 + (1/5)*sqrt(6)] = 0.9949 It's the positive root, because the angle is between 90 and 135 degrees. Check via calculator: B = arcsin(-0.2) = 191.54 degrees; B/2 = 95.77 degrees; sin(B/2) = sin(95.77 deg) = 0.9949. Yup...
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B^4+13b^2+36=0 (b^2+4)(b^2+9)=0 b^2=-4 b= sqrt -4=+/- 2i b^2=sqrt -9= +/- 3i
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B^3+49b = b(b^2+49) b^2+49 is prime so you can't factor that any further. What you have is the correct answer.
You could go as far as b(b^2+49) you can break this down further b(b-7)(b+7) reversing it using FOIL will give you b^2 - 7b + 7b + 49 (7b's cancel out and left with) B^2 + 49
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If A and B are roots to x^2 + 3x + 14, then by Viete's Formulas: A + B = -3 and AB = 14. Then, note that: (A - B)^2 = A^2 - 2AB + B^2 = (A^2 + 2AB + B^2) - 4AB = (A + B)^2 - 4AB. Plugging in our values obtained with Viete's Formulas gives: (A - B)^2 = (A + B)^2 - 4AB = (-3)^2 - 4(14) = 9 - 56 = -47. Taking the square root of both sides gives A - B = ±i√47. (The sign of A - B depends on if you take A > B or A < B) I hope this helps!
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Below is the "prime factorization" of A and B. A = a1*a2*a3*...*an B = b1*b2*b3*...*bm Assume the two are relatively prime. Then no factor of a is also a factor of b. GCF = 1 LCM = A*B GCF * LCM = A*B Assume the two are not relatively prime. Then we can set a subset of the factors of A equal to a subset of the factors of B. Let's call the product of all these factors C. A = a1*a2*a3*...*an*C B = b1*b2*b3*....*bm*C Now, GCF = C LCM = A*B/C GCF * LCM = (A*B/C)*C GCF * LCM = AB
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Sin (a) is negative. a is not in quadrant III. So, a must be in quadrant IV because sin (a) is positive in quadrant I and quadrant II. cot (b) is positive. b is not in quadrant III. So, b must be in quadrant I for cot (b) to be positive. cos (a) = sqrt (1 - sin^2 (a)) = sqrt (1 - (-7/25)^2) = sqrt (1 - (49/625)) = sqrt (576/625) = 24/25 [cos (a) is positive because a is in quadrant IV] sec^2 (b) = 1 + tan^2 (b) = 1 + (1/cot^2 (b)) = 1 + (1/(8/15)^2) = 1 + (1/(64/225)) = 1 + (225/64) = 289/64 = (17/8)^2 So, sec (b) = 17/8 [sec (b) is positive because b is in quadrant I] cos (b) = 1/sec (b) = 8/17 sin (b) = sqrt (1 - cos^2 (b)) = sqrt (1 - (8/17)^2) = sqrt (1 - (64/289)) = sqrt (225/289) = 15/17 [sin (b) is positive because b is in quadrant I] sin (a-b) = sin (a) cos (b) - cos (a) sin (b) = (-7/25) (8/17) - (24/25) (15/17) = (1/(25 × 17)) ((-7) (8) - (24) (15)) = (1/(25 × 17)) (-56 - 360) = (1/(25 × 17)) (-416) = -416/425
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B Vitamins increases blood circulation, tissue repair, and helps your body make protein and energy! B Vitamins is contained in many food, for example: tomatoes, brussel sprouts, asparagus, tuna, beans, peas. Mostly those dark leafy vegetables! Hope that helps!
Eh? you mean B-complex vitamins? a lot if things ranging from energy release in cells to prevention of some forms of mental illness.
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2ad/c-b = m Add b to both sides of the equation: 2ad/c = m + b Subtract m from both sides of the equation: 2ad/c - m = b (Q.E.D.)
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B minor (Or "Bm") is the relative minor of D Major, so it has 2 sharps, C# and F# B C# D E F# G A B (natural minor) B C# D E F# G A# B (harmonic minor) B C# D E F# G A# B A natural G F# E D C# B (melodic minor) It is not just like B because B is a major key with 5 sharps, F#, C#, G#, D#, and A#
A B minor is a scale in music. It's just a minor scale that starts on B. the notes are as follows B, C♯, D, E, F♯, G, A.
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B is obviously a subset of A (since every rational number is a real number). Therefore the topology generated by B is a subset of the topology generated by A. However, what's more interesting is that the two topologies are equal. For this, we show that every element of A is in the topology generated by B. Let (a,b) be an element of A, where a,b are real numbers. Now we let (a_n) be a sequence of rational numbers greater than a which converges to a. Similarly, let (b_n) be a sequence of rational numbers less than b which converges to b. Then (a,b) = U (a_n, b_n), where U denotes the union over all n. I've left some details for you to check, but that's the idea.
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Proof : 1) b > a (added b both sides) b+b > a + b 2b > a + b (divided by 2 both sides) b > (a + b)/2 2) b > a (added a both sides) b + a > a + a b + a > 2a (divided both sides by 2) (a + b)/2 > a 3) b > (a+b)/2 > a or a < (a+b)/2 < b
Prove it like this: (b+b)/2 > (a+b)/2 > (a+a)/2 2b/2 > (a+b)/2 > 2a/2 b > (a+b)/2 > a
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(a+b)(b+c)(c+a) = (a+b).(b.c + b.a + c.c + c.a) = (a+b).(a.b + b.c + c.a) + (a+b).(c.c) = (a+b+c).(a.b + b.c + c.a) - c.(a.b + b.c + c.a) + (a+b).(c.c) = (a+b+c).(a.b + b.c + c.a) - a.b.c - b.c.c - c.c.a + a.c.c + b.c.c = (a+b+c).(a.b + b.c + c.a) - a.b.c
Let a+b+c=S, abc=P (a+b)(b+c)(c+a)=(S-a)(S-b)(S-c) =S^3-S^2(a+b+c)+S(ab+bc+ca)-abc =S(ab+bc+ca)-P as required
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The definition of -(a/b) is that it is the unique number such that, when added to a/b, produces 0. So if we can show that (-a)/b also satisfies this property, it will follow by uniqueness that the two numbers are the same. In short, we only need to prove a/b + (-a)/b = 0 But recall that -a is the unique number such that, when added to a, produces 0. Therefore: a/b + (-a)/b = (a + (-a))(1/b) (distributive laws) = 0*(1/b) (since a + (-a) = 0) = 0 as desired. To be completely detailed you may want to prove that 0*x = 0 for any real number x. Proof: 0*x + 0*x = (0 + 0)x = 0*x subtracting 0*x from both sides gives 0*x = 0
If a,b ∈ R, -a,-b,a/b,-a/b,a/-b,-(a/b)∈ R. This much is evident.
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I only know one way to do it. Don't know if there's a shortcut. so, here goes: (a + b)(a + b) use FOIL: first, outer, inner, last (a + b)(a + b) = first a * a = a^2.....outer a*b = 1 ab.....inner b*a = 1 ba(same as 1ab) .....last b*b = ..........................b^2 ..........................now combine like terms: a^2 + 2ab + b^2 **************************************... (a + b)3: = (a + b)(a + b)(a + b).....use the answer to (a + b)^2 above, and multi by (a + b) (a + b)(a^2 + 2ab+ b^2) first multi each factor by a, then by b a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3....combine like terms a^3 + 3 a^2b + 3 ab^2 + b^3 **************************************... (a + b)^4 is either the answer to (a + b)^2(a+ b)^2 or the answer to (a + b)3(a + b) I would go on with the pattern we have (a + b)(a^3 + 3a^2b + 3 ab^2 + b^3) = a^4 + 3 a^3b + 3 a^2b^2 + ab^3 + a^3b + 3 a^2b^2 + 3 ab^3 + b^4 = a^4 + 4 a^3b + 6 a^2b^2 + 4 ab^3 + b^4 **************************************... for (a + b) ^5, I'm going to show you a different set-up. Set the problem up like a standard multiplication problem. ...................a^4 + 4 a^3b + 6 a^2b^2 + 4 ab^3 + b^4 ......................................... + b --------------------------------------... start multi. by b ...............a^4b + 4 a^3b^2 + 6 a^2b^3 + 4 ab^4 + b^5.........now multi by a .....a^5 +4a^4b + 6 a^3b^2 + 4 a^2b^3 + ab^4...................now add --------------------------------------... ..... a^5 + 5 a^4b + 10 a^3b^3 + 10 a^2b^3 + 5 ab^4 + b^5
Ok you should know that (a+b)^2=a^2+2ab+b^2 ok then (a+b)^3 is the same as (a+b)^2(a+b) so foil everything out (a+b)^4 is the same as (a+b)^2(a+b)^2 ithen foil everything out (a+b)^5 is the same as (a+b)^2(a+b)^2(a+b) foil everything out you can use wolfram alpha to check your work or use pascal's triangle as well to see the patern but all these problems are is seing that if you know one general papern (a+b)^2 then you can fit everything into that one form and then multiply everything out by using foil method
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C can not be zero, otherwise we could not have (a+b+c) c<0 (1) If c<0 or c>0 and a=0 then b can not be zero, otherwise (a+b+c) c = c^2 >0. So if b is not zero we will have b^2 > 0 and 4ac=0 so b^2 > 4ac (2) If c<0 and a>0 or c>0 and a<0 then 4ac <0 and therefore b^2 >=0 >4ac (3) If c<0 and a<0 then we must have a+b+c>0 so b > -(a+c) > 0 and so b^2 > (a+c)^2 = (a-c)^2 + 4ac > 4ac (4) If c>0 and a>0 then we must have a+b+c<0 so -b > a+c > 0 and so, as before, b^2 > (a+c)^2 = (a-c)^2 + 4ac > 4ac which completes the proof
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-b^2 ------ b = -b*b ----------- b cancel out one b from top and bottom = -b
When you have common variables all you have to do is subtract the expoinents. So with this question you would subtract 2-1 and get 1. So the answer is just -b.
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Given: 2 = 2/3 − 5/b Multiply everything by 3: 6 = 2 − 15/b Subtract 2: 4 = -15/b Solve for b: b = -15/4
2=2/3-5/b 2-2/3=-5/b 4/3=-5/b 4b=3(-5) b=-3.75
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Cos (a + b) = cos a cos b - sin a sin b given: sin a = - 7/25 cos a = - 24/25 do you really mean: cot b = - 8/15 ??? if so...then... sin b = 15/17 cos b = - 8/17 can you find the value now ? Why not ? What did you get ?? [cos (a + b) = (1944)/(17 * 25²) ]
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Sin(A+B) = sinAcosB + cosAsinB use pythagorean theorem to solve for the adjacent side of angle A and the opposite side of angle B. sin(A+B) = (-5/13)(8/17) + (-12/13)(-15/17) sin(A+B) = 140/221
Consider Chief SohCahToa Sin = opposite/Hypotenuse Cosine = adjacent/Hypotenuse Tangent = reverse/adjoining also memorize: Sec = 1/Cos Csc = 1/Sin Cot = 1/Tan sec(x) * cot(x) * sin^2(x) = H/A * A/O * O^2/H^2 = H/O * O^2/H^2 = O/H = tan(x) Do the opposite one the identical method.
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Log[b](3) * log[b](4) = log[b](12) log[b] (3 ^ log[b](4) ) = log[b](12) b^log[b](12) = 12 = 3^log[b](4) log(12) = log[b](4) * log(3) log(12)/log(3) = log[b](4) b^( log(12) / log(3) ) = 4 (log(12) / log(3) ) * log(b) = log(4) log(b) = log(4) * log(3) / log(12) log(b) = 0.31000087628424744261035268764194 b = 10^log(b) b = 2.0417420643272221346734427339768 Check it and see if it works. All I have is the windows calculator (my ti-89 is out in my car), so I can't verify it quickly right now.
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We are given that A⊆B. If x∈A∩B then x∈A and x∈B as every element in A∩B exists in A this means that A∩B⊆A Now , if x∈A then x∈B too as we are given A is a subset of B. this also means that x∈A∩B. so we conclude that A⊆A∩B by having A∩B⊆A and A⊆A∩B then it proves that A∩B=A ----------------------- for the second one it is quite the same: A⊆B If x∈A∪B then x∈A and x∈B. As every element in A∪B exists in B then A∪B⊆B Now, if x∈A then x∈B (for the reason given above) this leads to x∈A∪B. As every element in B exists in A∪B then B⊆A∪B by having A∪B⊆B and B⊆A∪B then it means that A∪B=B
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Ax^2 = ax + b Subtract ax from both sides ax^2 - ax = b Factor out a a(x^2 -x) = b Divide both sides by (x^2 -x) a = b/(x^2 -x) x cannot equal 0 or 1.
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F(x) = ((x - a) / (b - x))^(1/2) f'(x) = (1/2) * ((x - a) / (b - x))^(-1/2) * ((b - x) * (1) - (x - a) * (-1)) / (b - x)^2 f'(x) = (1/2) * ((x - a) / (b - x))^(-1/2) * ((b - x + (x - a)) / (b - x)^2 f'(x) = (1/2) * ((x - a) / (b - x))^(-1/2) * ((b - a)) / (b - x)^2 x = (a + b) / 2 f'((a + b) / 2) = (1/2) * ((b/2 - a/2) / (b/2 - a/2))^(-1/2) * (b - a) / (b/2 - a/2)^2 f'((a + b) / 2) = (1/2) * (1) * (b - a) / ((b - a)^2 / (4) f'((a + b) / 2) = (b - a) * 4 / ((b - a)^2 * 2) f'((a + b) / 2) = 2 / (b - a)
Y = √[ ( x - a ) / ( b - x ) ], .... ... a < x < b ....... (1) ......................................... At x = (a+b)/2, 2x = a + b, ... x - a = b - x, ... ( x-a ) / ( b-x ) = 1 ... ★★★ y = √1 = 1. Hence, at x = (a+b)/2, y = 1. ............................... (2) ......................................... Now, taking ' ln ' of eq(1), ln y = (1/2) · [ ln ( x-a ) - ln ( b-x) ]. Differentiating w.r.t. x, (1/y) · dy/dx = (1/2) · { [ 1 / (x-a) ] - [ 1 / ( b-x) ](-1) } dy/dx = (y/2) · { (b-a) / [ (x-a)(b-x) ] } ......................................... From (2), slope of the curve is dy/dx = (1/2) · { (b-a) / [ (a+b)/2 - a ][ b - (a+b)/2] } .........= (b-a) / { 2 [ (a+b-2a)/2 ][ 2b-a-b)/2 ] } .........= (b-a) / [ (b-a). (b-a)/2 ] .........= 2 / (b-a) ........................................... Q.E.D..
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B + b√b - 1 - √b = b - 1 + b√b - √b = (b - 1) + √b (b - 1) = (b - 1) ( 1 + √b) => (b + b√b - 1 - √b) / (b - 1) = 1 + √b => (b + b√b - 1 - √b) / (1 - b) = - (1 + √b). Edit: You asked a question in additional details which is sometimes missed by the answerer. It is better to send an email, if open. My email is open. Anyway, to answer your querry, (b - 1) + √b (b - 1) = 1 * (b - 1) + √b (b - 1) Taking (b - 1) common, = (b - 1) * (1 + √b).
I merely have been given (a)/(a+b) P.S. i think of logically it cant be a million. that's because of the fact for it to be a million, then (b/a) could desire to equivalent a million (So suitable and backside are the comparable). If (b/a) is a million, then (a/b) is likewise a million. which potential the denominator could be a million-a million making it 0, which it cant be.
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B' (8, 2) The translation moves the x-coordinate 3 units in the positive x direction, and it moves the y-coordinate 2 units in the negative y direction, so the image of B (5,4) is B' (8,2) .
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(a+b+c)[(a-b)²+(b-c)²+(a-c)²]= (a+b+c)[ 2a² + 2b² + 2 c² -2ab-2ac-2bc] =2(a+b+c)[ a²+b² +c²-ab -ac-bc] =2(a³+ab² +ac²- a²b-a²c-abc+a²b + b³+bc²-ab²-abc-b²c+a²c+b²c+c³-abc-ac²-bc...‡ =2(a³+b³+c³-3abc) Since a, b,c positive , then 2(a³+b³+c³-3abc)= (a+b+c)[(a-b)²+(b-c)²+(a-c)²] >=0 so (a³+b³+c³-3abc) >=0 so a³+b³+c³>=3abc
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10101 in base b is b^4 + b^2 + 1. That factorises as (b^2 + b + 1)(b^2 - b + 1) Since b >= 2, both of those are > 1, so it is composite. Re your additional question.. Firstly, I've done far too much maths to know have already seen the polynomial b^4 + b^2 + 1 before, so I knew what it factorised to from memory :P Option B would be just playing around with factors and seeing if they work. Theres not going to be a linear factor, as substituting b=1 and -1 in don't give you 0. So you're going to have something like: (b^2 + mb ± 1)(b^2 + nb ± 1). That expands to b^4 + (m+n)b^3 + (mn ± 2)b^2 ± (m+n)b + 1. So you want m+n = 0, and mn ± 2 to be 1: That tells you that m and n are 1 and -1. Finally, theres a nice trick, which is what you're really wanting. Complete the square: b^4 + b^2 + 1 = b^4 + 2b^2 + 1 - b^2 = (b^2+1)^2 - b^2 = (b^2 + 1 - b)(b^2 + 1 + b) since its a difference of two squares.
Stephen+ Assume a=b^2, then a^2+a+1=0; thus a1=-1/2 – sqrt(-3)/2 and a2= -1/2 + sqrt(-3)/2, isn’t it? So a1 = cos(2pi/3) - i*sin(2pi/3) = exp(-i*2pi/3) and a2 = cos(2pi/3) + i*sin(2pi/3) = exp(+i*2pi/3); hence b1=-sqrt(a1)=-exp(-i*pi/3) b2=+sqrt(a1)=+exp(-i*pi/3) b3=-sqrt(a2)=-exp(i*pi/3) b4=+sqrt(a2)=+exp(i*pi/3) and (b-b1)*(b-b3) * (b-b2)*(b-b4) = (b^2-(b1+b3)*b+1) * (b-(b2+b4)*b+1), where b1+b3 = -cos(pi/3)+i*sin(pi/3) - cos(pi/3)-i*sin(pi/3) = -1, and b2+b4 = +1; thus = (b^2-b+1) * (b^2+b+1);
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B=2a-8-10 2a=b+18 a=.5b+9 b=2(.5+9)-18 b=19-18 b=1 a=9.5 Proof b+10=2a-8 1+10=2(.5+9)-8 11=19-8
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(i) -2a = (ii) b / √4 = (iii) [8√b + a^2] / b - a (i) = (ii) : a = - b / 2. sqrt(4) = - b/4 => -4a = b (i) = (iii) -2a = [8 . sqrt(-4a) + a^2 ] / - 3a (assume a <0 or u cant solve the root (unless u work with complex numbers) => 6 a^2 = 8 sqrt(-4a) + a^2 => 5 a^2 = 8 sqrt(-4a) => 25 a^4 = - 64 a => a^3 = -64 / 25 => a= - 4/( 5^1/3) -> so assumed is true recall: -4a = b => b= 16/( 5^1/3)
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Suppose that all these sets are subsets of some universal set E. Then let x be an element in A U (B ∩ C) . Then x is an element of the set A or the set B ∩ C. If x is in A, then clearly x is in A U B and in A U C, and is therefore in the set (A U B) ∩ (A U C). If x is in B ∩ C, then x must be in both B and C, which implies that x is an element of A U B and of A U C. This, in turn, implies that x must be in (A U B) ∩ (A U C). So, the set A U (B ∩ C) ⊆ (A U B) ∩ (A U C). Now suppose that y is an element of the set (A U B) ∩ (A U C). Then y belongs to both A U B and to A U C. If y belongs to A U B, then it belongs to either A or B (and similary, if it belongs to A U C, then it belongs to either A or C). If y belongs to A, then, it clearly belongs to the set A U (B ∩ C). Now, if y is not in A, but is in B, then y must also belong to C. Let me elaborate: Since A ⊆ A U B and A ⊆ A U C and y is not in A, the only way that y is in the intersection of the two sets is if y is in both B and C. Since y belongs to both B and C, you have that y belongs to B ∩ C. So, once again y belongs A U (B ∩ C). This implies that (A U B) ∩ (A U C) ⊆ A U (B ∩ C). But, this implies the equality of the two sets, that is, A U (B ∩ C) = (A U B) ∩ (A U C), as desired.
Two approaches depending on what you can assume. This is from first principles. The next solution assumes we can use the derived laws of set theiry. First Principles: You need to say that if some element x belongs to the set on the Left Hand Side (LHS) of the equation it is also an element of the set on the Right Hand Side (RHS) . Then you need to do the same starting with the RHS and showing the element is in the set in the LHS. Part 1: Let x belong to A union (B intersect C) Then: x belongs to A OR x belongs to both B and C Consider each case. If x belongs to A then it belongs to (A union B) and also belongs to (A union C), so it belongs to the intersection of these two sets, which is the set on the RHS. If x belongs to both B and C, then it also belongs to both sets (A union B) and (A union C) and hence to their intersection, which is the set on the RHS. Part 2: Work the reverse. Suppose y belongs to the set on the RHS. Then : y belongs to (A union B) AND (A union C) So either (i) belongs to A or to B AND (ii) to A or to C. If x belongs to A then it belongs to the set on the LHS If x does not belong to A then it must (i) belong to B AND (ii) belong to C, so it belongs to (B intersection C) and hence the set on the LHS.
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Multiply out both sides: b² - b - 30 = b² + 3b + 2 Subtract b² from both sides: -b - 30 = 3b + 2 Add b to both sides: -30 = 4b + 2 Subtract 2 from both sides: -32 = 4b Divide both sides by 4: -8 = b
Okay, this is kind of long, so sorry if my explanation is hard to understand:) I will start with the left side of the equation (b-6)(b+5) you have to multiply the first B by the second B and 5, then multiply -6 by the second B and 5 it equals b^2+5b-6b-30. 5b and -6b are like terms, and they equal -1b. So you now have b^2-1b-30=(b+2)(b+1) Now i will do the right side (b+2)(b+1) Same as the first one, multiply everything together, so bXb, bX1, 2Xb, and 2X1 it comes out as b^2+b+2b+2. b and 2b are like terms, so i will add them. the whole equation is now b^2-1b-30=b^2+3b+2 Now i'm going to get the numbers on the right, and variables on the left b^2-1b-30=b^2+3b+2 +30 to both sides -b^2 to both sides -3b to both sides it is now -4b=32 32/-4=-8. So B=-8
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We need to find a and b such that a^2 - 9 = b and b^2 - 9 = a, where a does not equal b. Now (a^2 - 9) - (b^2 - 9) = b - a ----> a^2 - 9 - b^2 + 9 = b - a ----> a^2 - b^2 = b - a ----> (a - b)(a + b) = -(a - b). Now if a does not equal b then we can divide both sides by (a - b) without worrying about having divided by zero. This gives us a + b = -1, and so a = -b - 1 = -(b + 1). Now plug this into a^2 - 9 = b to get (-(b + 1))^2 - 9 = b ---> (b + 1)^2 - 9 = b ----> b^2 + 2b + 1 - 9 = b -----> b^2 + b - 8 = 0. Next, use the quadratic formula to solve for b: b = (-1 +/- sqrt(1^2 - 4*1*(-8))) / 2 = (-1 +/- sqrt(33)) / 2. So either b = (-1 + sqrt(33))/2 or b = (-1 - sqrt(33))/2. Now since a = -b - 1 we get respective values for a of a = (-1 - sqrt(33))/2 or a = (-1 + sqrt(33))/2. So either way, the two numbers are (-1 + sqrt(33))/2 and (-1 - sqrt(33))/2.
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B^2 / (b^2 - 4) - 4 / (b^2 + 2b) => b^2 / ((b - 2) * (b + 2)) - 4 / (b * (b + 2)) => (b^2 * b - 4 * (b - 2)) / (b * (b - 2) * (b + 2)) => (b^3 - 4b + 8) / (b * (b^2 - 4)) => (b^3 - 4b + 8) / (b^3 - 4b)
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(A B + B)' + ((B C)' D)' + (A B + B) + ((B C)' D)'' =(A B + B)' + (A B + B) + ((B C)' D)' + + ((B C)' D)" =(A B + B)' + (A B + B) void each other. it will always be high, If one is low the other will be high. You're adding so you will always get a high. Because of this the circuit would always be high, you can stop here. If you keep going anyway ((B C)' D)' + ((B C)' D)" ((B C)' D)"= NOTNOT or a mistype? If that is not not they void each other too and this would always be high. ((B C)' D)"= ((B C)' D) To answer the question no you dont need all of the variables to appear. You want to reduce and remove when you can. If you're in class and this is a digital class most of the math will have worthless variables.
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If you draw a sketch of "A(cap) - B(cap)" and then of "A(cap) + B(cap)" using tip to tail vector construction you will find that A(cap) & B(cap) are legs of a right triangle. The hypoteneuse of each right triangle is (one drawn for vector addition and one drawn for vector subtraction) the same magnitude. They just point in different directions. The ratio you gave is the ratio of the magnitudes of the two hypotenuses. You can also apply the dot product to these vectors; C^2 = (A(cap) - B(cap)) dot (A(cap) - B(cap)) = A^2 + B^2 -2ABCos() D^2 = (A(cap) + B(cap)) dot (A(cap) + B(cap)) = A^2 + B^2 + 2ABCos() The absolute values are just the sq. roots and "()" is the angle between A(cap) & B(cap).
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... (2x+a) (x+3) ≡ 2x² + 4ax + b or 2x² + (a + 6)x + 3a ≡ 2x² + 4ax + b (a + 6) = 4a → a = 2 3a = b → b = 6 (a,b) = (2,6)
It's very easy... (2x+a)(x+3)= 2x^2+6x+ax+3a=2x^2+4ax+b Then.. 6x+ax=4ax => a=2 3a=b => b=6 good luck
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B = -i -j -k = ( -1, -1, -1) ==> | b | = sqrt( (-1)^2 + (-1)^2 + (-1)^2 ) = sqrt( 1+1+1 ) = sqrt(3) Or if b is a vector in a 4D space (e.g. a quaternion, where i, j and k are independent imaginary axes): b = -i -j -k = (0, -1, -1, -1) ==> | b | = sqrt(0^2 + (-1)^2 + (-1)^2 + (-1)^2) = sqrt(0+1+1+1) = sqrt(3)
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A/(a+b)^2 - b/(a^2-b^2) a/[(a+b)(a+b)] - b/[(a+b)(a-b)] Your common denominator is: (a+b)(a+b)(a-b) so you need to multiply each term by [(a+b)(a+b)(a-b)]/[(a+b)(a+b)(a-b)] You will get: a(a-b)/[(a+b)(a+b)(a-b)] - b(a+b)/[(a+b)(a+b)(a-b)] Further: [a(a-b) - b(a+b)] / [(a+b)(a+b)(a-b)] [a² - ab - ab - b²] / [(a+b)(a+b)(a-b)] [a² - 2ab - b²] / [(a+b)(a+b)(a-b)] ← Final answer since numerator cannot be factored further
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A U B = B A ∩ B = A A - B = (empty set) Proofs are below: Let < denote subset property so that A < B means A is a subset of B. 1) If A < B, then A U B = B. Proof: Let A < B. Need to show that A U B < B and B < A U B. First let x be in A U B. Then x is in A or x is in B. If x is in A, then x is in B since A < B. In all cases x is in B. Thus, A U B < B. Now let x be in B. Then clearly, x is in A U B so that B < A U B. Thus, A U B = B. 2) If A < B, then A ∩ B = A. Proof: Let A < B. Need to show that A ∩ B < A and A < A ∩ B. First let x be in A ∩ B. Then x is in A and x is in B. Thus, x is in A so that A ∩ B < A. Now let x be in A. Then x is in B since A < B. Thus, since x is in A and x is in B we see that x is in A ∩ B so that A < A ∩ B. Thus, A ∩ B = A. 3) If A < B, then A - B = (empty set) Proof: Let A < B and assume by contradiction that A - B is nonempty. Thus, there exists an x such that x is in A - B. Thus, x is in A but x is not in B. Since x is in A and A < B we see that x is in B. But this is a contradiction since x is not in B. Thus, A - B = (empty set).
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B=the median. Proof: Minimize Int (x=-int to +inf) |x-b| p(x) dx = Int(x=-inf to b) (b-x)p(x)dx + Int(x=b to inf) (x-b)p(x)dx Now differentiate with respect to b and equate to zero. Using fundamental theorem of calculus, there are two terms for each integral (1) replace x by b in the integrand - this gives zero in each case and (2) differentiate by b inside the integral. Only (2) gives a nonzero result, as follows: Int (x=-inf to b)p(x)dx + Int(x=b to inf)(-p(x))dx=0 Int (x=-inf to b)p(x)dx = Int(x=b to inf)p(x)dx But this just says b is the median (area to the left equals area to the right)
The link has a proof for E(x^k) Let k=1, E(x) = mean = np Var(x)= E(x^2) - mean^2 Let k=2 and find E(x^2) and subtract n^2p^2
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Yes , we can conclude that A=B. And the easiest way to do so is by proving that A is a subset of B and B a subset of A. I will prove this in to ways and you'll chose the best for you. Way l:By contradiction Assume that A is not a subset of B then x∈A but x∉B. If x∉C then x∈A∪C but x∉B∪C. that contradicts the fact that A∪C=B∪C. On the other hand , if x∈C then x∈A∩C but x∉B∩C that contradicts the fact that A∩C=B∩C By those contradictions this means that A⊆B. --------------- Now prove by contradiction too that B⊆A. If we assume that B is not a subset of A then x∈B but x∉A. If x∉C then x∈B∪C but x∉A∪C that contradicts the fact that A∪C=B∪C. If x∈C then x∈B∩C but x∉A∩C that contradicts the fact that A∩C=B∩C. So by the contradictions we conclude that B⊆A. ------------- We know that A⊆B and B⊆A make A=B. Hence approved! :) -------------------------------------- Way ll: Direct way: Let x∈A. Then by having A∪C=B∪C we can state that x∈B no matter if x is in C or not. If x has elelments in both A and B it means that A⊆B. Let x∈B.Then by having A∩C=B∩C we can say that x∈A no matter if x is in C or not.If x has elements in both B and A it means that B⊆A. By the written above A⊆B and B⊆A implies A=B Hence approved in 2 ways! :)
If f(f'(x)) = x then f'(x) = the inverse of f. A pt. of the curve f(x) is given by (x,f(x)) I recommend drawing some graphs of f(x) and it truly is mirrored image over the y=x line. i'm no longer going to provide each and every of the information, even with the indisputable fact that the most important concepts are: f(x0) and f'(x0) = x0 for some fee of x0 >= 0. f(x>x0) can't dip below x0. There won't be able to be any maxima/minima (except on the starting up and end). You suggested you needed generalizations of the styles of applications. The above 3 are real by pondering once you reflect f(x) over the y=x line you get f'(x), which won't be able to be defined for any values outdoors of f(x). you're precise about there being fxns that could fulfill this even with the reality that. I nevertheless imagine it facilitates to attempt to visualise it.
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B = -4(c-2d) so c-2d = -b/4 2d-c = b/4 2d = b/4 + c d = b/8 + c/2
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For A+B the vectors A, B and A+B form a right angled triangle with the longest side being A+B using pythagoras' theorem mag A+B = sqrt((magA)^2 + (magB)^2) = sqrt(5^2 + 10^2) =sqrt(125) = 11.2m for A-B coincidentally, the triangles formed with the vectors A, B and A+B , A, B and A-B and A, B and B-A are all congruent (equal) so the magnitudes of A+B, A-B and B-A are the same, but as they are all vectors they have different directions
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1) To prove that two sets are disjoint, you must prove that none of the members of the first set can belong to the second, and vice versa. Try a proof by contradiction: assume A ∩ B and A \ B are not disjoint. In this case, there is some element x such that x is in A ∩ B and x is in A \ B. But if x is in A ∩ B, by definition, x must be in B; and if x is in A \ B, by definition, x must not be in B. Hence we have a contradiction, hence our assumption was false. Hence the sets are disjoint. 2) NB for this question I assume you mean A = (A ∩ B) U (A \ B), not A = (A ∩ B) ∩ (A \ B), since if A ∩ B and A \ B are disjoint, then their intersection is zero, so that (A ∩ B) ∩ (A \ B) = 0 (the empty set.) To prove A = (A ∩ B) U (A \ B), as you say, we must prove both A ⊆ (A ∩ B) U (A \ B) [1] and (A ∩ B) U (A \ B) ⊆ A [2]. To prove a set is a subset of another set, we just need to show that all the elements of the first set can be found in the second. Take [1]. Let x be any element in A. Now we ask, is x in B? We have two mutually exclusive and exhaustive choices; either x is in A and in B - so it is in A ∩ B - or it is in A and not in B - so it is in A \ B. Therefore every element of A is also an element of (A ∩ B) U (A \ B). Hence A ⊆ (A ∩ B) U (A \ B). Now for [2]: we have to prove that every element of (A ∩ B) U (A \ B) can be found in A. Well, (A ∩ B) U (A \ B) is made up of 2 parts, so we just need to show we can find all elements from both parts in A. It is clear that A ∩ B ⊆ A, if an element is both in sets A and B, then it must be in A (x in A ∩ B implies x in A); and it is clear that A \ B ⊆ A, since A \ B is all elements that are in set A but not in B (x in A \ B implies x in A), and hence all elements in A \ B are in A. Hence (A ∩ B) U (A \ B) ⊆ A. Therefore, using the principle (A ⊆ B) and (B ⊆ A) -> A = B, we have (A ∩ B) U (A \ B) = A.
You're actual -- an exclamation factor, amazingly sufficient, designates exclamation. and because "exclamation" and "yelling" are not synonymous, people who confuse the two are to be ostracized from well mannered society till they get their heads at the same time, ideally after years of therapy.
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A*sin(θ+α)=b*sin(θ+β) b*sin(θ+β) - a*sin(θ+α) = 0 b sin(θ)cos(β) + b sin(β)cos(θ) - a*sin(θ)cos(α) -a*sin(α)cos(θ) = 0 sin(θ)[ b cos(β) - a*cos(α)] + cos(θ)[ b sin(β) -a*sin(α)] = 0 sin(θ)[ b cos(β) - a*cos(α)] = cos(θ)[a*sin(α) - b sin(β) ] if [ b cos(β) - a*cos(α)] not 0 divide by [ b cos(β) - a*cos(α)] and by cos(θ) if not zero sin(θ)/cos(θ) = [a*sin(α) - b sin(β) ] / [ b cos(β) - a*cos(α)] multiply by -1 tan(θ) = [ b sin(β) - a*sin(α) ] / [ a*cos(α) - b cos(β) ]
I've got been advised that there are some examples that bettas would desire to be knowledgeable to do some stuff. individually i won't confirm that. My betta basically does despite it likes, the two i attempt to play and save it busy or not. looks that it would not provide a damn ;-) i've got been additionally advised to place up against it a mirror so as that it would desire to confirm it somewhat is reflection and flare, yet additionally to be careful that it would not save bumping on the glass. i've got tried that -> result : it flares for basically one sec and then basically turns it somewhat is returned swimming away ... clarify that... i think of you shouldn't attempt to "coach" them a lesson. attempt greater advantageous a tank separator or a different tank, so as which you will relish the two certainly one of your fish with none undesired effects !
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B^2 / a^3 a negative exponent puts it on the bottom.
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|ax + b| > c This can be written as two inequalities thus: (ax + b) > c and -(ax+b) > c => ax > c - b and -ax - b > c => x > (c-b)/a and -ax > c + b which gives x < -(c+b)/a Note the reversal of the inequality sign because we have divided through by -a Solution: x >(c-b)a and x < -(c+b)/a 2 |ax + b| < c (ax + b) < c and -(ax+b) < c x < (c-b)/a and -ax - b < c which gives x > -(c-b)/a Note change of inequality sign (3) |ax + b| = c (ax + b) = c and -ax -b = c x = (c-b)/a and x = -(c+b)/a
1) |ax+b|>c ax+b<-c or ax+b>c ax<-b-c or ax > c-b x<-(b+c)/a or x> (c-b)/a 2) |ax+b|<c -c<ax+b<c -b-c<ax< c-b -(c+b)/a<x< (c-b)/a 3) |ax+b|=c ax+b= +/-c ax= -b+/-c x= (-b+/-c)/a x= (-b+-c)/a or x= (-b-c)/a
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If A and B are independent, then P(A and B) = 0 P(A U B) = 0.4 + 0.23 = 0.63
P(A ∪ B) = P(A) + P(B) - P(A ∩ B) Since A and B are independent, P(A ∪ B) = P(A) + P(B) - P(A)P(B) P(A U B) = 0.4+0.23 - (0.4)(0.23) =0.538
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Using the two-point formula for a straight line: (y-y1)/(x-x1)=(y2-y1)/(x2-x1) where (x1,y1)=(a,b) and (x2,y2)=(a^2,b^2) (y-b)/(x-a)=(b^2-b)/(a^2-a) is the equation of the intersecting line y-b=[(b^2-b)/(a^2-a)](x-a) y=[(b^2-b)/(a^2-a)]x - a[(b^2-b)/(a^2-a)] + b y = [(b^2-b)/(a^2-a)]x - [(b^2-b)/(a-1)] + b is the line simplified The x-intercept is where the line crosses the x-axis. This corresponds to the point where y=0. y = 0 = [(b^2-b)/(a^2-a)]x - [(b^2-b)/(a-1)] + b Solving for x you get: x = [[(b^2-b)/(a-1)] - b] / [(b^2-b)/(a^2-a)] = [[(b-1)/(a-1)] - 1] / [(b-1)/(a^2-a)] x = a(a-1)[[(b-1)/(a-1)] - 1] / (b-1) = a - a(a-1)/(b-1) = a[1 - (a-1)/(b-1)] So the x-intercept is a[1 - (a-1)/(b-1)]
I believe my argument should not be nicely won. I say the possibility is 50% enable a ??, enable b ?? and randomly go with the values for a and b. As already stated, for a ? 0, P( a < b²) = a million, that's trivial. in straightforward terms quite much less trivial is the theory that P(a < 0 ) = a million/2 and subsequently P( a < b² | a ? 0) = a million and P( a < b² ) ? a million/2 Now evaluate what happens whilst a > 0 For a > 0, at a similar time as that's user-friendly to tutor there's a non 0 possibility for a finite b, the shrink, the possibility is 0. a < b² is such as saying 0 < a < b², bear in mind we are in straightforward terms observing a > 0. If this a finite era on an infinite line. The possibility that a is an element of this era is 0. P( a < b² | a > 0) = 0 As such we've an entire possibility P( a < b² ) = P( a < b² | a ? 0) * P(a ? 0) + P( a < b² | a > 0) * P(a > 0) = a million * a million/2 + 0 * a million/2 = a million/2 bear in mind, that is using infinite instruments. no count number what form of era you draw on paper or on a working laptop or laptop you will stumble on a finite possibility that seems to suggestions-set a million. yet that's because of the finite random form turbines on the laptop and if we had this question asked with finite values there may be a a answer greater advantageous than 50%. i do no longer recommend to be condescending, yet please clarify why utilizing the Gaussian to approximate a uniform distribution is a robust theory? are not infinite numbers exciting. Cantor whilst mad working with them! :)
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B^A is the set of functions f:A → B, so: B^A = {f: f:A → B} But a function f:A → B is just a set of ordered pairs: f = {(a,b) ∈ A×B: ∀a∈A∀b,c∈B(b = f(a)∧c = f(a) ⇒ b = c)} Therefore f ∈ A×B ⇒ B^A ⊂ P(A×B)
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Z = (A+(not)B).(B+C).((not)A+B+D) = (not A)(not B)C + AB +CD(A + not B) Let Z = YW Where Y is Y = (AB +AC +(not B)C) since (not B)(B) = 0 Y = AB + C(A + not B) and W is W = (not A + B + D) Z =(AB +AC +(not B)C)(not A +B + D) Z = ( not A)(AB +AC) + (not A )( not B)C + ABB + ABC + C(not B)(B) + ABD + ACD + (not B)CD Z = (not A)(not B)C + AB + ABC + ABD + ACD + (not B)CD since ABB = AB and (not A)(AB +AC) = 0 Z = (not A)(not B)C + AB +CD(A + not B) since AB = AB(1 +C +D) =
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B' = {1, 2, 6, 7, 8, 9, 10} Thus A intersect B' = {2, 7} Edit: B' usually means B complement, referring to set of elements not in B
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Log_b (log_b (a)) Assuming b > 1 When 0 < a < 1, then log_b(a) < 0, so log_b (log_b(a)) is not defined When 1 < a < b, then log_b(a) > 0, so log_b (log_b(a)) is defined Assuming b < 1 When 0 < a < b < 1, then log_b(a) = ln(a)/ln(b) Since a and b are both < 1, then ln(a) < 0, ln(b) < 0 and ln(a)/ln(b) > 0 Therefore log_b(a) > 0, so log_b (log_b(a)) is defined So log_b (log_b(a)) is not always defined for positive values of a
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(a*b)*c=(a+b/2)*c=a+b/2+c/2 a*(b*c)=a*(b+c/2)=a+(b+c/2)/2=a+b/2+c/... (a*b)*c=a*(b*c) only if c/2=c/4=>c=0 So in my opinion the correct answer is B BUT if you ment that q*r=(q+r)/2 than you're right the correct answer is E. (a*b)*c=((a+b)/2)*c=((a+b)/2+c)/2=a/4+... a*(b*c)=a*((b+c)/2)=(a+(b+c)/2)/2=a/2+... (a*b)*c=a*(b*c) only if a/4+b/4+c/2=a/2+b/4+c/4 which means that a+2c=2a+c=> a=c -Ana-Maria
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Since A and B are countably infinite, we may enumerate A and B: A = {a(1), a(2), ...} and B = {b(1), b(2), ...}. Note that since A and B are disjoint, {a(1), b(1), a(2), b(2), a(3), b(3), ...} is an enumeration of A U B. Thus, we conclude that A U B is countably infinite. ------------------------------ Use a Cantor Diagonal Argument from the grid: (a(1), b(1)), (a(1), b(2)), (a(1), b(3)), ... (a(2), b(1)), (a(2), b(2)), (a(2), b(3)), ... (a(3), b(1)), (a(3), b(2)), (a(3), b(3)), ... ......................................... So, {(a(1),b(1)), (a(1),b(2)), (a(2),b(1)), (a(1),b(3)), ...} is an enumeration of A x B. Thus, A x B is countably infinite. I hope this helps!
In view that A and B are countably limitless, we are in a position to write A={a_0,a_1,a_2,...} and B={b_0,b_1,b_2,...}. Then mapping mapping a_n to 2n and b_n to 2n+a million for each organic extensive type n, we get carry of a bijection between A?B and the organic numbers, meaning A?B is countably limitless.
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A ∩ (B U C) ⊂ (A ∩ B) U (A ∩ C). Let x ϵ A ∩ (B U C) then x ϵ A & x ϵ B U C. x ϵ B U C implies xϵ B or xϵ C. Suppose wlog xϵ B, then x ϵ (A ∩ B) hence x ϵ (A ∩ B) U (A ∩ C). ===================================== Now we show that( A∩ B) U (A ∩ C).⊂A ∩ (B U C). Let xϵ (A ∩ B) U (A ∩ C) then xϵ (A ∩ B) or xϵ (A ∩ C). Suppose now, wlog, that xϵ (A ∩ B), hence xϵ A & xϵ B, xϵ B implies xϵ(B U C), hence Quindi xϵ A ∩ (B U C). ==================== We show A ∩ (B U C) ⊂ (A ∩ B) U (A ∩ C). and (A ∩ B) U (A ∩ C).⊂A ∩ (B U C) so A ∩ (B U C) = (A ∩ B) U (A ∩ C) ================== Sorry for english ^^'', I'm italian
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Suppose sqrt(a+b)=sqrt(a) + sqrt(b) then squaring both sides: a + b = a + 2sqrt(a)sqrt(b) + b Cancel a+b: 2sqrt(a)sqrt(b) = 0 so sqrt(a)= 0 or sqrt(b) = 0 hence a=0 or b=0. On the other hand if a=0 or b=0 then sqrt(a+b) = sqrt(a)+ sqrt(b). Thus we get the following result: sqrt(a+b)=sqrt(a)+sqrt(b) if and only if a=0 OR b=0.
If you square both sides of the equality, you have a + b on the left, and a + b + 2squareroot(ab) Therefore, if the equality is to be satisfied, then squareroot(ab) must equal zero. Hence, the only way for this to work, is if either a, b, or both are zero. Fib
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(((a)/(b))+c)/(((a)/(b))-c) Remove the parentheses around the expression (a)/(b). ((a)/(b)+c)/(((a)/(b))-c) To add fractions, the denominators must be equal. The denominators can be made equal by finding the least common denominator (LCD). In this case, the LCD is b. Next, multiply each fraction by a factor of 1 that will create the LCD in each of the fractions. ((a)/(b)+c*(b)/(b))/(((a)/(b))-c) Complete the multiplication to produce a denominator of b in each expression. ((a)/(b)+(bc)/(b))/(((a)/(b))-c) Combine the numerators of all expressions that have common denominators. (((a+bc)/(b)))/(((a)/(b))-c) Reorder the polynomial a+bc alphabetically from left to right, starting with the highest order term. (((bc+a)/(b)))/(((a)/(b))-c) Remove the parentheses around the expression (a)/(b). (((bc+a)/(b)))/((a)/(b)-c) To add fractions, the denominators must be equal. The denominators can be made equal by finding the least common denominator (LCD). In this case, the LCD is b. Next, multiply each fraction by a factor of 1 that will create the LCD in each of the fractions. (((bc+a)/(b)))/((a)/(b)-c*(b)/(b)) Complete the multiplication to produce a denominator of b in each expression. (((bc+a)/(b)))/((a)/(b)-(bc)/(b)) Combine the numerators of all expressions that have common denominators. (((bc+a)/(b)))/((a-bc)/(b)) Reorder the polynomial a-bc alphabetically from left to right, starting with the highest order term. (((bc+a)/(b)))/((-bc+a)/(b)) To divide by ((-bc+a))/(b), multiply by the reciprocal of the fraction. (b)/(-bc+a)*(bc+a)/(b) Remove the common factor of b from the numerator of the first expression and denominator of the second expression. (1)/(-bc+a)*(bc+a) Multiply the rational expressions to get ((bc+a))/((-bc+a)). (bc+a)/(-bc+a)
Multiply it by b/b (which would equal 1) (a/b) + c becomes a+bc and (a/b)-c becomes a-bc so you get... (a+bc) / (a-bc)
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I'll use "sqrt()" in place of a radical sign. Given: sqrt(36a^5/ b^7) Simplify: sqrt(36a^5) / sqrt(b^7) sqrt(6^2 * a^4 * a) / sqrt(b^6 * b) Remove the square terms from within sqrt(): 6*a^2 * sqrt(a) / b^3 * sqrt(b) To rationalize the denominator, multiply numerator and denominator by "sqrt(b)": 6*a^2 * sqrt(a)*sqrt(b) / b^3 * sqrt(b) *sqrt(b) 6*a^2 * sqrt(a)*sqrt(b) / b^3 * b 6*a^2 * sqrt(a)*sqrt(b) / b^4
The gist of the discussion out there (sorry, there are too many pages to provide all the links) is that wands help to focus the magic resident in the witch or wizard, but wandless magic is still possible, particularly for very talented people like Snape and Dumbledore. Some examples ... Apparition Assuming one's Animagus form (or Metamorphpagus) Accio (the Summoning Charm) Elves can do magic without using wands Lumos
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B^2-13b+30=0 (b-10)(b-3)=0 b=10 or b=3
Carry the 13b and the -b^2 to the other side b^2 - 13b + 30 = 0 Factor it by reverse foiling (b-10)(b-3)= 0 then calculate what b could equal b= 10 b= 3
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A) vA.vB/B If by vA and vB you mean vectors A and B, by "." scalar product, and by B the magnitude of vector vB. vB/B is unit vector in the direction of vB, that's what you want to use for projecting vector vA in direction of vB.
B-a/|b|
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P(A|B') = P(A given not B) = P(A and not B) / P(not B) = [P(A and not B) + P(A and B) - P(A and B)] / (1 - P(B)) = [P(A) - P(A and B)] / (1 - P(B)) = (0.5 - 0.3) / (1 - 0.7) = 0.2/0.3 = 2/3. Lord bless you today!
P(A and not B)= P(A)-P(A and B) ---Use venn diag whenever confused =0.2 P(not B)= 1-P(B)=0.3 Get answer=2/3
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(a) A=[[T]]_B = [1 -2;1 4] (rows separated by ;) (b) The (column) vector [1 ; 0] are the coordinates of {1,1} in base B'. The (column) vector [0 ; 1] are the coordinates of {1,2} in base B'. Then, the transition matrix from B' to B is M = [[Id]_B'→B = [1 1 ; 1 2] (its columns are the vectors of B', expressed in B) (c) A ' = M⁻¹ A M M⁻¹ = [2 -1;1 -1] (construct the matrix with the diagonal elements interchanged, and the off-diagonal elements with signed inverted, then divide this matrix by the determinant of A). A' = [-7 -15 ; 6 12] (d) [T(v)]_B' = A' v_B' = [9, -6]^T (e) [T(v)]_B = M [T(v)]_B' = [3, -3]^T v_B = M v_B' = [1, -1]^T [T(v)]_B = A v_B = [3, -3]^T OK.
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(A∪B)∪C=(A∪C)∩(B∪C) is FALSE for let A = {1} B ={2} C = {3} then (A∪B) = {1,2} and (A∪B)∪C= = { 1, 2, 3} Also (A∪C) = {1,3} and B∪C) = { 2, 3 } then (A∪C)∩(B∪C) = {3} NOT EQUAL TO (A∪B)∪C
Sorry, but I don't NEED to prove that. Perhaps YOU should prove it, if it's so important. Besides, aren't you supposed to do your OWN homework so you can actually LEARN something? (Perhaps I'm wrong on that, but that's the way it used to be. Perhaps it's true what they say, that our country is going down the tubes with a generation of people who don't know how to learn. I sure hope I'm wrong!) N
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It is a cube in R^3. All of its sides have length b - a. The vertices of the cube are (a,a,a), (a,a,b), (a,b,a), (a,b,b), (b,a,a), (b,a,b), (b,b,a), and (b,b,b). It may be instructive to consider the example a = 0, b = 1 (which people sometimes call "the unit cube" in R^3). The picture for other a and b is just a translated and stretched version of this picture. It may be instructive to consider the picture for [a,b]^2 = [a,b] x [a,b] as a subset of R^2. It is a square. All of its sides have length b - a. The vertices of the square are (a,a), (a,b), (b,a), and (b,b).
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B-flat (the musical note). It's normally written "B♭", but if you don't have the flat symbol (♭) it's traditional to write "Bb".
Upper case and lower case.
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(b^4-x^4)/(b^5-x^5) (When x approaches b?). When x → b, x will equal b. =(b^4 - b^4)/(b^5 - b^5) =(b^4 - b^4)/(b^5 - b^5) = 0 / 0 = 1
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.Q.Assume that Pr [A∪B]=0.5 and Pr [A]=0.3 . Find is Pr [B] if A and B are independent event . Ans : Given : Pr [A∪B]=0.5 Pr [A]=0.3 According to the Question: A and B are independent event Therefore : Pr(A∩B) = Pr[A]*Pr[B] Apply : Pr [A∪B] = Pr[A] + Pr[B] - Pr(A∩B) Substitute the values in : => Pr [A∪B] = Pr[A] + Pr[B] - Pr[A]*Pr[B] =>0.5 = 0.3 + Pr[B] - (0.3)*Pr[B] => (1-0.3)*Pr[B]= 0.5 - 0.3 =>(0.7)Pr[B] = 0.2 => Pr[B] = 2/7 => Pr[B] = 0.28. I wish you will appreciate my answer.
P(A U B) = P(A) + P(B) - P(AB) for any events A and B. For independent events, P(AB) = P(A) P(B) So P(A U B) = P(A) + P(B) - P(A) P(B). Plug in the known values of P(A) and P(AUB) and then P(B) is the only unknown
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B1, B 2, B 3, B6, B 12....NIACIN, BIOTIN, EAT LIVER I SAY....YA GOT IRON POOR BLOOD? EAT YOUIR SHARE OF LIVER,.,,
B1-thiamine, b2- riboflavin,b6- pyridoxine, b12-cyancobalamin
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Plug in the point (4, -7) x = 4 y = -7 Solve for b: -7 = -3(4) + b -7 = -12 + b -7 + 12 = b 5 = b Answer: The y-intercept (b) of that line is 5.
- 7 = - 12 + b b = 5
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Let (x,y) ∈ (AxB)∩(BxA) => {(x,y) ∈ (AxB)} and {(x,y) ∈ (BxA)} => {x ∈ A and y ∈ B} and {x ∈ B and y ∈ A} => {x ∈ A and x ∈ B} and {y ∈ A and y ∈ B} => {x ∈ (A∩B)} and {y ∈ (A∩B)} => (x,y) ∈ (A∩B)x(A∩B) => (AxB)∩(BxA) ⊆ (A∩B)x(A∩B) ... (1) Now let, (x,y) ∈ (A∩B)x(A∩B) => x ∈ (A∩B) and y ∈ (A∩B) => (x ∈ A and x ∈ B) and (y ∈ A and y ∈ B) => (x ∈ A and y ∈ B) and (x ∈ B and y ∈ A) => (x,y) ∈ (AxB) and (x,y) ∈ (BxA) => (x,y) ∈ (AxB)∩(BxA) => (A∩B)x(A∩B) ⊆ (AxB)∩(BxA)...(2) (1) and (2) implies, (A∩B)x(A∩B) = (AxB)∩(BxA)
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You can't . Only matrix A is given. What is B? More info is needed. All I can say is the components of A + B will be 1 + b_11 , b_12 , 4 + b_13 -1 + b_21 , 2 + b_22 , 2 + b_23 b_31 , -2 + b_32 , -3 + b_33
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Sin b = sin(pi -c) = sin pi cos c - cos pi sin c = 0 - (-1) sin c = sin c. cos b = cos(pi -c) = cos pi cos c + sin pi sin c = -1 * cos c + 0 = -cos c. So, 2(1-sin b sin c) = 2(1-sin b sin b) = 2 (1- sin^2 b) = 2 cos^2 b = cos^2 b + cos^b = cos^2 b + cos^2 c. This last step follows from squaring both sides of cos b = -cos c.
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From this equation find Cos A sin square A + cos square A = 1 ( -9/11) square + cos square A = 1 cos A = (square root of 40) / 11 same way, Find sin B sin square B + cos square B = 1 Sin B = (square root of 33) / 7 After finding Cos A and sin B plug these following values in the formulas listed below: sin A= -9/11, cos B= -4/7 , cos A = (square root of 40) / 11, Sin B = (square root of 33) / 7 (a) sin(A+B)=sin A cos B + cos A sin B (b) cos(A-B)=cos A cos B + sin A sin B i have no idea how to find tan(a+b) and i don't understand part d of your question Enjoy !
Ans of(a)and (b) are given by shivam to find tan(a+b),first find cos(a+b)=cos a cos b-sin a sin bthen tan(a+b)=sin(a+b)/cos(a+b) when u know the value of (a+b),then u will know which quadrant it is in
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(a) Let P(A) = 0.65, P(B) = 0.55, P(A ∩ B) = 0.15 P(A U B) = P(A) + P(B) - P(A ∩ B) P(A U B) = 0.65 + 0.55 - 0.15 P(A U B) = 1.05 No, that is not possible. No probability can be greater than 1. (b) For the least possible P(A ∩ B), let P(A U B) = 1. P(A) + P(B) - P(A ∩ B) = 1 0.65 + 0.55 - P(A ∩ B) = 1 P(A ∩ B) = 0.20 (c) Event (A ∩ B) is a subset of event B, so P(A ∩ B) cannot be greater than P(B). No, P(A ∩ B) = 0.6 is not possible. (d) The greatest possible P(A ∩ B) is 0.55. That is equal to P(B) and still less than P(A), so there are no conflicts. This would be the case if B were a subset of A, but that is not necessarily the case.
The smallest possible value of P(A and B) is Max(0, (P(A) + P(B)) - 1). The largest possible value of P(A and B) is Min(P(A), P(B)). (A) no, see above. (B) 0.20 (C) no, see above. (D) 0.55 _____ Consider a Vern diagram such that the area of the named region matches its probability. The total area of two named regions (counting their overlap only once) cannot exceed 1, but one may overlap the other completely.
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( (b+2) / 3) - ( (b-2) / 7 ) = 2 ( 7(b+2) / 21) - ( 3(b-2) / 21 ) = 2 Find the common denominator ( (7b+14) / 21) - ( (3b-6) / 21 ) = 2 (4b+20)/21 = 2 4b + 20 = 42 4b = 22 b = 11/2
Cross multiply to get the same denominator. divide both sides by the same denominator. distribute and solve for b.
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P(B' | (A' v B)) = P(B' ^ (A' v B)) / P(A' v B) [1] P(B' ^ (A' v B)) = P[(B' ^ A') v (B' ^ B)] = P(B' ^ A') [2] Combining [1] and [2]: P(B' | (A' v B)) = P(B' ^ A') / P(A' v B) [3] Numerator of [3]: P(B' ^ A') = 1 - P(A v B) = 1 - 0.9 = 0.1 [3.5] Denominator of [3]: P(A' v B) = P(A') + P(B) - P(A' ^ B) [4] P(A' ^ B) = P(A v B) - P(A) = P(A v B) - [1 - P(A')] = 0.9 - (1 - 0.3) = 0.2 [5] Combining [4] and [5]: P(A' v B) = 0.3 + 0.5 - 0.2 = 0.6 [6] Combining [3], [3.5] and [6]" P(B' | (A' v B)) = 0.1 / 0.6 = 1/6
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Let C = A + B. If one of the sets A and B , say A, is not bounded above, then, sup A = ∞. If we fix some b ∈ B, then, for every M > 0, there is a ∈ A such that a > M - b and, therefore, c = a + b > M. Since c ∈ C, C is not bounded above, so that sup C = ∞ = ∞ + sup B = sup A + sup B Now, supose A and B are bounded above, so that they have supremums in R. For every c ∈ C, there are a ∈ A and b ∈ B such that c = a + b. Since a ≤ sup A and b ≤ sup B, then c ≤ sup A + sup B, which shows sup A + sup B is an upper bound for C. To show sup A + sup B is the supremum of C, pick ε > 0. By the definition of supremum, sup A - ε/2 is not an upper bound for A and sup B - ε/2 is not one for B. Hence, there are a ∈ A and b ∈ B such that a > sup A - ε/2 and b > sup B - ε/2 It follows that c = a + b ∈ C and c > sup A + sup B - ε. Since ε is arbitrary and sup A + sup B is an upper bound for C, then sup C = sup A + sup B.
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Let f(x) = sqrt(1+x^2). f is continuous and differentiable on (a,b). By the Mean Value Theorem, there exists a number c in (a,b) where f(b) - f(a) = f'(c) (b-a). Take modulus on both sides:|f(b) - f(a)| = |f'(c)| (b-a). Note that |b-a|=(b-a) since (b-a) is >0. Now for any real number d, |f'(d)|=2|d|/2sqrt(1+d^2) = |d|/sqrt(1+d^2). Clearly |f'(d)|<1 since the square of the numerator is one less than that of the denominator. In particular, for d=c, |f'(c)|<1 where c is the number we obtained from the mean-value theorem. Hence |f(b) - f(a)| = |f'(c)| (b-a) < (b-a). Proved.
I see you keep asking this question so I will give you some input although I cannot do this The mean value theorem states that f'(x) = (f(b)-f(a))/(b-a) If we let f(x) = sqrt(1+x^2) then [sqrt(1+b^2)-sqrt(1+a^2)]/[b-a] = f'(x) so sqrt(1+b^2)-sqrt(1+a^2) = f'(x)(b-a) --------------------------------------... from here I am not sure if I am correct sqrt(1+b^2)-sqrt(1+a^2) = f'(x)(b-a) if we divide both sides by f'(x) then we can say sqrt(1+b^2)-sqrt(1+a^2) < (b-a) because when we divide the number gets smaller I hope that is right. As i was doing my input I had an epiphany and I hope this is right.
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(A \ B ) \ C =? (A \ B) n (A \ C) be x an element in Ω if x is in A \ B but not in C then its in the left group and also in the right one if x is in A \ B and in C its neither in the left nor in the right group if x is in A \ C but not in B then its in the left group and also in the right one if x is in A \ C and in B its neither in the left nor in the right group if x is not in A its not in the left and also not in the right group and so on therefore (A \ B ) \ C = (A \ B) n (A \ C)
A,b,c, c are units then given c? a = set a comprise each member of set c(c belongs to a). d? b = set b comprise each member of set d(d belongs to b). evidence :- c? d? a? b c? d =those individuals that comprise the two units c and d(C Intersection d ). a? b =those individuals that comprise the two units a and b.(a Intersection b). then c? d 's individuals belongs to a? b 's individuals so as that has proofed c? d? a? b it fairly is evidence only
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Expand: (6a + b) dot (a-2b) = (6a) dot a + b dot a + (6a) dot (-2b) + (b) dot (-2b) = 6 (a dot a) + (b dot a) - 12 (a dot b) - 2 (b dot b) = 6 (a dot a) - 11 (a dot b) - 2 (b dot b) ------------------------------(1) Now a, b are unit vectors (a dot a = length of a = 1 b dot b = length of b = 1) And a dot b = (length of a)*(length of b) * cos (angle between a and b) So here a dot b = 1*1*cos(60) = 1/2 So from (1) (6a + b) dot (a-2b) = 6 - 11/2 - 2 = -1.5
{ 6 | a |² + 11 a dot b + 4 | b |² } = { 10 + 11 cos 60 }
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What do you mean by " A,B" ? Does the comma stand for ' and ( ∩ ) ' ? ......................................... P [ ( A∩B ) | C ] = P ( A∩B∩C ) / P(C) ................... (1) P (A | C ) = P( A∩C ) / P(C) ................................... (2) P [ B | ( A∩C ) ] = P [ B ∩ ( A∩C ) ] / P( A∩C ) . . . . . . . . . . . . = P ( A∩B∩C ) / P ( A∩C ) ............ (3) Multiplying equations (2) and (3), P ( A | C ) • P [ B | ( A∩C ) ] = { P ( A∩C ) / P (C) } • { P ( A∩B∩C ) / P ( A∩C ) } = P( A∩B∩C ) / P(C) = P [ ( A∩B ) | C ] ............ from (1) Hence, the result you have stated is CORRECT ! ......................................... Happy To Help ! .........................................
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1 / [1 + x^(-b) ] = [ x / (1 - x) ]^(1/b) [I've been trying for a while now something like this; but it really doesn't lead me anywhere... :] z := x - 1/2 1 / [1 + (1/2 + z)^(-b) ] = [ (1/2 + z) / (1/2 - z) ]^(1/b) invert 1 + (1/2 + z)^(-b) = [ (1/2 - z) / (1/2 + z) ]^(1/b) 1 = [ (1/2 - z) / (1/2 + z) ]^(1/b) - 1 / (1/2 + z)^b multiply by [ (1/2 + z) / (1/2 - z) ]^(1/b) (1 + 1 / (1/2 + z)^b) [ (1/2 + z) / (1/2 - z) ]^(1/b) = 1 .... now what...
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Here is the question in full, I'm not sure where to begin with my answer, so a step by step approach would be appreciated, thank you Let a, b and c be three vectors such that a + b + c = 0. (i) Show that a × b = b × c = c × a. (ii) Explain what this means geometrically. (Your answer should refer to a triangle.)
Let : a + b + c = 0. ∴ ( a + b + c ) x b = 0 x b ∴ a x b + b x b + c x b = 0 ∴ a x b + 0 - b x c = 0 ∴ a x b = b x c ....................... (1) Similarly, we can show that ... b x c = c x a ........................ (2) Combining (1) and (2), ... a x b = b x c = c x a ............................ Q.E.D. _______________________ Now, in Δ ABC, let : BC = a, CA = b and AB = c. ∴ AB + BC + CA = AA = 0 ∴ a + b + c = 0. If (Δ↑) is the Vector area of ABC, then a x b = bx c = c x a = 2 ( Δ↑ ) ∴ vector area of a triangle is perpendicular ... to the plane of the triangle. ___________________________________ Happy To Help ! ____________________________________
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(a−c)(b−c) = ab−c(a+b)+c² But a+b=−q, ab=1 and c²=−1−pc so this becomes 1+cq−1−pc = c(q−p) (a+d)(b+d) = ab+d(a+b)+d² and d²=−1−pd so this is 1−dq−1−pd = −d(p+q) ∴ (a−c)(b−c)(a+d)(b+d) = cd(p²−q²) = p²−q² since cd=1
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Draw two circles that intersect (part of one circle passes into the second circle). Call one circle A and the other circle B The part in the centre that belongs to both circles is A∩B This is the "intersection" of both circles; in that part (the intersection) write the numbers 3, 6 and 9 The part of A that is NOT inside B is called A-B The part of B that is NOT inside A is called B-A write the appropriate elements into each section. Once you have finished, the set A contains all the elements inside the circle called A (even those elements that are also in B). The set B contains all the elements inside the circle called B (even those elements that are also in A). The elements 3, 6 and 9 will be in both sets. Otherwise A∩B = {3, 6, 9} would be false. A - B
A =1, 3, 5, 6, 7, 8, 9 B = 2, 3, 6, 9, 10
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Let f: [a, b] --> R be continuous on [a, b] and differentiable on (a, b). Let f ' be differentiable on (a, b). Now assume for all x in (a, b) that f ' (x) < 0. Need to show that f is strictly decreasing on [a, b]. Apply to Mean Value Theorem to f on [a, b]. Then there exists a number c in (a, b) such that [f(b) - f(a)] / [b - a] < f ' (c). Since c belongs to (a, b) we see that f ' (c) < 0. Thus, [f(b) - f(a)] / [b - a] < 0. Clearly, b - a > 0 so we can multiply both sides by b - a to get f(b) - f(a) < 0 f(b) < f(a) Thus, if a < b, then f(a) > f(b). By definition, f is strictly decreasing on [a, b] as needed to show.
Use the definition of differentiability: f'(x) = [f(x+h) - f(x)]/h < 0 => f(x+h) < f(x) Edit: this is just the broad strokes of the proof. Clearly, the defintion of the derivative is a limit and not a straight equation. You should probably use the definition where: f'(x) = lim as x -> c of [f(c) - f(x)]/(x-c) It will be easier to settle the details of c being in [a, b].
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See: S(x)=ax/(1+(x/b)^c) S '(x)=[ax'.(1+(x/b)^c)-ax.(1+(x/b)^c)']/ (1+(x/b)^c)^2 S '(x)=[a.(1+(x/b)^c)-acx/b. (1+(x/b)^(c-1))]/ (1+(x/b)^c)^2
Use the quotient rule: S'(x) = (a(1 + (x/b))^c - acx(x/b)^(c-1))/(1 + (x/b)^c)^2
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=(b^2+9b+20)-(2b+8)b^2-9 =(b^4+9b^3+20b^2-20b+8-9b^2) =(b^4+9b^3+11b^2-20b+8)
I suspect that question is presented incorrectly. Brackets required. Please check.
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A + b + c = 1, ab + bc + ca = 2, abc = 3 P = 1/(a + bc) + 1/(b + ca) + 1/(c + ab) a + bc = (1 - b - c) + bc = (1 - b)(1 - c) b + ca = (1 - c - a) + ca = (1 - c)(1 - a) c + ab = (1 - a - b) + ab = (1 - a)(1 - b) P = 1/(1 - b)(1 - c) + 1/(1 - c)(1 - a) + 1/(1 - a)(1 - b)  = [(1 - a) + (1 - b) + (1 - c)] / (1 - a)(1 - b)(1 - c)  (1 - a)(1 - b)(1 - c) = 1 - (a + b + c) + (ab + bc + ca) - abc = 1 - 1 + 2 - 3 = - 1  (1 - a) + (1 - b) + (1 - c) = 3 - (a + b + c) = 3 - 1 = 2 P = 2/(-1) = - 2
A + b + c = 1 (a + b + c)² = 1² (a² + b² + c²) + 2 (ab + bc + ca) = 1 (a² + b² + c²) + 2(2) = 1 a² + b² + c² = −3 ab + bc + ca = 2 (ab + bc + ca)² = 2² (a²b² + b²c² + c²a²) + 2 (a²bc + ab²c + abc²) = 4 (a²b² + b²c² + c²a²) + 2abc(a+b+c) = 4 (a²b² + b²c² + c²a²) + 2(3)(1) = 4 a²b² + b²c² + c²a² = −2 1/(a+bc) + 1/(b+ca) + 1/(c+ab) = [(b+ca)(c+ab) + (a+bc)(c+ab) + (a+bc)(b+ca)] / [(a+bc)(b+ca)(c+ab)] = [(a²bc+ab²c+abc²) + (a²b+a²c) + (b²c+ab²) + (ac²+bc²) + (ab+bc+ac)] / [(a³bc+ab³c+abc³) + a²b²c² + (a²b²+b²c²+c²a²) + abc] = [abc (a+b+c) + (a²b+a²c+abc) + (b²c+ab²+abc) + (ac²+bc²+abc) + (ab+bc+ac) − 3abc] / [abc(a²+b²+c²) + (abc)² + (a²b²+b²c²+c²a²) + abc] = [abc (a+b+c) + a(ab+ac+bc) + b(bc+ab+ac) + c(ac+bc+ab) + (ab+bc+ac) − 3abc] / [abc(a²+b²+c²) + (abc)² + (a²b²+b²c²+c²a²) + abc] = [abc (a+b+c) + (ab+ac+bc)(a+b+c) + (ab+bc+ac) − 3(abc)] / [abc(a²+b²+c²) + (abc)² + (a²b²+b²c²+c²a²) + abc] = [3(1) + (2)(1) + 2 − 3(3)] / [3(−3) + (3)² + (−2) + 3] = −2
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{1 + (1/a)}{1 + (1/b)}{1 + (1/c)} = 2 {1 + (1/c)}³ ≥ 2 1/c ≥ ∛2 - 1 > 1/4 c ≤ 3 (1) c = 1 {1 + (1/a)}{1 + (1/b)} = 1.........no (2) c = 2 {1 + (1/a)}{1 + (1/b)} = 4/3 (3a + 3)(b + 1) = 4ab 3ab + 3a + 3b + 3 = 4ab ab - 3a - 3b = 3 (a - 3)(b - 3) = 12 (a,b) = (7,6),(9,5),(15,4) (3) c = 3 {1 + (1/a)}{1 + (1/b)} = 3/2 {2a + 2}{b + 1} = 3ab 2ab + 2a + 2b + 2 = 3ab ab - 2a - 2b = 2 (a - 2)(b - 2) = 6 (a,b) = (5,4),(8,3) in all, (7,6,2), (9,5,2), (15,4,2), (5,4,3), (8,3,3) □
(a,b,c) = (9,5,2) (7,6,2) (15,4,2) (5,4,3) (8,3,3) are the only solutions
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My guess is that the question is: If ab = ba, then (ab)^(|a| |b|) = e. The proof uses two facts: (1) If ab = ba, then (ab)^n = a^n b^n. (2) a^(mn) = (a^m)^n = (a^n)^m Both of these follow by induction on n or m. Then (ab)^(|a| |b|) = a^(|a| |b|) b^(|a| |b|) = (a^|a|)^|b| (b^|b|)^|a| = e^|b| e^|a| = e
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A = c × e where c = gcf and e is uncommon factor of a b = c × f where c = gcf and f is uncommon factor of b e and f being uncommon factors of a and b their LCM = 1 suppose e and f have common factor g then GCD of e & f = g then e = g × factor1 and f = g × factor2 Then a = c × e = c × g × factor1 & b = c × f = c × g × factor2 Then gcf of a and b = c × g i.e. other than c which is cotradictory. Hence gcf of e & f must be 1 No. are a and b, GCF = c and uncommon factors of a is e and that of b is f LCM = c × e × f a × b = c × e × c × f LCM × GCF = c × e × f × c = ab LCM = ab/c = ab/ gcf (a,b) LCM =
Lcm(a,b)*hcf(a,b)=a*b
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A) (A∪B) ⊆ (A∪B∪C) Spose p ∊ (A U B). Since (A∪B) ⊆ (A∪B∪C), p ∊ (A U B U C) also. This is the easiest statement to prove. b) (A∩B∩C) ⊆ (A∩B) Spose p ∊ (A∩B∩C). Then, by referring to the inclusion-exclusion principle, p must be in either A, B, C or some intersections of the set [see: http://en.wikipedia.org/wiki/Inclusion%E... ] I leave this exercise to you. Suggestion: Consider some cases in which p belongs to A and B. Good luck!
C) (A-C) ∩ (C-B) = Ø
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[b + (1/2)] * [b + (1/4) ] = b^2 + 3b/4 + 1/8 id est
I would first multiply (b+.5) by 2 and (b+.25) by 4. This results in multiplying (2b+1) by (4b+1) which is 8b^2+6b+1.
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B = a is one solution. The conditions are like the assumptions in a Chinese Remainder Theorem problem, if you look at it correctly. That will lead directly to a simple class of numbers that serve for "b". a+1 | b+1 means b+1 = 0 mod a+1 or b = -1 = a mod a+1 a+2 | b+2 means b+2 = 0 mod a+2 or b = -2 = a mod a+2 etc. So you have 1396 congruences to satisfy, all of the form b = a mod a+k, for k = 0 to 1395. We already have one solution, b, and the CRT says all other solutions are b + n(LCM(a, a+1, ... , a+1395)), where n is any integer. We don't need to actually find the LCM of those numbers, any multiple will do, e.g. (a+1395)! So if b = a + n(a+1395)!, where n is any natural number, we have, for 0 <= k <= 1395 b+k = a+k + n(a+1395)! = a+k + (a+k)n(a+1395)!/(a+k) = (a+k)[1 + n(a+1395)!/(a+k)] = a multiple of a+k, so a+k | b+k Just in case it's not clear, that is a multiple of a+k because the expression in square brackets is an integer, because a+k divides (a+1395)!, because k is <=1395. So b + n(a+1395)! represents an infinite number of natural numbers that satisfy the requirements.
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(b - 6) = (c + 6) + 18 b - 6 = c + 24 c - b = - 6 - 24 c - b = - 30 -
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Think of it as 2 overlapping circles. The area of one of them is 0.7, of the other 0.8. P(A and B) is the area of overlap. I assume that your notation of (A|B) does not mean A or B, because the minimum value of P(A or B) is the value of B. Assuming then that (A|B) is the exclusive OR, meaning A or B but not both, on the diagram this would be represented by the portions of A and B which do not overlap. That is sufficient for you to calculate the portion which does overlap.
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(dQ/dt) = b-Q (dQ/dt) + Q = b This is a linear first order differential equation. Caluculate the Integrating Factor which is e^(∫1dt) = e^t (e^t)Q = ∫e^(t)*b dt = e^(t)*b + c Q = b + c*e^(-t)
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B/a=2 and c/b=3 so b = 2a and c = 3b a+b = a + 2a = 3a b/c = 2a/3b = 2a /6a 3a:1/3 a:1/9
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Sin (a+b)=sin a cos b+cos a sin b sin a=2/3 right triangle with h=3 & 1 leg=2 the other leg is l^2=3^2-2^2=9-4=5 l=√5 cos a=√5 /3 cos b=3/15=1/5 right triangle with h=5 & 1 leg=1 the other leg id l^2=5^2-1^2=24-1=24 l=2√6 sin b=2√6 /5 sin (a+b)=sin a cos b+cos a sin b sin (a+b)=2/3 * 1/5 + √5 /3 * 2√6 /5 sin (a+b)=2/15+2√30 /15 sin (a+b)=2(1+√30)/15
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Determine the relationship between set A and set B (that is, A ⊂ B, B ⊂ A, A = B, A and B are disjoint) if the following is true: 1. A ∩ B = B 2. A ∪ B = B 3. A ∩ B = ∅ 4. A ∪ B = A 5. A ∩ B = A 6. A ∩ B = A ∪ B I'm not asking anyone to solve this for me, I just want someone to help me understand what I'm supposed to be doing here.
1) A n B = B means that B is subset of A 2) A U B = B means that A is subset of B 3) An B = null set means that A and B are disjoint 4) A U B = A means that B is a subset of A 5) A n B = A means that A is a subset of B 6) A n B = A U B means that A = B
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4/(b-3) + 3/b = -2b/(b-3) (4+2b)/(b-3) + 3/b=0 Multiply by b(b-3) b(4+2b) + 3(b-3)=0 4b+2b^2+3b-9=0 2b^2+7b-9=0 (2b+9)(b-1)=0 b=1 or b=-9/2
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A = -4 ; b = 3 a - b = (- 4) - (3) = - 4 - 3 = - 7 (- a) - (b) = (- (-4)) - (3) = 4 - 3 = 1 a - (- b) = ( - 4) - (- 3) = - 4 + 3 = - 1 - a - (- b) = - (- 4) - (- 3) = - 4 + 3 = - 1 :o)
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First of all, this is a biconditional proof. This means that P <=> Q. To prove a biconditional proof, you can prove it in two steps, thanks to the tautology (P <=> Q) <=> (P => Q) ^ (Q => P). Treat "a+1 | b and b | b+3" as P, and "a=2 and b=3" as Q. (i) This is the simple part, (Q => P). Assume a=2, and b=3. Inserting the assumed values into "a+1 | b and b | b+3" gives us "3|3 and 3|6," both of which are true. Thus, Q => P. (ii) We'll prove this one through contrapostion ([P => Q] <=> [~Q => ~P]). Since Q is "a=2 and b=3," ~Q is simply inserting other values for a and b. Let them equal 4 and 5, respectively. Thus, we can rewrite "a+1 | b and b | b+3" as "(4+1) | 5 and 5 | 8." This is not true; thus, ~P. Therefore, ~Q => ~P, and P => Q. P => Q and Q => P is equivalent to P <=> Q.
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I think like this . . . . . . . . . . sec(a)sec(b) sec(a - b) = ▬▬▬▬▬▬▬ . . . . . . . . . .1 + tan(a)tan(b) Prove : sec(a - b) . . . . 1 = ▬▬▬▬ . .cos(a - b) . . . . . . . . . 1 = ▬▬▬▬▬▬▬▬▬▬▬▬ . .cos(a)cos(b) + sin(a)sin(b) . . . . . . . . . . .1 . .▬▬▬▬▬▬▬▬▬▬▬▬ . . . . . . .cos(a)cos(b) = ▬▬▬▬▬▬▬▬▬▬▬▬ . .cos(a)cos(b) + sin(a)sin(b) . .▬▬▬▬▬▬▬▬▬▬▬▬ . . . . . . .cos(a)cos(b) .. . sec(a)sec(b) = ▬▬▬▬▬▬▬ . .1 + tan(a)tan(b)
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Since angle B is acute sin(90 - B) = cos B Sin^2 B + cos^2 B = 1 (12/13)^2 + cos^2 B = 1 cos^2 B = 1 - (12/13)^2 cos^2 B = 1 - 144/169 cos^2 B = (169 - 144) / 169 cos^2 B = 25 / 169 cos B = 5 / 13 sin(90 - B) = 5 / 13 I hope this helped Kia
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Remember, it helps if you draw out the triangles. Given sec (A) = -13/12 and A is in Q2, so cos (A) = -12/13 = (adjacent/hypotenuse) opposite = √(13² - 12²) = 5 sin (A) = opposite/hypotenuse = 5/13 cos (A) = -12/13 tan (A) = -5/12 Given tan (B) = 7/24 = (opposite/adjacent) and B is in Q3, so hypotenuse = √(7² + 24²) = 25 sin (B) = -7/25 cos (B) = -24/25 tan (B) = 7/24 (a) sin (A + B) = sin (A) cos (B) + cos (A) sin (B) = (5/13) (-24/25) + (-12/13) (-7/25) = -36/325 (b) tan (A - B) = [tan (A) - tan (B)] / [1 + tan (A) tan (B)] = [(-5/12) - (7/24)] / [1 + (-5/12) (7/24)] = -204/253
Cos A = -12/13 = sinA = 5/13. sin B = -7/25, cos B = -24/25. Sin (A + B) = sinAcosB + cosAsinB = -24/65 + 84/325 = -36/325.
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Log_b (1 - 3x) = 3 + log_b (x) log_b (1 - 3x) - log_b (x) = 3 log_b (1 - 3x)/x = 3 <--- log m - log n = log (m/n) b^3 = ( 1 - 3x ) / x <--- express in exponential form b^3 = 1/x - 3 <--- separate into 2 fractions b^3 + 3 = 1/x <--- isolate the x 1/(b^3 + 3) = x <--- take the reciprocal of both sides x = 1/(b^3 + 3)
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If (a,b) is an element of LHS, then since b ∈ B=B1 U B2, either b∈B1 or b ∈ B2. Then (a,b) belongs to either (AxB1) or (AxB1) so it belongs to the RHS, showing the LHS is contained in the RHS. Conversely, if (a,b) is an element of the RHS then it belongs to either (A x B1) or (A x B2), so a∈A and b∈B1 or B2. Then b∈B (in either case), so (a,b) belongs to the LHS and hence the RHS is contained in the LHS, and equality holds. Hope this helps!
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You are using properties of logs ....loga^b= bloga in any base 1) logb(5^(1/2))= (1/2) logb(5)= (1/2)c or c/2 Log ab = loga + logb 2)logb(15)^(1/2)= (1/2)[ logb3 + logb5]= (1/2)(b+c) or (b+c)/2 Log (a/b)= loga - logb 3)logb (.4^(1/3))= (1/3)logb .4= (1/3) logb(2/5)= (1/3)[logb2 - logb5] = (1/3)[a-c] or (a-c)/3 4) logb (60^(1/4))= (1/4) logb (60)=(1/4)[ logb(2*2*3*5)] = (1/4)[logb2+ logb2 + logb 3 + logb 5] = 1/4)[a+a+b+c] = (2a+b+c)/4 Hoping this helps!
Hey log and Ln are 2 diffrent words. base of log is 10 and base of Ln is e(exponential). for Ln (y) – Ln (y-4) = Ln (2) => Ln [y/(y-4)] = Ln (2) taking antilog the two factors we get y/(y-4)=2 => y=2y-8 => y=8
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It can't be proven, it's not logically sound. The C comes from nowhere! It does work out you meant (C or ~B) instead of (C and ~B). Or if C was also a premise. It seems most likely that it is meant to say or: ~(A ^ B) > (C v ~B). This is how you should Derive that: Show ~(A ^ B) > (C v ~B) 1. ~(A ^ B) (assume antecedent) 2. ~A v ~B (1, DeMorgan's Law) 3. A (Premise) 4. ~~A (3, forget) 5. ~B (2,4, forget) 6. C v ~B (5, forget) The key here is that given ~A and ~(A ^ B), we get ~B from DeMorgan's Law. Once you know something you can add "or anything" because you are not gaining any information you didnt already know, infact you are losing information. If C had been a premise it would be done basicly the same way, just add C at 6. Note - > is ⊃, ^ is •, v is or "forget" means I forget the name of the rule, check the back of your Symbolic Logic textbook.
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18/a - 18/b = 18 ab(18/a - 18/b) = ab(18) 18b - 18a = 18ab 18b - 18ab - 18a = 0 18b - 18ab = 18a 18b(1 - a) = 18a b(1 - a) = a b = a/(1 - a)
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Let (b_n_k) be any subsequence of (b_n) that has a limit in the extended real system. Then, since (a_n_k + b_n_k) is a subsequence of (a_n + b_n) and a and b are in R, (a_n_k) has a limit and lim a_n_k= lim(a_n_k+ b_n_k) - lim b_n_k <= a + b - liminf b_n <= a + b - b = a It follows that limsup a_n <= a. And since, by assumption, liminf a_n >= a, then limsup a_n <= a <= liminf a_n. From this, it follows that liminf a_n = limsup a_n = lim a_n = a. It's immediate this implies that lim b_n = b. Another proof, maybe a bit simpler: limsup a_n < limsup(a_n + b_n) + limsup(-b_n) = a + b - liminf(b_n) <= a + b - b = a. As we've seen, this leads to the desired conclusion. To see this may fail if at least one of the numbers a and b is not finite, put a_n = n, b_n = (-1)^n. Then, a = oo, b = -1, lim (a_n + b_n) = oo = a + b. But (b_n) does not have a limit.
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Let |A|,|B| < ∞ and f:A→B. a. If f is injective, then for x,y ∈ A, x = y ⇔ f(x) = f(y). That is, there are at least as many elements in B as there are in A. However, there may exist y ∈ B such that f(x) ≠ y for ∀x ∈ A. So |A| ≤ |B|. b. If f is surjective, then for all y ∈ B, ∃x ∈ A such that f(x) = y. However, there may also exist z ∈ A such that f(z) = y. So |B| ≤ |A|. c. As has been shown, the fact that f is injective implies |A| ≤ |B| and the fact that g is injective implies |B| ≤ |A|. So |A| = |B|. Hence, there exists no y ∈ B such that f(x) ≠ y for ∀x ∈ A. So f is surjective, and therefore bijective. The argument for g's bijectivity is identical.
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See: S(x)=ax/(1+(x/b)^c) S'(x)=[ax'(1+(x/b)^c)- ax.(1+(x/b)^c)']/ (1+(x/b)^c)^2 S'(x)=[a(1+(x/b)^c)- acx.(1+(x/b)^(c-1)).1/b]/ (1+(x/b)^c)^2 S'(x)=a(1+(x/b)^c)[1-cx/ b(1+(x/b)]/(1+(x/b)^2c) S'(x)=a[1+(x/b)-cx]/ b(1+(x/b)(1+(x/b)^2) S'(x)=a[1+(x/b)-cx]/ b(1+(x/b)^3)
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B + (B*0.17*1.5) + (B*0.1) = $ 7,867,000 B + 0.255B + 0.1B = $ 7,867,000 1.355B = $ 7,867,000 B = $ 5,805,904.06
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9/10 = -3/2b - 3/4b ---------------Explained 9/10 = -b ( 3/2 + 3/4) ------- -b is common to both fractions 9/10 = -b ( (6+3)/4) -------- Evaluate within brackets 9/10 = -b x 9/4 (9/10)/(9/4) = -b ------------Divide across by 9/4 (9/10) X (4/9) = -b ----------Invert divisor fraction 4/10 = -b ---- Evaluate product of fractions, 9s cancel b = -4/10 or 0.4 = Answer
B=−2/5
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1. By definition of a negative number a + (-a) = 0 a*(b + (-b)) = a*0 = 0. Multiply out to get a*b + a*(-b) = 0 But by definition of a negative number a*b + -(a*b) = 0 Comparing the last two lines, a*(-b) = -(ab) To prove (-a)*b = -(a*b) start with (a + (-a))*b and go in a very similar way. 2. (-a)*(b + (-b)) = (-a)*0 = 0. Multiply out to get (-a)*b + (-a)*(-b) = 0 ----> -(a*b) + (-a)*(-b) = 0 (The last part comes from the proof in the part 1.) But by definition of a negative number -(a*b) + a*b = 0 Comparing the last two lines, (-a)*(-b) = a*b
Both are true.
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If you mean 21st century math. No. A lot of the math considered in math B is at least 300 years old. Modern mathematics builds on concepts learned in Math B.
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Let B' be the complement of B. Then, (A ∪ B) - B = (A ∪ B) ∩ B' = (A ∩ B') ∪ (B ∩ B') = (A ∩ B') ∪ ∅ = A ∩ B' If A and B are disjoint, then A ⊂ B', so that A ∩ B' = A If A and B are not disjoint, then, there's an element x that belongs to A and doesn't belong to B' (for x is in B). Then x is not in A ∩ B', which implies that A ∩ B' ≠ A.
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Let x represent an arbitrary element of A. If A ⊆ (B∩C) then x, being an element of A, is also an element of B∩C (by definition of subsets). Therefore, x is an element of B and x is an element of C (by definition of intersection). This holds for every element of A. Thus, every element of A is an element of B, so A ⊆ B, and every element of A is an element of C, so A ⊆ C. So if A ⊆ (B∩C), then A ⊆ B ∧ A ⊆ C. Conversely, if A ⊆ B ∧ A ⊆ C, then x, being an element of A, is an element of B (by definition of subsets), and x is an element of C (also by definition of subsets), therefore x is an element of B∩C (by definition of intersection). This holds for every element of A, so A ⊆ (B∩C) Thus, if A ⊆ B ∧ A ⊆ C, then A ⊆ (B∩C) We have shown that if A ⊆ (B∩C), then A ⊆ B ∧ A ⊆ C, and if A ⊆ B ∧ A ⊆ C, then A ⊆ (B∩C). Thus, A ⊆ (B∩C) is equivalent to A ⊆ B ∧ A ⊆ C by the definition of equivalence.
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A) A b) A union B Explanation for a: "Everything in both A and B" combined with "Everything in A but not in B" yields "everything in A" (regardless of B) Explanation for b: "Everything in a or in a or b" is just redundant.
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3( 1 - sin² B) = sin B 3 - 3sin² B = sin B 0 = 3sin² B + sin B - 3 3sin² B + sin B - 3 = 0 By quadratic formula: sin B = {-1 ± √[1 - 4(3)(-3)]}/2(3) sin B = {-1 ± √37}/6 sin B = (- 1 - √37)/6 = Discard < -1.0 sin B = (- 1 + √37)/6 B = arcsin((- 1 + √37)/6) = 57.9° or 122.1°
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B. sometimes.. if a = b = c, then yes a/b = (a+c) / (b+c) if c = 0, then yes.. a/b = (a+c) / (b+c) but.. for any other situations. a/b ≠ (a+c) / (b+c)
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6(b - 4) > 30 --- distribute through the parenthesis 6b - 24 > 30 -- add 24 to both sides 6b > 30 + 24 6b > 54 -- divide both sides by 6 b > 54/6 b > 9 or, if it is easier, do it this way.. 6(b - 4) > 30 -- divide both sides by 6 b - 4 > 30/6 b - 4 > 5 -- add 4 to both sides b > 5 + 4 b > 9 either way, you will arrive at the same answer :)
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(1-2i)/(2-i) + a + bi = b - ai Multiply the fraction by (2+i)/(2+i): ((1-2i)(2+i)) / ((2-i)(2+i)) = (2 - 4i + i - 2i^2) / (4 - i^2) = (4 - 3i) / 5 = 4/5 - 3/5i Therefore, 4/5 - 3/5i + a + bi = b - ai a - b = -4/5 ai + bi = 3/5i (so divide through by i) a + b = 3/5 a - b = -4/5 a + b = 3/5 2a = -1/5 a = -1/10 -1/10 + b = 3/5 b = 7/10 Answer: a = -1/10 and b = 7/10
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If d divides a and b then it also divides r (= a - bq) and b. In particular since HCF(a,b) divides a and b it also divides r and b ==> HCF(a,b) divides HCF(r,b). On the other hand if d divides b and r then it also divides a(=bq + r) and b. In particular since HCF(r,b) divides r and b it also divides a and b ==> HCF(r,b) divides HCF(a,b). From this we see that HCF(a,b) = HCF(r,b). So k = 1/2.
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Your least common denominator is (b)(b + 4), so multiply EACH term on BOTH sides of the equation by this. In each case, one of the factors will cancel out and you will now have: b(b) - 1(b+4) = 2(b) or b^2 - b - 4 = 2b ***be sure you distributed the NEGATIVE sign to both terms!*** This becomes b^2 - 3b - 4 = 0 At this point, either factor or use the quadratic formula, so (b - 4)(b + 1) = 0 so b = 4 or -1
(b / b+4) - (1 / b ) = 2 / b+4 -(1 / b) = 2/ (b+4) - (b/ (b + 4)) -(1/b) = (2 - b) / (b + 4) b(2 - b) = -1 ( b+4) 2b - b^2 = -b - 4 b^2 - b3 - 4 = 0 (b + 1) (b - 4) = 0 b = 4, b = -1
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Given that a and b are positive numbers, assume: sqrt(a^2+b^2)=a+b. Then squaring both sides, we have: a^2 + b^2 = a^2 + 2ab + b^2 Subtracting (a^2 + b^2) from both sides, we have: 2ab = 0 Clearly, 2ab=0 is only true when a=0 and/or b=0, in which case a and/or b are not positive. Since it is given that a and b are positive, we have a contradiction, so our assumption, sqrt(a^2+b^2)=a+b, is incorrect. Therefore, if a and b are positive numbers, then sqrt(a^2+b^2) does not equal a+b. I hope I helped!
Evaluate a semicircle with diameter AB and a element on the circumference C From C drop a perpendicular to the circumference to fulfill it at D and allow advert = a and BD = b Now evaluate ?'s ACD and CBD considering the fact that <ACD is a acceptable perspective (perspective in a semicircle is a acceptable perspective) <ACD is the supplement of <BCD yet <CAD is likewise the supplement of <ACD for this reason <CAD = <BCD additionally < CDA = <BDC = ninety° (by shape) for this reason ?'s ACD and CBD are comparable (equiangular) for this reason, considering the fact that corresponding factors are in share So advert/CD = CD/BD as a effect CD² = advert.BD = ab as a effect CD = ?(ab) Now CD ? radius of the semicircle AB = diameter = advert + DB = a + b So radius = ½(a + b) for this reason ?(ab) ? ½(a + b) and of direction this says that the geometric propose of two numbers is below or equivalent to their arithmetic propose. The equality holds whilst the two numbers are equivalent. (ie whilst D coincides with the centre of the circle) Euclid became so stylish along with his geometric proofs!!!!!!
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Here's one way: Square both sides. |A + B|^2 = |A - B|^2 The magnitude squared of a vector is the same as the dot product with itself. |A|^2 = A dot A. |A + B|^2 = (A + B) dot (A + B) = (A dot A) + 2(A dot B) + (B dot B) |A - B|^2 = (A - B) dot (A - B) = (A dot A) - 2(A dot B) + (B dot B) Set those equal and the (A dot A) and (B dot B) terms cancel. You should come to the conclusion that A dot B = 0, which will answer the question. Here's another approach if you're not familiar with the identities I used there. Add A and B graphically by drawing A and B head to tail. They form the two sides of a triangle. Call the third one C = A + B. By the law of cosines C^2 = A^2 + B^2 - 2AB cos(theta) where theta is the angle between them drawn head to tail. But "the angle between them" is not this theta, it's 180 - theta, which you can see if you actually draw the picture. Call that angle theta2. So C^2 = A^2 + B^2 + 2AB cos(theta2). This is my first equation above. If you draw the other triangle for A - B with the same C, then the law of cosines will give you C^2 = A^2 + B^2 - 2AB cos(theta2). Again setting them equal gives you cos(theta2) = 0, which is the same conclusion.
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I have to find P(a Intersection B) given P(a) = 0.2 P(b)=0.4 and p(a|b) + p(b|a) = 0.75 Now I've done a page of working out, but I ended up with a loop and couldn't solve it for real numbers, I was using p(a|b) = p(a intersect b) / p(b) but kept going around in circles.... is there an exact answer for this... or a formula that I need to know about intersection??
P(a|b) = p(a intersect b) / p(b) similarly p(b|a) = p(a intersect b) / p(a) now add these 2 equations p(a|b) = p(a intersect b) / p(b) p(b|a) = p(a intersect b) / p(a) --------------------------------------... p(a|b)+p(a|b) = p(a intersect b) / p(b) +p(a intersect b) / p(a) .75 = [p(a intersect b) /.4] +[p(a intersect b) / .2] taking lcm in the right hand side lcm is .4 .75 = [p(a intersect b) +2p(a intersect b)] /[.4] .75 = [3p(a intersect b) ] /[ .4] .75 = [3p(a intersect b) ] /[ .4] .75*.4=3p(a intersect b) .3=3p(a intersect b) .3/3=p(a intersect b) .1=p(a intersect b)
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A+b+c=11 a+b=11-c b+c=11-a c+a=11-b (11-c)(11-a)(11-b)+abc = (121-11a-11c+ac)(11-b)+abc= 1331-121b-121a+11ab-121c+11bc+11ac -abc+abc= 1331-121(a+b+c)+11(ab+bc+ca) => 1331-121(11)+11(20) =>220 (a+b)(b+c)(c+a)+abc= 220
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B=e^a taking log on both sides, ln(b)=ln(e^a) since by property of logarithms, ln(x^y) = yln(x) Therefore, ln(b)=aln(e) ln(e)=1 Hence, ln(b)=a
Check online or your textbooks
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I was looking up Engineering courses in different universities, and I would like to know the difference between a B.Sc (Bachelor of Science) in Engineering, B.E. (Bachelor of Engineering), and B.Tech (Bachelor of Technology). I'd also like to know which one would offer a higher chance in getting a job , and basically what kind of stuff in taught in each course. ~ :)
B.sc is not engineering...B.E & B.Tech comes under engineering category...B.tech basically has higher chance of getting job
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It's clearly true when k is 0 or 1. Assume its true for some integer k>=0, and consider: b^[m(k+1)]. From the distributive and power-of-a-sum properties you have: b^[m(k+1)] = b^(mk + m) = b^(mk) + b^m From the assumption, the first term is b^(mk) = (b^m)^x, so b^[m(k+1)] = (b^m)^k * b^m = (b^m)^k * (b^m)^1 = (b^m)^(k + 1) So, if the statement is true for m,k then it's true for m,(k+1). Since the statement is clearly true for k=0 and any m, then it's true from all m,k where k>=0 is an integer. I've tacitly assumed that b is nonzero. Otherwise, there may be problems when m or k is 0, since 0^0 is undefined.
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1) n = 1 (b^(1+1)-1)/(b-1) = (b^2-1)/(b-1) = (b-1)(b+1)/(b-1) = b+1 --> Ok, we have this! 2) Let n = k 1+b+b^2 + ... + b^k = (b^(k+1)-1)/(b-1) 3) k+1 1+b+b^2+ ... + b^k + b^(k+1) = (b^(k+1)-1)/(b-1) + b^(k+1) = [b^(k+1)-1 + b^(k+1)*(b-1)]/(b-1) = [b^(k+1) - 1 + b^(k+2) - b^(k+1)] / (b+1) = [b^(k+2) - 1] / (b+1) = [b^(k+1+1) - 1]/(b+1) ---> So, we proved that if we put k+1 on the left side we will eventually get k+1 on the right side!
Upload the fractions interior the denominator first, so your denominator is (a + b) / (ab) Now, we are dividing by this FRACTION so we multiply by the reciprocal. answer = (ab) / (a + b)
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If A is similar to B then we can find an invertible matrix D such that: A = D^(-1) B D therefore A^2 = (D^(-1) B D)(D^(-1) B D) = D^(-1) B^2 D so A^2 is similar to B^2. It follows that A^n is similar to B^n for all positive integers n since A^n = D^(-1) B^n D. On the other hand det(A) = det(D^(-1) B D) =det(D^(-1)) det(B) det(D) = (1/det(D)) det(B) det(D) = det(D)/det(D) * det(B) = 1* det(B) = det(B). So det(A) = det(B).
Are you supposed to prove this? Looks like an induction proof. A similar to B means A = inv(P)*B*P for some invertible P. Multiply by A: A^2 = inv(P)*B*P * inv(P)*B*P = inv(P) * B * [P * inv(P)] * B * P which gives you the first part. Now can you do a similar trick for the induction step? Suppose you know A^k ~ B^k where ~ means "is similar to". Can you use straight multiplication as I did to show that A*A^k ~ B*B^k or B^k*B? For the second part: What do you know about the determinant of products of matrices? What do you know about the determinant of inverses of matrices? Apply that to the expression for A in terms of B.
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Log_b(x) + log_b(x+2) - log_b(8) = 0 log_b(x * (x+2)) - log_b(8) = 0 log_b(x * (x+2)) = log_b(8) raise both sides to the power of b: b^log_b(x * (x+2)) = b^log_b(8) x * (x+2) = 8 x^2 + 2x - 8 = 0 (x+4)(x-2) = 0 x = -4,2 PS. the generally accepted notation for subscript is " _<subscript> ". Just an FYI! Using brackets is fine too since you explained what you were using the brackets for.
Log(b)x+log(b)(x+2)-log(b)8 = log{x(x+2)/8} because log (ab) = log a + log b and log(c/d) = log c - log d..................Ans
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Note (a) : P(X) + P(X') = 1. Note (b) : If X and Y are both events in the same sample space S, then they are independent iff : P( X ∩ Y ) = P(X) · P(Y). ............... (1) __________________________________ To Prove : A, B are ind. iff A', B are ind.. __________________________________ Part - 1 : Let : A and B be ind., i.e., P(A∩B) = P(A) · P(B) ................ (2) Then, we have P(A' ∩ B ) = P(B) - P(A∩B) . . . . . . . .= P(B) - P(A) · P(B) ............ from (2) . . . . . . . .= [ 1 - P(A) ] · P(B) . . . . . . . . .= P(A') · P(B) Hence, from (1), A' and B are ind.. _______________________________ Part - 2 : Let : A' and B be ind., i.e., P( A' ∩ B ) = P(A') · P(B) ...... (3) Again, we have : P(A'∩B) = P(B) - P(A∩B). Then, we have : P(A∩B) = P(B) - P(A'∩B) . . . . . .= P(B) - P(A') · P(B) ...... from (3) . . . . . .= [ 1 - P(A') ] · P(B) . . . . . .= P(A) · P(B) Hence, from (1), A and B are ind.. __________________________________ The above two parts prove the ' iff ' part, i.e, prove the result ' biconditionally '. Hence, the result. ____________________________________ Happy To Help ! ____________________________________
A and B are independent if and only if P(A and B) = P(A)P(B) if and only if P(A and B) + P(A comp. and B) - P(A comp. and B) = P(A)P(B) if and only if P(B) - P(A comp. and B) = P(A)P(B) if and only if P(A comp. and B) = P(B) - P(A)P(B) if and only if P(A comp. and B) = (1 - P(A))P(B) if and only if P(A comp. and B) = P(A comp.)P(B) if and only if A complement and B are independent.
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If √a e √b are rational, then the conclusion is immediate. For the converse, suppose at least one of the number √a and √b, say √a without loss of generality, is irrational. √a + √b and √a - √b are roots of the polynomial P(x) = x² - 2√a x + a - b Since, by assumption, √a is irrational and a and b are rational, then 2√a is irrational and a - b is rational. It follows that, for rational x, P(x) is irrational and, therefore, nonzero. Since √a + √b and √a - √b are roots of P, then they must be irrational. By contraposition, it follows that, if √a + √b (or√a - √b) is rational, then so are √a and √b.
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We are looking for k(a-1)(b-1)(c-1) = abc-1 such that k is an integer. Therefore k = (abc-1)/(a-1)(b-1)(c-1) < [a/(a-1)][b/(b-1)][c/(c-1)]. The largest possible value for k occurs for the smallest possible values of a, b=a+1, c=a+2 so k < (a+2)/(a-1). When a=2 then k < 4/1 = 4. The smallest possible value for k occurs for large values of a,b,c. Then k > 1. So k = 2 or 3. When k=2 then a+2 >= 2(a-1), a <=4. So a = 2, 3 or 4. k=2 : 2(a-1)(b-1)(c-1) = abc-1. To make the RHS even each of a, b, c must be odd. a can only be 3. Then b>=5 and c>=7. 4(bc-b-c+1)=3bc-1 bc-4b=4c-5 b(c-4)=4(c-4)+11. 11 must be a multiple of c-4, so c=11+4=15 and b=4+11/11=5. k=3 : 3(a-1)(b-1)(c-1) = abc-1. If a=2 then 3(bc-b-c+1)=2bc-1 bc-3b=3c-4 b(c-3)=3(c-3)+5. 5 must be a multiple of c-3, so c=3+5=8 and b=3+5/5=4. If a=3 then 6(bc-b-c+1)=3bc-1 3bc-6b=6c-7 3b(c-2)=6(c-2)+5 b=2+5/3(c-2). 5 is not a multiple of 3(c-2) for any value of c. No solution. If a=4 then 9(bc-b-c+1)=4bc-1 5bc-9b=9c-10. b(5c-9)=(5c-9)+(4c-1). 4c-1 must be a multiple of 5c-9 : 4c-1>=5c-9, 8>=4c, c<=2 which contradicts c>a. No solution. ANSWER : the only solutions are (a,b,c) = (2,4,8) and (3,5,15).
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You may be slightly confused by the notation here; T(a,b) is a function from X to X. The cases you write out are defining the value of the function T(a,b) at an element x in X. So it's not "T(a,b)" that equals x, or b, or a, but "the value of the function T(a,b) at x" which if you use the most common notation for function application would be written T(a,b)(x). But this may look confusing, as the parentheses surrounding (a,b) are part of the name of the function and only the parentheses surrounding x denote function application. So in answering I am going to denote the function T(a,b) by f. Namely, you have fixed elements a and b of x, and defined a function f from X to X by f(x) = x if x is not in {a,b}, f(b) = a, f(a) = b. We are told to compute f o f. Well, if x is not an element of {a,b} we have f(f(x)) = f(x) [by the definition of f: because x isn't in {a,b}, f(x) = x] = x [by the definition of f, again, because x isn't in {a,b}, f(x) = x]. We also have f(f(a)) = f(b) [by the definition of f: f(a) = b] = a [by the definition of f, again: f(b) = a]. and f(f(b)) = f(a) [by the definition of f: f(b) = a] = b [by the definition of f, again: f(a) =b]. What this shows is that f(f(t)) = t for all t in X. In all cases--- it doesn't matter whether or not t is in {a,b}--- the calculation shows that when we apply f twice to t, we get t back. This shows that f is one-to-one (if f(x) = f(y), applying f to both sides you deduce f(f(x)) = f(f(y)) and because f(f(x)) = x and f(f(y)) = y, you deduce that x = y) and onto (given x in X, we see that x is in the range of f because f(f(x)) = x). Slightly more glibly, f is one-to-one and onto because the fact that f(f(x)) = x for all x in X shows that f is invertible: in fact, f is its own inverse. In any case, that's why f is a permutation of X.
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First, what are the backslashes (\) for? Are they supposed to denote division? If so, you should use forward slash (/). Second, you need to use parentheses to make expression clearer. Is 31+b/2a-b-3a-b/2a+b supposed so be: 31 + b/(2a) - b - 3a - b/(2a) + b (31+b)/(2a) - b - 3a - b/(2a+b) (31+b)/(2a) - b - (3a-b)/(2a+b) (31+b)/(2a) - b - (3a-b)/(2a) + b (31+b)/(2a) - (b-3a-b)/(2a) + b (31+b)/(2a) - (b-3a-b)/(2a+b) 31 + b/(2a-b) - 3a - b/(2a) + b 31 + b/(2a-b-3a) - b/(2a) + b (31+b)/(2a-b) - 3a - b/(2a) + b (31+b)/(2a-b-3a) - b/(2a) + b . . . And is 2a/4a^2-b^2 supposed to be 2a/(4a^2) - b^2 or 2a/(4a^2-b^2) ? ____________________ Using parentheses then, we get: [(3a+b)/(2a−b) − (3a−b)/(2a+b)] / [2a/(4a²−b²)] = [(3a+b)/(2a−b) − (3a−b)/(2a+b)] * (4a²−b²)/(2a) = [(3a+b)/(2a−b) − (3a−b)/(2a+b)] * (2a−b)(2a+b)/(2a) = [(3a+b)/(2a−b) * (2a−b)(2a+b)/(2a)] − [(3a−b)/(2a+b) * (2a−b)(2a+b)/(2a)] = [(3a+b) * (2a+b)/(2a)] − [(3a−b) * (2a−b)/(2a)] = [(3a+b)(2a+b) − (3a−b)(2a−b)] / (2a) = [(6a² + 5ab + b²) − (6a² − 5ab + b²)] / (2a) = 10ab / (2a) = 5b Mαthmφm
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Cos(A+B)cos(A-B) =>[(cosAcosB - sinAsinB)(cosAcosB + sinAsinB)] =>[(cosAcosB)^2 - (sinAsinB)^2] =>[cos^2(A)cos^2(B) - sin^2(A)sin^2(B)] =>[cos^2(B)(1 - sin^2(A)) - sin^2(A)(1 - cos^2(B)] =>[cos^2(B) - sin^2(A)cos^2(B) - sin^2(A) + sin^2(A)cos^2(B)] =>[cos^2(B) - sin^2(A)] => - [sin^2(A) - cos^2(B)] => -(sinA + cosB)(sinA - cosB)
I really don't know what you 're trying to prove, so I"m just doodling. c(A+B) = cA cB - sA sB c(A-B) = cA cB + sA sB c(A+B) * c(A-B) = (cA cB)^2 - (sA sB)^2 = (1 - sA^2) cB^2 - sA^2 * (1 - cB^2) = cB^2 - sA^2 = (cB + sA) * (cB - sA)
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Sin(a)cos(b) + cos(a)sin(b) + cos(a)cos(b) + sin(a)sin(b)= ---> group first with third and second with fourth [sin(a) + cos(a)]*cos(b) + [cos(a) + sin(a)]*sin(b) = [sin(a) + cos(a)]*[sin(b) + cos(b)] cos(a)cos(b) - sin(a)sin(b) - sin(a)cos(b) + cos(a)sin(b) = ---> group first with fourth and third with second cos(a)*[cos(b) + sin(b)] - sin(a)*[sin(b) + cos(b)] = [sin(b) + cos(b)]*[cos(a) - sin(a)] When you put them all together [sin(b) + cos(b)] simplifies and you get: [sin(a) + cos(a)] / [cos(a) - sin(a)] You can still make some rationalizing with sqrt(2)/2 and put sin(45) or cos(45) to get some formulas. I mean 45 degrees = pi/4. [sin(a) + cos(a)] / [cos(a) - sin(a)] = [sqrt(2)/2]*[sin(a) + cos(a)] / [sqrt(2)/2]*[cos(a) - sin(a)] = ( [sqrt(2)/2]*cos(a) + [sqrt(2)/2]*sin(a) ) / ( [sqrt(2)/2]*cos(a) - [sqrt(2)/2]*sin(a) ) = ( sin(45)*cos(a) + cos(45)*sin(a) ) / ( cos(45)*cos(a) - sin(45)*sin(a) ) = sin(a+45) / cos(a+45) = tan(a+45) Best regards!
B
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Log[b](x^(1/2)) = sqrt(log[b](x)) (1/2) * log[b](x) = sqrt(log[b](x)) log[b](x) = t (1/2) * t = sqrt(t) (1/4) * t^2 = t t^2 = 4t t^2 - 4t = 0 t * (t - 4) = 0 t = 0 t - 4 = 0 t = 4 t = 0 , 4 log[b](x) = 0 , 4 b^0 , b^4 = x 1 , b^4 = x I don't understand what you mean by "b>0 add to 17"
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B. A/B c. A*B Addition and subtraction require both terms to have the same dimensions, while division and multiplication do not.
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A^-1 - b^-1 / a^-2 - b^-2 = [(1/a) – (1/b)] / [(1/a²) – (1/b²)] = [(b/ab – a/ab)] / [(b²/a²b²) – (a²/a²b²)] = [(b-a)/ab] / [(b²-a²)/a²b²] = (a²b²(b-a)) / ab(b²-a²) = a²b²(b-a) / ab(b+a)(b-a) = ab/(b+a) → (c)
A^-1 - b^-1 / a^-2 - b^-2 (1/a - 1/b) / (1/a^2) - (1/b^2) (b-a/ab) / [(b^2-a^2)/(a^2b^2)] (b-a/ab) / [(b-a)(b+a)/(a^2b^2)] (b-a/ab) * (a^2b^2)/(b-a)(b+a) ab/b+a answer is c.
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Sqrt(7)*(a + b) + sqrt(9)*b = 4 - a sqrt(7)*a + sqrt(7)*b + sqrt(9)*b = 4 - a b(sqrt(7) + sqrt(9)) = 4 - a - sqrt(7)*a b = (4 - a - sqrt(7)*a) / (sqrt(7) + sqrt(9)) sqrt(7)*a + a = 4 - sqrt(7)*b - sqrt(9)*b a(sqrt(7) + 1) = 4 - sqrt(7)*b - sqrt(9)*b a = (4 - sqrt(7)*b - sqrt(9)*b) / (sqrt(7) + 1) You have one equation but two unknowns. That's all you can do without more information.
You cannot find a unique solution for two variables with only one equation.
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X-a/x-b + x-b/x-a = a/b + b/a Multiplying across by ab(x-a)(x-b)... ab(x-a)² +ab(x-b)² = (a²+b²)(x-a)(x-b) ab(x²-2ax+a² +x²-2bx+b²) = (a²+b²)(x²-(a+b)x+ab) ab(2x²-2(a+b)x+a²+b²) = (a²+b²)(x²-(a+b)x+ab) Making the algebraic substitution e=(a+b), f=ab to make it more tractable, and noting (a²+b²)=e²-f: f(2x²-2ex+e²-f) = (e²-f)(x²-ex+f) 2fx² -2efx +f(e²-f) = (e²-f)x² - e(e²-f)x +f(e²-f) 2fx² -2efx = (e²-f)x² - e(e²-f)x Dividing across by x = 0 which is one solution (as Raichu comments) (2f + f-e²)x = 2ef - e(e²-f) x = e(f-e²) / (3f-e²) x = e(e²-f) / (e²-3f) Substituting back for e=(a+b), f=ab x = (a+b)((a+b)²-ab) / ((a+b)²-3ab) x = (a+b)(a²+ab+b²) / (a²-ab+b²) Looks as good as you can get it. If you try the formula you get painful experiences like below: ______________________________________... So applying: x = [ -B ± √(B² - 4AC) ] / 2A with A = 2 B = -2(a+b) C = (a²+b²)(ab-1)/ab... x = [ 2(a+b) ± √(4(a+b)² - 8(a²+b²)(ab-1)/ab)... ] / 4 Crunching out the determinant D, or D² for convenience D = √(4(a+b)² - 8(a²+b²)(ab-1)/ab) D² = 4(a²+2ab+b²) - 8(a²+b²)(ab-1)/ab = [ 4a³b+8a²b²+4ab³ -8a³b -8ab³ +8a² +8b² ] /ab = 4[ -a³b+2a²b²-ab³ +2a² +2b² ] /ab If you make some algebraic substitution like u=(a+b), v=ab then maybe you get something more tractable: A = 2 B = -2u C = (u²-v)(v-1)/v
I presume you mean (x-a)/(x-b) + (x-b)/(x-a) = a/b + b/a. Multiply both sides by (x-a)(x-b) and you get (x-a)^2 + (x-b)^2 = (a/b + b/a)(x-a)(x-b). Rewrite as (x-a)^2 + (x-b)^2 - (a/b + b/a)(x-a)(x-b) = 0. Expand the brackets, collect like terms and simplify. You'll end up with a quadratic of the form Ax^2 + Bx + C = 0 which is solved by the quadratic formula (-B +or- sqrt(D)) / 2A where D= B^2 - 4AC. In fact, I tried solving it and it simplifies so that you don't even need the formula. You can see it because in fact x=0 is a solution which makes C=0, so the other solution is x=-B/A.
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Log_b(2)=A and log_b(5)=B and log_b(11)=c? Log(2)/logb=A Log(2)=A(logb) a) log_(5b)2 log2/log(5B) A(logb)/log(5B) (logb)^A/log(5B) Answer: Log[5B]b^A log_b(11)=c log11/logb=c log11=c(logb) log11=logb^c log_b(5)=B log5/logb=b log5=b(logb) log5=logb^b b) log_(5b)55 log55/log(5B) log11+log5/log(5B) we know that log11 is log11=logb^c and log5=logb^b logb^c+logb^b/log(5B) log(b^c*b^b)/log(5B) log(b^c+b)/log(5B) Answer: log[5B](b^c+b)
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Expand cos (a+b) c o s ( a + b ) NOT THAT SORT OF EXPAND!!! Cos (a+b) = cos a cos b - sin a sin b Now draw two right angled triangles, and on the first mark angle a and the opposite as 7cm and the hypotenuse as -25, making the adjacent sqrt(625-49) = 24 so (CARE: it must be in the third quad) cos a = -24/25. On the second triangle mark angle b and the opposite 15 and the adjacent -8, making the hypotenuse 17 so sin b = 15/17 and cos b = -8/15. You now have the four values to substitute in. -24/25 x -8/15 - -7/25 x 15/17 =
Sin(a) = -7/25, cos(a) = -24/25. cot(b) = -8/15, sin(b) = 15/17, cos(b) = -8/17. cos(a+b) = cos(a) cos(b) - sin(a) sin(b) = +192/425 +105/425 = 297/425.
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Well, this a circular permutation, and it goes like this ! I will assume a, b and c are positive let s = a+b+c then we have : s(a+b) = 21 s(c+b) = 42 s(a+c) = 98 add them all : s(a+b + c+b + a+c) = 21+42+98 s(2a + 2b + 2c) = 161 2s^2 = 161 s = sqrt(161/2) PS: isn't there any error among the numbers : 21, 42, 98 ? for example, if we would have a sum = 162 instead of 161 then we would obtain s = sqrt(162/2) = sqrt81 = 9 hope it' ll help !!
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A/1 = b/(d+c) (turned upside down both fractions) 1/a = (d + c)/b b/a = d + c d + c = b/a d = b/a - c (answer)
A million. and 2 could genuine be solved as quickly as you recognize that the diagonals of an isosceles trapezoid are equivalent. 3. EF is the conventional of BC and advert 4. comparable approach. (2x + 2)/2 = 12 5. advert is parallel to EF
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Vector A: ------> vector B; <---------- choices are: A) Vector A-vector B B) A-B C) Vector B-vector A D) B-A E) Vector A + vector B F) A+B G) |vector B- Vector A| I think the answer is D, but im not sure. Also I think there is more than one answer. Can you guys please help me out. Thanks.
Because vector B is 180 degrees to Vector A, The length of Vec A - Vec B = |A| + |B| A - B ----->---------->
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[ 1/B^2 - 1/A^2 ] / (B - A) [ (A^2 - B^2)/A^2B^2 ] / (B - A) [ (A + B)(A - B)/A^2B^2 ] / (B - A) [ -1(A + B)(-A + B)/A^2B^2 ] / (B - A) -1(A + B)/A^2B^2 For your enjoyment .. instanteous rate of change at x = A lim[B ~> A] -1(A + B)/A^2B^2 lim[B ~> A] -1(A + A)/A^2A^2 = -2/A^3 .. instanteous rate of change at x = B lim[A ~> B] -1(A + B)/A^2B^2 lim[A ~> B] -1(B + B)/B^2B^2 lim[A ~> B] -2/B^3
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(ax+b)/c <=b .... Given ax + b >= bc .... (since c is negative, flip the inequality when multiplying) ax >= bc - b ..... (no change for subtraction) x <= (bc - b) / a .(flip it again because a is negative too)
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A - b > 0 a - b + b > 0 + b a > b Let's go through each possible answer: -ab > a -ab / a > a / a -b > 1 not necessarily Let a = -2 Let b = -5 -ab > a -(-2)(-5) > -2 -10 > -2 FALSE b + 5 > a - 5 b + 5 - 5 > a - 5 - 5 b > a - 10 not necessarily Let a = 100 Let b = 10 b + 5 > a - 5 10 + 5 > 100 - 5 15 > 95 FALSE b^4 > a^4 FALSE as a > b Let a = 4 Let b = 2 b^4 > a^4 2^4 > 4^4 16 > 256 FALSE b^3 > a^2 not necessarily as b could be negative and thus b^3 would be negative whereas a^4 will always be positive Let a = 2 Let b = -3 b^3 > a^2 (-3)^3 > 2^2 -27 > 4 FALSE 2a > b not necessarily as a could be negative Let a = -5 Let b = -7 2a > b 2(-5) > -7 -10 > -7 FALSE None of those MUST be TRUE as I just demonstrated cases where each is FALSE
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(2a + b) * (a - 3b) = 2(a * a) - 5(a * b) - 3 (b * b), since dot product is commutative = 2 |a|^2 - 5 (a * b) - 3 |b|^2 = 2 (2|b|)^2 - 5 (a * b) - 3 |b|^2, since |a| = 2|b|. = 5 |b|^2 - 5(a * b). Since (2a + b) * (a - 3b) = 0, by hypothesis, we have that a * b = |b|^2. However, a * b = |a| |b| cos t, where t is the angle between a and b. ==> |b|^2 = |a| |b| cos t ==> |b|^2 = 2|b|^2 cos t. If b is nonzero (note this also implies that a is nonzero), 1 = 2 cos t ==> t = pi/3. I hope this helps!
Vector silenced with pink dot sight is powerful.. you dont want maths for call of accountability remeber mutually as aiming purpose at a 2 degree suggestions-set for a suited headshot why in the international might you think of of of you p.c.. decimals and prendicular for u and v? in case you p.c.. to degree the gun get a ruler and degree it
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P(x)= x³ - (a+b)x² + ab·x -17 ,, a+b>0 and ab>0 --> p(x)<0 if x<0 --> If c integer root,, f(c)=0 --> c|17 and c>0 --> c=1 or c=17 * If c=1 --> (1-a)·(1-b) = 17 ,, a,b≠1 --> a,b>1 b-1=17/(a-1) --> a=2, b=18 or a=18, b=2 --> a+b+c=18+2+1=21 * If c=17 --> 17·(17-a)·(17-b)=17 --> (17-a)·(17-b)=1 --> a,b <17 or a,b>17 --> a=b=16, a+b+c=49 // or a=b=18 and a+b+c=53 Saludos.
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[sin(A + B) + sin(A - B)] / [cos(A + B) + cos(A - B)] = tan(A) Recall the following two facts: sin(A)cos(B) = (1/2)[sin(A + B) + sin(A - B)] cos(A)cos(B) = (1/2)[cos(A + B) + cos(A - B)] Sub those into the left side: [2 sin(A)cos(B)] / [2cos(A)cos(B)] tan(A) Thus LS = RS and QED. Alternatively, you could expand everything out in the left side with the sine and cosine addition and subtraction formulas. ********* sin(α) + sin[α +(2π)/(3)] + sin [α + (4π)/(3)] = 0 Apply sin(a + b) = sin(a)cos(b) + sin(b)cos(a) in the left side: sin(α) + sin(α)cos(2π / 3) + sin(2π / 3)cos(α) + sin(α)cos(4π / 3) + sin(4π / 3)cos(α) Use the following facts (from the unit circle or otherwise): sin(2π / 3) = √(3) / 2 sin(4π / 3) = -√(3) / 2 cos(2π / 3) = -1 / 2 cos(4π / 3) = -1 / 2 sin(α) + (-1/2)sin(α) + (√(3) / 2)cos(α) - (1/2)sin(α) + (-√(3) / 2)cos(α) 0 Thus LS = RS and QED. Done!
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1) x^3 - y^3 = (x - y)(x^2 + xy + y^2) (<== formula) a^3 - b^3 = (a - b)(a^2 + ab + b^2) 2) (a - b)^3 = (a - b)(a - b)(a - b) = (a*a - b*a - ab + b*b)(a - b) = (a^2 - ab - ab + b^2)(a - b) = (a^2 - 2ab + b^2)(a - b) = a^2*a - 2ab*a + b^2 * a - a^2*b + 2ab*b - b^2*b = a^3 - 2a^2b + ab^2 - a^2b + 2ab^2 - b^3 = a^3 - 2a^2b - a^2b + ab^2 + 2ab^2 - b^3 = a^3 - 3a^2b + 3ab^2 - b^3
I think that would be factorisation for the first? (a-b)(a^2 + ab + b^2).
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Remember: loga(b) = c => b = a^c Hence: logb(M) = a => M = b^a logb(N) = b => N = b^b logb(P) = c => P = b^c logb[sqrt(P)(N^3/M)] =logb[sqrt(b^c)(b^3b / b^a)] =logb[b^0.5c(b^3b / b^a)] =logb(b^1.5cb / b^a) =logb(b)^(1.5cb - a) =1,5cb - a
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B. 1 or -1 ax + b = bx + a (a-b)x = a-b x = 1 or ax + b = -bx - a (a+b)x = -a - b x = -1 Hope that helps! (Oh, and thanks for explaining that those "|" were absolute value. I thought they were lower case L's!)
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P(A U B) = P(A) + P(B) - P(A n B) where 'n' represents intersection Solving for P(A n B): P(A n B) = P(A) + P(B) - P(A U B) P(A n B) = 3/4 + 2/5 - 4/5 = 7/20
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S = {(4a + b)/(a + 4b)} + {(4b + c)/(b + 4c)} + {(4c + a)/(c + 4a)} (1) a......the biggest number. a = b, a = c ........possible. (4a + b)/(a + 4b) = {(a + b) + 3a}/{(a + b) + 3b} ≥ 1 (4b + c)/(b + 4c) > (4b + c)/(16b + 4c) = 1/4 (4c + a)/(c + 4a) > (4c + a)/(16c + 4a) = 1/4 add three 3/2 < S b or c......the biggest number. in the same way. (2) a......the smallest number. a = b, a = c ........possible. (4a + b)/(a + 4b) = {(a + b) + 3a}/{(a + b) + 3b} ≤ 1 (4b + c)/(b + 4c) < (4b + 16c)/(b + 4c) = 4 (4c + a)/(c + 4a) < (4c + 16a)/(c + 4a) = 4 add three S < 9 b or c......the smallest number. in the same way.
A is a really nice letter, according to my grandmother.
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According to the definition I found here https://en.wikipedia.org/wiki/Symmetric_...AΔB is "the set of elements which are in either of the sets and not in their intersection" AUB is the set of elements which are in either A and B A∩B is the intersection P - Q is the set of elements which are in P and not in Q therefore AΔB = (AUB) - (A∩B) Let A and B be any two sets. If A∩X = B∩X= Ø and AUX = BUX for some set X. Show that A = B. Let p∈A then if A∩X = Ø, p∉X and if AUX = BUX this means that AUX = BUX - X = B, so p∈B p∈A → p∈B is equivalent to say that A⊆B Let now p∈B. If B∩X= Ø then p∉X and if BUX = AUX but p∉X, this means that BUX = AUX - X = A p∈B → p∈A means B⊆A A⊆B and B⊆A means A = B. QED
To show A=B Let a be in A. Then a is in AUX, so a is in BUX But since A∩X is empty, a is not in X. Therefore a is in B Now pick b in B. Then b is in BUX, so b is in AUX. But since B∩X is empty, b is not in X, therefore b is in A Therefore A and B are subsets of each other, hence equal
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Hi a) (1) 2x + y = a (2) 3x + 6y = b 6 *(1) - (2) 12 x - 3 x = 6a - b x = (1/9) (6a - b) x = (2/3) a - (1/9) b 3*(1) - 2*(2) 3y - 12 y = 3a - 2 b y = -(1/3) a + (2/9) b Verification 2x + y = 2((2/3) a - (1/9) b) + (-(1/3) a + (2/9) b) = (4/3) a - (2/9)b - (1/3)a + (2/9)b = a 3x + 6y = 3((2/3) a - (1/9) b) + 6(-(1/3) a + (2/9) b) = 2 a - (1/3) b - 2a + (4/3)b = b ==================================== b) (1) x1 + x2 + x3 = a (2) 2 x1 + 2 x3 = b (3) 3 x2 + 3 x3 = c 2*(1) - (2) 2x2 = 2a - b x2 = a - (1/2) b ---------------------------- 3*(1) - (3) 3 x1 = 3a - c x1 = a - (1/3) c ---------------------------- From(3) x2 + x3 = (1/3) c x3 = (1/3)c - x2 = -a + (1/2)b + (1/3) c ---------------------------- ==================================== Verification (a - (1/3) c) + (a - (1/2) b) + (-a + (1/2)b + (1/3) c) = a + a - a - (1/3) c - (1/2) b+ (1/2)b + (1/3) c = a 2(a - (1/3) c) + 2(-a + (1/2)b + (1/3) c) = 2a - 2a - (2/3) c + b + (2/3) c = b 3(a - (1/2) b) + 3(-a + (1/2)b + (1/3) c) = 3a - 3a - (3/2) b + (3/2) b + c = c regards
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L{F(t)} = int[e^(-st) cos(at - b)*dt] (0 < t < ∞) Integrate by parts: diff....... ...... ........int cos(at - b) ........ e^(-st) -asin(at - b) ..... -e^(-st)/s -a^2*cos(at - b) e^(-st)/s^2 int[e^(-st)*cos(at - b)*dt] = -e^(-st) cos(at - b)/s + ae^(-st) sin(at - b)/s^2 - int[a^2 e^(-st)cos(at - b)/s^2*dt] (1 + a^2 / s^2)int[e^(-st) cos(at - b)*dt] = -e^(-st) cos(at - b)/s + ae^(-st) sin(at - b)/s^2 Evaluate from 0 < t < ∞: [(s^2 + a^2) / s^2]int[e^(-st) cos(at - b)*dt] = 0 + cos(-b) / s - a*sin(-b) / s^2 int[e^(-st) cos(at - b)*dt] = [s^2 / (s^2 + a^2)][cos(b) / s + a*sin(b) / s^2] L{F(t)} = int[e^(-st) cos(at - b)*dt] = [s*cos(b) + a*sin(b)] / (s^2 + a^2)
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If we factor b^2 - 1, we get, by difference of two squares: b^2 - 1 = (b - 1)(b + 1) Can you see the common denominator in the equation? Every term's denominator divides (b - 1)(b + 1), so we multiply through by (b - 1)(b + 1): (b - 1)(b + 1) / (b + 1) + (b - 1)(b + 1) / (b - 1) = 2(b - 1)(b + 1) / (b^2 - 1) b - 1 + b + 1 = 2 2b = 2 b = 1 If we try back-substituting, we see that this produces an invalid result. Our chain of logic has been solid going forward, that is, in every step, the previous equation has implied that the next equation is correct. We have proved that if there exists a 'b' such that: 1 / (b + 1) + 1 / (b - 1) = 2 / (b^2 - 1) then it must be true that: b = 1 but this is a contradiction. Therefore there exists no b satisfying the above equation. The equation has no solution.
Well, lets see an addition of fractions go as follows A / B + C / D = ( AD + CB ) / BD so, for the left side of your equation 1 / ( b + 1 ) + 1 / ( b - 1 ) ( ( b - 1 ) + ( b + 1 ) ) / ( ( b + 1 ) ( b - 1 ) ) ( b - 1 + b + 1 ) / ( b^2 - 1 ) 2b / ( b^2 - 1 ) equating to the right side 2b / ( b^2 - 1 ) = 2 / ( b^2 - 1 ) as you can see, you can simplify the denominators as they are the same 2b = 2 b = 1 however, when you substitute 1 in the equation you get 1 / 2 + 1 / 0 = 2 / 0 which cannot be solved
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First, if z^2+az+b has two real roots, then a and b are real, and the above equation is satisfied. But if a and b are real and a^2-4b < 0, the equation does not have real roots. So either you misstated the problem, or the problem itself was stated poorly. Now let's assume z^2+az+b has exactly one real root. Then so does (I'm going to let * denote conjugate) z^2+a*z+b*--in fact, it has the same real root. What is it? Subtract the second equation from the first to get: (a-a*)z+(b-b*) Now if a is real, then we must have b is real for this linear polynomial to have a 0. But this isn't the situation we're in. So a must be nonreal, and we have the real root is (b*-b)/(a-a*). Plug this in to the polynomial and we get: (b*-b)^2/(a-a*)^2+a(b*-b)/(a-a*)+b = 0 Multiply through by a-a*, group things appropriately, and we see: (b*-b)^2+a(a-a*)(b*-b)+b(a-a*)^2 = 0 (b*-b)^2+(ab*-ab+ba-ba*)(a-a*) = 0 (b*-b)^2+(ab*-a*b)(a-a*) = 0 (b-b*)^2 = (a-a*)(a*b-ab*) This is the identity you have above. Conversely, if you have this identity, and assuming that a ≠ a*, work through the algebra above backward to find that (b*-b)/(a-a*) is a root of the polynomial. Note that this is real.
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A x B = ||A|| ||B|| sin(A,B) by the definition of cross product. Then ||A x B||^2 = ||A||^2 ||B||^2 sin^2(A,B) But sin^2 = 1 - cos^2 So ||A x B||^2 = ||A||^2 ||B||^2 ( 1 - cos^2(A,B) ) = ||A||^2 ||B||^2 - |||A||^2 ||B||^2 cos^2(A,B) = ||A||^2 ||B||^2 - ( A . B)^2 The last step uses the definition of the inner product A.B = ||A|| ||B|| cos(A,B) .
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B(a) = 2a/10x - 7....and if ' x ' is an operation symbol the B(a) =[a / 5 ] - 7
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X1=2 y1=0 x2=0 y2=5 where (x1,y1) and (x2,y2) are points. slope=(y2-y1)/(x2-x1)=5/-2=-5/2 y=mx+b , m being the slope is the equation y=(-5/2)x+b 0=(-5/2)(2)+b (choose x=2 y=0- pair 1) 0=-5+b b=5
To find the slope, which is m, do the first Y minus the second Y all divided by the first x music the second x. (0-5)/(2-0)= -5/2 the plug in with one of the coordinates: 0=-5/2(2) +b 0=-5 +b b=5
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(b+5)/(b-2) / (4b+20)/(b-6) people say to multiply by the inverse of the denominator what you are doing is multiplying by '1' in the form of (b-6)/(4b+20) / (b-6)/(4b+20) this turns the denominator into 1 so it is dismissed leaving you with (b+5)/(b-2) * (b-6)/(4b+20) (b+5)(b-6) / (b-2)(4b+20) 4b+20=4(b+5) so the (b+5)'s cancel out (b-6)/(b-2)*4 (b-6)/4(b-2)
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I'll show (a)=>(b)=>(c)=>(a). "(a)=>(b)": Let a belong to A. Then a belongs to B. So a doesn't belong to B^c. Since a was arbitrary, this shows that no elements of A belong to B^c. I.e. A ∩ B^c = ∅. "(b)=>(c)": Let x belong to X. We need to show that x belongs to A^c or to B. Well, if x doesn't belong to A^c, then x belongs to A. By (b), x doesn't belong to B^c, so x belongs to B. This finishes the proof. "(c)=>(a)" Suppose for contradiction that (a) is false. Then some a belongs to A and not to B. But then a belongs to neither A^c nor to B, contradiction (c), since a belongs to X = A^c U B. I hope this helps!
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(b-234)^2 = 49 Take the square root of both sides b - 234 = ±7 b = 234 ± 7 b = {227, 241} m05292013
Well first take the square root of each side. now its: (b-234)=7 or take away the parentheses. b-234=7 add 234 to both sides. b=241
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Proof by contradiction: Assume a/b+c/d = (ad+bc)/bd is an integer. Then bd | (ad+bc) (read: b (evenly) divides ad+bc) and more specifically, b | (ad+bc) and since b | bc, then b | ad and since GCD(a,b)=1 then b | d Likewise, d | (ad+bc) and since d | ad, then d | bc and since GCD(c,d)=1 then d | b. Since b | d and d | b then b = d. This is a contradiction of the problem's conditions, so a/b + c/d is not an integer.
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B^2/(b^2 - 1) * (5b^2 - 5b)/25 = b^2/(b^2 - 1^2) * 5(b^2 - b)/25 = b^2/(b + 1)(b - 1) * b(b - 1)/5 = b^2/(b + 1) * b/5 = b^3/5(b + 1)
You can times across the top and bottom for this one = b^2(5b^2 - 5b) / [25(b^2 - 1)] = (b^4 - b^3) / (5b^2 - 5)
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Apply the Mean Value Theorem to the interval [a, b]. The Mean Value Theorem says that if f is continuous and differentiable on [a, b], then there exists a number c in (a, b) such that f'(c)(b - a) = f(b) - f(a). In this case, f(x) = sin (x) and f'(x) = cos (x), so this means there exists a number c in (a, b) such that (cos c)(b - a) = sin (b) - sin(a). Taking absolute values gives |(cos c)||b - a| = |sin (b) - sin(a)|. But |cos c| <= 1 everywhere; so this means |b - a| >= |sin (b) - sin(a)|. To deduce that sin(x) <= x for x > 0, plug in a = 0 and b = x above: |b - a| >= |sin (b) - sin (a)| |x - 0| >= |sin (x) - sin (0)| |x| >= |sin x| x >= |sin x|. Since |sin x| >= sin x, then x >= sin x as well.
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Linear or variable separable dR/ ( aR + b) = dt integrating {ln(aR+b) }/ a = t +k ln(aR+b) = at +ak aR+b =e^(at +ak) aR +b =C e^(at) Solve for R
A
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A*b*a^(-1)=b^4 Now raise both sides to the 4th power (a*b*a^(-1))^4=(b^4)^4 now on the left side, write out the four terms multiplied by each other: a*b*a^(-1)*a*b*a^(-1)*a*b*a^(-1)*a*b*a... = b^16 the terms a^(-1)*a cancel and you get a*b^4*a^(-1) = b^16 But if you take the original condition a*b*a^(-1)=b^4 and conjugate both sides by a you get: a^2*b*a^(-2) = a*b^4*a^(-1) So putting the two things together: b^16 = a*b^4*a^(-1) = a^2*b*a^(-2)
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F(x)avg = [1/(b-a)] ∫f(x)dx, with the integral evaluated from a to b. First, ∫ (8 + 6x - 3x^2) dx = 8x + 3x^2 - x^3, So, f(x)avg = [(1/(b-0)] [(8b + 3b^2 - b^3)-(8*0 + 3*0^2 - 0^3)] = 9; 8 + 3b - b^2 = 9; b^2 - 3b +1 = 0; so b = (3±√5)/2, or b ≈ 0.382 and 2.618.
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Given that A U B = A ∩ B, we need to prove A ∩ B' = empty set. Assume temporarily that A ∩ B' is not the empty set. Then there exists some element x in A ∩ B'. So x is in A and x is not in B. Since x is in A, clearly x is in A or x is in B, and so x is in A U B. Then since A U B = A ∩ B and x is in A U B, x is in A ∩ B. Since x is in A ∩ B, x is in A and x is in B. This statement that x is in B contradicts the earlier statement that x is not in B. So our temporary assumption that A ∩ B' is not the empty set is false. We conclude that A ∩ B' = empty set. Lord bless you today!
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You do not say what A intersection B equals. A intersection B (or, in set notation, A ⋂ B) is a binary operation on two sets, not a comparison. In other words, there must be a result giving some subset C such that A ⋂ B = C, even if C is the empty set ∅. As it happens I think you meant to say: "If A is a subset of B, then A - B = the empty set" or A ⊆ B => ∅. If so then yes....... if A = B. However, if A is a proper subset of B (ie A ⊂ B), then A - B is meaningless, since |A| < |B| and you cannot have sets of negative cardinality.
If set A is contained in B... supplement of B is composed of no B and so no A for this reason intersection is empty additionally in opposite, if A intersection supplement B is empty each and every a in A ought to belong to B
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For any events A and B with P(B) > 0, show that P(A|B) + P(A'|B) = 1. I have gotten as far as stating that P(A|B) + P(A'|B) = [P(A ∩ B) + P(A' ∩ B)]/P(B). But I have gotten a little stuck with rest of the solution. I guess I shouldn't really say "stuck" because I did find a solution by changing P(A ∩ B) to P(A)*P(B) and obtained the correct answer this way. However, I was under the impression that this property only applied to independent events, and the question asks for the solution to "ANY" events. Any advice would be appreciated.
Well, P(B) > 0 and by definition : P(A | B) = P(A ∩ B)/P(B) the same way : P(A ' | B) = P(A ' ∩ B)/P(B) therefore : P(A|B) + P(A'|B) = P(A ∩ B)/P(B) + P(A ' ∩ B)/P(B) = [ P(A ∩ B) + P(A ' ∩ B)] /P(B) and as (A ∩ B) and (A ' ∩ B) are disjoint : P(A|B) + P(A'|B) = [ P[ (A ∩ B) U (A ' ∩ B)] ] /P(B) = [ P[ (A U A ') ∩ B ] ] /P(B) using the property of distribuvity of ∩ wrt U : P(A|B) + P(A'|B) = [ P[ Ω ∩ B ] ] /P(B) where Ω is the global superset and as : Ω ∩ B = B P(A|B) + P(A'|B) = P[B] /P(B) = 1 finally : P(A|B) + P(A'|B) = 1 in other words : P( . | B) is a law of probability hope it' ll help !!
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(a + b + c)^2 (a + b + c)(a + b + c) (distribute terms in first group to the second group) a(a + b + c) + b(a + b + c) + c(a + b + c) (distribute) a^2 + ab + ac + ab + b^2 + bc + ac + bc + c^2 (combine like terms) a^2 + b^2 + c^2 + 2ab + 2ac + 2bc
(a+b+c)^2 is equal to (a+b+c)(a+b+c) then multiply out a^2 + ab + ac + ba + b^2 + bc + ca + cb + c^2 then combine like terms a^2 + 2ab + b^2 + 2bc + c^2 + 2ac Thats the same as your answer, even though its ordered different
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(1) Prove a ≤ b Assume a > b, then a - b > 0 Let c = a - b > 0 Then we can choose ε such that 0 < ε < c ε < a - b b + ε < a This is a contradiction. Therefore our assumption (that a > b) must be false ∴ a ≤ b (2) a = b < b + ε a < b + ε So when a = b, then for every ε > 0, we have a < b + ε So we cannot state with certainty that a < b.
(1) By way of contradiction, suppose a > b. Let ε = (a - b)/2. Then ε > 0 and a - b = 2ε > ε, contrary to the hypothesis. (2) 1 < 1 + ε for all ε > 0, but 1 ≮ 1.
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Nope. Let A be the emptyset, and let B and C be any two different sets. Then AxB and AxC are equal (since they're both the emptyset), but B and C aren't equal. However, the claim does hold if we require that A can't be empty. To see this, let b be an arbitrary element of B, and pick an element a of A. Then (a,b) belongs to AxB. Hence (a,b) belongs to AxC. So b belongs to C. So B is contained in C. By symmetry, C is contained in B, and we have B = C as required.
(a) A = {a million, 2} B = {a million, 2, 3} A - B = ? yet A ? B. (b) A = {a million, 2} B = {3, 4} A ? B = ? yet A ? B. (c) A = {a million. 2} B = {2, 3} C = {2, 4} A ? C = B ? C yet A ? B. (d) A = {a million. 2} B = {2, 3} C = {a million, 2, 3} A ? C = B ? C yet A ? B.
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If (a+b) | (a^n+b^n) with a and b coprime, then n must be odd. Indeed (a^n+b^n) = (1 + (-1)^n)a^n mod (a+b). Now if n is odd non prime, say n = pq, then a^n = (a^p)^q and b^n = (b^p)^q. Hence a^p + b^p | (a^n+b^n) and a + b | a^p + b^p so that (a^n+b^n)/(a+b) is not a prime By contraposition, we get your statement
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(b^1/2-a^1/2)/b-a = (√b-√a)/(b-a) =(√b-√a)/ {(√b )²-(√a) ²} { using the formula x²-y²=(x+y)(x-Y) } =(√b-√a)/(√b+√a)(√b-√a) =1/(√b+√a) ; this is your answer but if you want to write it without division sign then write, (√b+√a) ^-1 Good luck...!!!
I will enable you to guess what the movies are. even though it is going to no longer be stressful. mom, son, and a robotic from the destiny attempt to outrun yet another robotic from the destiny. woman, toddler, and marine combatants attempt to get off an alien planet. woman seeks help from a cannibal with the intention to seize a serial killer. guy enables a pregnant woman for the duration of u . s . a . in a destiny the place women folk are infertile. A boy who see ghosts consults a toddler therapist. BQ: They have been indexed so as from #a million to #5 of favorites. BQ2: Alien franchise, even although T2 is my well-liked action picture ever. BQ3: Brokeback Mountain. It became in simple terms shattering. I in no way mandatory to apply tissue whilst staring at a action picture, that became basically a great love action picture. in no way cried greater in any action picture. Rewatched eternal Sunshine of the Spotless concepts.
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Do you mean (b + 2)/(b² - 9) - 9 - (b + 5)/(b² - 6a + 9)? By factoring, we have: ==> (b + 2)/((b - 3)(b + 3)) - 9 - (b + 5)/(b - 3)² Note that the LCD is (b - 3)²(b + 3) ==> (b + 2)/((b - 3)(b + 3)) * (b - 3)/(b - 3) - 9(b - 3)²(b + 3)/((b - 3)²(b + 3)) - (b + 5)/(b - 3)² * (b + 3)/(b + 3) ==> (b + 2)(b - 3)/((b - 3)²(b + 3)) - 9(b - 3)²(b + 3)/((b - 3)²(b + 3)) - (b + 5)(b + 3)/((b - 3)²(b + 3)) Finally, combine the expressions altogether and simplify (This is going to be the long one! Watch carefully!)! ==> (b² - b - 6 - 9(b² - 6b + 9)(b + 3) - (b² + 8b + 15))/((b - 3)²(b + 3)) ==> (b² - b - 6 - (9b² + 54b - 81)(b + 3) - b² - 8b - 15)/((b - 3)²(b + 3)) ==> (-9b - 21 - (8b³ + 27b² + 54b² + 162b - 81b - 243))/((b - 3)²(b + 3)) ==> (-9b - 21 - (8b³ + 81b² + 81b - 243))/((b - 3)²(b + 3)) ==> (-9b - 21 - 8b³ - 81b² + 81b - 243)/((b - 3)²(b + 3)) ==> (-8b³ - 81b² + 72b - 264)/((b - 3)²(b + 3)) I hope this helps! (Damn! I took very long with the computation! I better do a little bit quicker than before! :]) ~Tsugara
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Cosine rule states that, given a triangle with sides a, b, c, with A being the angle opposite the side length a, we have: a^2 = b^2 + c^2 - 2bc cos(A) So, if we have vectors u and v, which are both rooted at the origin, then the vector u - v (and v - u) must form a triangle with vectors u and v. The angle between them is alpha (which will correspond to A), and the lengths of u, v and u - v are the sides (so, b = |u|, c = |v|, a = |u - v|). We therefore have: |u - v|^2 = |u|^2 + |v|^2 - 2|u|.|v|,cos(A) If u = (x, y, z) and v = (a, b, c), we have: |(x - a, y - b, z - c)|^2 = |(a, b, c)|^2 + |(x, y, z)|^2 - 2|u|.|v|.cos(A) (x - a)^2 + (y - b)^2 + (z - c)^2 = a^2 + b^2 + c^2 + x^2 + y^2 + z^2 - 2|u|.|v|.cos(A) x^2 - 2ax + a^2 + y^2 - 2by + b^2 + z^2 - 2cz + c^2 = a^2 + b^2 + c^2 + x^2 + y^2 + z^2 - 2|u|.|v|.cos(A) -2ax - 2by - 2cz = -2|u|.|v|.cos(A) ax + by + cz = |u|.|v|.cos(A) u.v = |u|.|v|.cos(A) Hope that helps!
Let A, B, A-B be director vectors of a triangle, and alpha the angle between A and B |A-B|^2=|A|^2+|B|^2-2|A||B|cos(alpha) 2|A||B|cos(alpha)=|A|^2+|B|^2-|A-B|^2 =x^2+y^2+z^2+a^2+b^2+c^2-(x-a)^2-(y-b)^2... =2(ax+by+cz) Knowing that A.B=ax+by+cz 2|A||B|cos(alpha) =2A.B
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((2a-b)/(a^2+ab)) - ((2b-a)/(b^2+ab)) + ((a-b)/(ab)) = (2a-b)/a(a+b) + (a-2b)/b(b+a) + (a-b)/ab Make the denominators all the same. The common denominator is: ab(a+b) = (b(2a-b))/(ab(a+b)) + (a(a-2b))/(ab(a+b)) + ((a-b)(a+b)) / (ab(a+b)) = (2ab - b^2 + a^2 - 2ab + a^2 - b^2) / (ab(a+b)) = (-b^2 + a^2 + a^2 - b^2) / (ab(a+b)) = (-2b^2 + 2a^2) / (ab(a+b)) = -2(b^2 - a^2) / (ab(a+b)) = (-2(b-a)(b+a)) / (ab(a+b)) = (-2(b-a)/(ab)
Ok :) i'm answering ur question .. so u have -- 2a-b / (a^2+ ab) - (2b - a ) / (b^2 + ab) + (a - b) /ab = (2a-b)(b^2 + ab) - (2b -a) ( a^2 + ab) / (a^ + ab)(b^2 +ab) + a-b /ab on multiplying and simplifying u gets ..merely multiply and cancel out the equivalent and opposite words! = a^b + a^3 - b^3 - ab^/ (a^+ab)(b^2 + ab) + a-b /ab = a^2 ( a+b) - b^2 ( a +b ) / ( a^ +ab )(b^2 + ab) + a-b/ab = (a^2- b^2 ) (a+b) / a(a+b) b (a+b) + a-b/ab = (a-b)(a+b).(a+b) / ab(a+b)(a+b) + a-b/ab as a^2 - b^2 = (a+b)(a-b) cancel out the priority-free words .. = a-b / ab + a-b/ab = 2(a-b)/ab ANSSSSSS yay !!!!!! u have been on condition that dude!
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[This is proved with the assumption that a, b & c all are real and greater than zero]. Solution: i) Dividing and multiplying by abc, left side is: (abc)*{(a/c) + (a/b) + (b/c + b/a) + (c/a + c/b)} = (abc)*{(a/c + c/a) + (a/b + b/a) + (b/c + c/b)} ---------- (1) ii) (a/c + c/a) is the sum of a number and its reciprocal. A sum of a positive and its reciprocal is ≥ 2 Thus the sum (a/c + c/a) ≥ 2 Similarly (a/b + b/a) ≥ 2 and (b/c + c/b) ≥ 2 Adding all the above, {(a/c + c/a) + (a/b + b/a) + (b/c + c/b)} ≥ 6 ---------- (2) iii) Thus from (1) & (2), Left side is ≥ 6abc ==> a²(b + c) + b²(a + c) + c²(a + b) ≥ 6abc [Proved] If you are satisfied with the above solution kindly acknowledge, need not be BA - just with a simple word of Thanks. More than 90% of the persons who ask question are not even obliged to acknowledge with a simple thanks, which I believe keeps many from providing solutions. REQUEST: Those who wish to rate this solution with TD, you are welcome; but only one request, kindly give the reasons for the same in the comments column, which will make me to correct mistake and learn more. Thanks for your cooperation.
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If sin(B)=x and -pi/2<B<0, Find exact values, in terms of x, for each of the following: cos(b), tan(-B)? ]a.) cos(B) = –√(1 – x²) b.) tan (-B) = – tan (B) = – sinB/cos B = – x/(–√(1 – x²)) = x/√(1 – x²) c.) sin (pi-B) = sin B = x d.) sec (B-pi) = 1/cos (B – pi) = 1/(sin A cos pi + cos A sin pi) ......................................... = 1/(sin A (–1) + cos A (0)) ......................................... = 1/(–sin A) ......................................... = –1/sin A ......................................... = –1/x
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P(B|A) = P(A & B) / P(A), so P(A & B) = P(B|A).P(A) P(A|B) = P(A & B) / P(B) = P(B|A).P(A) / P(B) -----------------------------
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L1=cos2A + cos2B + cos 2C =2cos(A+B).cos(A-B)+2cos^2(C) -1 =-2cos(C).cos(A-B)+2cos^2(C) -1 =cos(C){-2cos(A-B)-2cos(A+B)}-1 =cos(C){-2cos(A)cos(B)-2sin(A)sin(B)-2... =cos(C){-4cos(A)cos(B)}-1 =-4cos(A)cos(B)cos(C)-1 =-1-4cos(A)cos(B)cos(C) that's it am a genius right???
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..A..B...C ..0...0...0 ..0...1...0.....C = A ..1...0...1 ..1...1...1 ..A..B...C ..0...0...1 ..0...1...0.....C = B' ..1...0...1 ..1...1...0 ..A..B...C ..0...0...1 ..0...1...0.....C = (A + B' ) ..1...0...1 ..1...1...1 ..A..B...C ..0...0...0 ..0...1...1.....C = (A + B' )' ..1...0...0 ..1...1...0 ..A..B...C ..0...0...1 ..0...1...1.....C = A' ..1...0...0 ..1...1...0 ..A..B...C ..0...0...0 ..0...1...1.....C = A'B ..1...0...0 ..1...1...0
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√(a + b) = 1/√(a - b) √(a + b)√(a - b) = 1 ...multiply both sides by √(a - b) √((a + b)(a - b)) = 1 (a + b)(a - b) = 1 ...square both sides a² - ab + ab - b² = 1 a² - b² = 1 E is the correct answer. Hope this helps you!
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P(B|A) = .77 = P(Intersection A and B)/ P(A) = P(Intersection A and B)/ 0.40 P(Intersection A and B) = 0.40*0.77 = 0.308...............(i) P(A|B) =0 .36 = P(Intersection A and B)/ P(B) = 0.308/ P(B) P(B) = 0.308/0.36 =77/90= 0.85555555555556................(ii) P(AUB) = P(A)+P(B) - P(Intersection of A and B) = 0.40 +0.85555555555556 - 0.308 = 0.9476............................Ans
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B(2b-5)(b-4)-1(2b-5)(b-4) =(2b^2-5b)(b-4)-2b(b-4)+5(b-4) =2b^2(b-4)-5b(b-4)-2b^2+8b+5b-20 =2b^3-8b^2-5b^2+20b-2b^2+13b-20 =2b^3-15b^2+33b-20
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Log₁₁(2)+ 2 log₁₁ (a-b) = log₁₁(a) + log₁₁(b) log₁₁(2) + log₁₁(a - b)^2 - log₁₁(a) - log₁₁(b) = 0 ..........2(a - b)^2 log₁₁[------------------] = 11^0 .............ab 2(a - b)(a - b) -------------------- = 1 ....ab ....2(a^2 - 2ab + b^2) ----------------------------- = 1 ..........ab 2a^2 - 4ab + 2b^2 = ab 2a^2 - 4ab - ab + 2b^2 = 0 2a^2 - 5ab + 2b^2 = 0 (2a-b)(a-2b)=0 2a - b = 0, a - 2b = 0 2a = b, a = 2b a/b = 1/2 or a/b = 2 a/b = 1/2, 2 answer//
Log base(11) 2+ 2 log base(11) (a-b) = log base(11) a + log base(11) b lets take log base(11) x=log x thus log2+2log(a-b)=loga+logb log(2*(a-b)^2)=log(ab) removing log from both side, (2*(a-b)^2)=ab 2a^2+2b^2=5ab 2(a/b)^2-5(a/b)+2=0 (a/b-2)(a/b-1/2)=0 thus a/b=2,1/2 Hope this helps :)
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You can indeed get a B.A. or a B.S. in finance. It depends on the college. And it depends on what you intend to do with your degree. If you want a more concentrated curriculum and more courses in finance, for an analytical position like a financial analyst, take the B. S. If you want a broader education and more liberal arts courses, for a job in financial administration, for an organization, like a hospital or non-profit where you are dealing with budgets and fiscal policy, then take the B.A.
How can you get a B.A. in Finance what do they teach you, creative financing? There is no BA for Finance since it is not at "Art" you can only get a BS in Finance.
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A. A ∪ C = B ∪ C In this case, you can't. Consider the case A = {1,2} B = {1,3} C = {2,3} A and B in this case are different, but their unions with C are the same. b. A ∩ C = B ∩ C Again, in this case, you can't. Consider: A = {1,2} B = {1,3} C = {1} A and B are different, but they have the same intersection with C. c. A ∪ B = B ∪ C and A ∩ C = B ∩ C Again, you can't. Consider: A = {1} B = {1,2} C = {1} But in this case, I suspect a typo: I think this should have been A ∪ C = B ∪ C and A ∩ C = B ∩ C and from these, we can indeed prove that A = B. Let x be an arbitrarily chosen element of A. If x is in C, then A ∩ C contains x, therefore B ∩ C contains x (because the two intersections are equal), and therefore x is in B. x is also in A ∪ C, and therefore in B ∪ C. If it is not in C, it must therefore be in B. Because x is an arbitrarily chosen element of A, this proves that all elements of A are in B. Similarly, we can prove that any arbitrarily chosen element of B is in A, therefore all elements of B are in A. Therefore A = B (if my assumed typo and fix are correct).
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For events A and B we have P(A) = 0.3 P(B) = 0.8 P(A U B) = 0.9 (a) Find P(A|B), P(A' ∩ B) and P(B' U A') (b) Are A and B independent? Why? I would like to know how to do it, so please don't just give me the final answer, show your steps if you could also tell me what the three symbols (| ∩ U) mean in this context that would be wonderful as well. Thanks a lot.
P(A U B) = P(A)+P(B) -P(A'∩ B) .9 = .3+.8 - P(A'∩ B) P(A'∩ B) = .2 a) P(A/B)= P(A'∩ B)/P(B) = .2/ .8 = 2/8 = 1/4 P((A' ∩ B) +P(A'∩ B)=P(B) P((A' ∩ B) +.2 = .8 P((A' ∩ B) =.6 P(B' U A') = P( A' U B') = P( [(A'∩ B]' ) --- De Morgan's law = 1- P((A ∩ B) = 1-.2 = .8 b) If A and B are independent, P(A/B) = P(A) P(A/B)=1/4 --- from previous problem P(A) = .3 Since they are not equal, A and B are not independent.
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If A and B are acute angles such that : csc (A) = 13/5 and cot (B) = 24/7 , Find sin A = 1/csc A = 5/13 cos A = √(1 - sin² A) = √(1 - (5/13)²) = 12/13 tan A = 5/12 tan B = 7/24 sin B = tan B/√(1 + tan² B) = (7/24)/√(1 + (7/24)²) = 7/25 cos B = 1/√(1 + tan² B) = 1/√(1 + (7/24)²) = 24/25 (a) sin (A + B) = sin A cos B + cos A sin B ..................... = (5/13)(24/25) + (12/13)(7/25) ..................... = 120/325 + 84/325 ..................... = 204/325 (b) tan (A + B) = (tan A + tan B)/√(1 - tan A tan B) ..................... = ((5/12) + 7/24)/√(1 - (5/12)(7/24)) ..................... = ((10/24) + 7/24)/√(1 - 35/288) ..................... = (17/24)/√(253/288) ..................... = (17/24)√(288/253) ..................... = (17/24)(12√(2/253) ..................... = (17/2)√(2/253)
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It you are talking about vectors, then draw the triangle OAB where O is the origin of the coordinate system. The lengthof OA and OB anre |A| and |B| and the length of AB is |B - A| (which is the same as | A - B |). Two sides of a triangle must be >= the third side. The equality holds when the triangle collapses into a straight line. You can prove it for numbers a different way. First notice that a^2 always equals |a|^2. Then |a-b|^2 = (a-b)^2 = a^2 - 2ab + b^2 = |a|^2 - 2ab + |b|^2 >= |a|^2 - 2|a||b| > |b|^2 To see tor the last step: If the product ab >= 0, then ab = |a||b| If ab < 0, then |a||b| >= 0 > ab or -ab > -|a||b| = ( |a| - |b| ) ^2 So we have shown that ( |a-b| )^2 >= ( |a| - |b| )^2 To take the sqaure root of this inequality, we need to be careful about the signs. |a-b| is always >= 0. If |a| - |b| is also >= 0, we can the square root and get |a-b| >= |a| - |b| If |a| - |b| < 0, then we have |a-b| >= 0 > |a| - |b| So the result is true for any values of a and b.
Triangle inequalities. The magnitude of the sum of two vectors must be greater than the difference in magnitudes and less than the sum of the magnitudes. Probably the easiest proof would be to use the law of cosines. |A+B| = sqrt (A^2 + B^2 + 2AB cos(theta) ), where theta is the angle between the vectors. If the vectors are in line, cos(theta)=1 and |A+B|=|A|+|B| If the vectors are opposite, cos(theta) = -1, and |A+B| = | |A|-|B| | Cos(theta) varies continuously between -1 and 1, so the magnitude of the sum will always fall between the sum and the difference of the magnitudes. I don't believe there is anyplace the inequality does not hold. Note that the way your inequality was written, you can just replace B with -B and get the usual form of the triangle inequality.
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Y = a*[x*sin(b*x) + z*sin(b*z)] y/a = x*sin(b*x) + z*sin(b*z) now remember the double angle identity sin(2x)=2sin(x)cos(x) <==you have to either memorize this or it is given. Looking at the identity, instead of sin(2x) think of it as sin(bx), meaning we substitute b for 2, this means according to the identity; sin(bx)=b*sin(x)cos(x) and sin(bz)=b*sin(z)cos(z) Now using the two identities we derived above, substitute the identities into the equation; y/a = x*sin(b*x) + z*sin(b*z) y/a=x*[b*sin(x)cos(x)] + z*[b*sin(z)*cos(z)] y/a=x*b*sin(x)cos(x) + z*b*sin(z)*cos(z) Now factor out b because it is a common factor y/a=b*[x*sin(x)cos(x) + z*sin(z)cos(z)] Now divide y/a by [x*sin(x)cos(x) + z*sin(z)cos(z)] and isolate b (y/a) / (x*sin(x)cos(x) + z*sin(z)cos(z)) = b Now simplify y/[a(x*sin(x)cos(x) + z*sin(z)cos(z))] = b
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|a| ≥ |b+c| means a^2 - (b+c)^2 ≥ 0 So (a-b-c)(a+b+c) ≥0 Using other two relations we get, (b-a-c)(a+b+c) ≥0 and ((c-a-b)(a+b+c) ≥0. First let us consider inequality >. Its natural that if the products are positive, each pair must be of same sign. But a+b+c is a common factor in each. So a-b-c, b-c-a, c-a-b are of same sign (as of a+b+c), which cannot happen. One of the 3 numbers must be of opposit sign. So >0 is impossible. Means that the products must be zero and a+b+c is the common factor so this itself must be zero.
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A(x - 1) + B(x + 1) = 2x - 3 First distribute Ax - A + Bx + B = 2x - 3 Rearrange to get the x's and the numbers together Ax + Bx - A + B = 2x - 3 Factor (A + B)x - (A - B) = 2x - 3 As you can see, the left and right have x parts and number parts, so A + B = 2 (since the x's on each side need to be the same) A - B = 3 Now, we can solve these two equations Solve the first equation for A A = 2 - B Plug this into the second equation A - B = 3 (2 - B) - B = 3 2 - B - B = 3 2 - 2B = 3 Subtract 2 from both sides -2B = 1 Divide both sides by -2 B = -1/2 Since A = 2 - B A = 2 - (-1/2) = 2 + 1/2 = 5/2 Answer: A = 5/2, B = -1/2 Hope this helps you! Please email anytime for further questions.
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Assuming B and b are used interchangeably (they shouldn't be), we get this. cos(B)[sec(B) - cos(B)] = cos(B)sec(B) - cos²(B) = 1 - cos²(B) = sin²(B), B ≠ (2k + 1)π/2 for any integer k Choice D is probably the one that is expected by whoever wrote the question, but that would allow B to take any real value, so it is not actually equivalent to the given expression.
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For the first equation, take the loga of both sides of the equation. Since loga(a^x) = x is the definition of a logarithm, you get this: loga (a^loga b) = loga(b) loga b = loga b For the second one, do exactly the same thing, except take the log base e (ln) of both sides: ln (e^ln b) = ln(b) ln b = ln b Alternatively, you can immediately see that both statements are true simply by remembering that the exponential function and the logarithm function are inverses of each other. In both of these examples you take a number, take the logarithm of it, and then take the exponential of the result. This will always give you back the same number you started with. It's exactly like taking a number, doubling it, and then taking half of the result. You will end up right where you started.
(a^loga) b = b a^loga is equal to 1
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P = (b²+c²)/(b+c) + (c²+a²)/(c+a) + (a²+b²)/(a+b) = b+c - 2bc/(b+c) + c+a - 2ac/(c+a) + a+b - 2ab/(a+b) = 2(a+b+c) - (2bc/(b+c) + 2ac/(a + c) + 2ab/(a + b)) 2bc/(b + c) is the harmonic mean of b and c The harmonic mean ≤ the arithmetic mean The harmonic mean > 0 (a, b and c are all positive numbers) 2bc/(b+c) + 2ac/(a+c) + 2ab/(a+b) ≤ (b+c)/2 + (a+c)/2 + (a+b)/2 (The sum of three harmonic means must be less than or equal to the sum of the corresponding arithmetic means) 2bc/(b+c) + 2ac/(a+c) + 2ab/(a+b) ≤ 2(a+b+c)/2 2bc/(b+c) + 2ac/(a+c) + 2ab/(a+b) ≤ (a+b+c) 2bc/(b+c) + 2ac/(a+c) + 2ab/(a+b) > 0 (Each harmonic mean is greater than 0, so the sum must be greater than 0) P < 2(a+b+c) P ≥ 2(a+b+c) - (a+b+c) P ≥ (a+b+c) (a+b+c) ≤ P < 2(a+b+c) D
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You're thinking of drop B as being C standard, with the lowest string dropped by one half-step. Usually drop B refers to C# standard, with the lowest string dropped two half-steps. C# standard is C# - F# - B - E - G# - C# So drop B is B - F# - B - E - G# - C# Alternatively, you could think of it as drop D, with everything lowered three half-steps - it works out the same.
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