how do you answer this equation?  a=(1 3 2)/( 2 1 -3)find a^t?
26943
It appears that the question period has expired.  If you have received an answer that meets your needs, please choose one of those as a 'best answer.'  If you haven't received a good answer for your question, you may want to consider the following,\n\n1) Re-post your question.  Newer questions get more activity on Yahoo! Answers than old ones.\n2) If you do re-post your question, consider why it wasn't answered the first time.  Could it be more specific?  Could it be worded better? Were there grammatical or spelling errors? Was it in the best category?\n\nIf it doesn't seem likely that re-posting your question will help you, then here's a listing of my favorite 'answer sites'.  Maybe one of them will help you.\n\nAnswers.com http://www.answers.com/\nBartleby http://www.bartleby.com/\nYahoo Reference http://education.yahoo.com/reference/\nHowStuffWorks http://www.howstuffworks.com/\nWikipedia http://en.wikipedia.org/wiki/Main_Page\n\nSince I really haven't answered your question, it is not necessary to give me any points.  Regards.
1046507
[2x/(x+3)(x-3)]*[(x-3)/4(x)]\ncancelling 2 and x-3 we will get\n=x/(x+3)2(x)\nsimplify we will get\n=x/(2x^2+6x)
589702
I think this means:\n\n(t^2 - 2t - 15)/(t-5) * (t+5)/(t+3)\n\nIf that's right, then\n\n(t^2 - 2t - 15)/(t-5) * (t+5)/(t+3) = (t-5)(t+3)/(t-5) * (t+5)/(t+3)\n= t+5
529729
(x2)-3 (x-2) over (x2) -4?\nIs this what you mean?\n(x^2)^(-3)*x^(-2)/(x^2)^(-4)\n= x^(-6)*x^(-2)/x^(-8)\n= x^(-6-2-(-8))\n= x^(-8+8)\n= x^0 \n= 1
929562
∫ax(2 -x)dx\n= ∫(2ax - ax²)dx\n= 2a * ½x² - a * ⅓x³ + c\n= ax² - ⅓ax³ + c
699436
(2^x-2^-x) / 3 (the whole thing is divided by 3)\n\nThat is not an equation to be solved.   So I don't know where you are getting an equals sign (=).??\n\nYou may be able to simplify it, but you cannot say x= something unless you were provided with an EQUATION.\n\nI will do some work on it.\n\nAfter further review, try to input your value for x into the statement\n(2^x-2^-x) / 3 and what you get is another question.   Your value then EQUALS what?   Does this answer your question?    All you will get is a value.
814891
a^2 - 2ab + b^2 = (a - b) ^2\n\nthen...\n\n(a - b)^2 / (a + b) = (a -b) (a - b) / (a + b)
300751
Use the chain rule and the power rule.\n\nIf y = x^a, then y' = a * x^(a-1).\n\nSo, y = (2x+3)^(.5)\ny' = .5 * 2 * (2x+3)^(-.5) = (2x+3)^(-.5)\ny'' = -.5 * 2 * (2x+3)^-1.5 = -(2x+3)^(-1.5)
268855
we have a^(-b)=1/(a^b), with this rule, the expression will be :\n[m^3/(m.n^5)]^(-3)\n=(m^2/n^5)^(-3)\n=1/(m^2/n^5)^3\n=n^15/m^6\nhope helped ^^
301328
Homework.\n\nLook at B=A. You should find something interesting about tr(A^tA).
1031456
g(1-t) = 3(1-t)^2 - 2(1-t) + 1\nFirst expand (1-t)^2\n(1-t)^2 = (1-t)(1-t) = 1 - t - t + t^2\n= 1 - 2t + t^2 \nnow multiply by 3\n3(1-t)^2 = 3(1 - 2t + t^2 )\n= 3 - 6t + 3t^2\n\nNow move on to the next term:\n-2(1-t) = -2 + 2t\n\nPut everything together and combine like terms.\ng(1-t) = 3(1-t)^2 - 2(1-t) + 1\n= 3 - 6t + 3t^2 -2 + 2t + 1\n= 3t^2 - 4t +2
892759
g(t)=1/3(t+2)^2-12 (1<t<4) \nDomain is 1 < t < 4\nRange is -9 < g(t) < 0\n\nSo for inverse\nt = 1/3((g^(-1)(t) + 2)² -12\nSo 3t = ((g^(-1)(t) + 2)² - 36\n((g^(-1)(t) + 2)² = 3t + 36\n(g^(-1)(t) + 2) = √(3t + 36) (NOTE: NOT -√(3t + 36))\n\nSo g^(-1)(t) = √(3t + 36) - 2\nDomain: -9 < t < 0\nRange: 1 < g^(-1)(t) < 4
381253
3^(2x) + 3^(x) - 2 = 0\n(3^(x))^2 + 3^(x) - 2 = 0\ny = 3^x\ny^2 + y - 2 = 0\neither y = -2 or y = 1\n\ny = 3^x, then y > 0, then y = 1\n3^x = 1\n\nx = 0
605867
(x^2)/2 - 3(x^(3/2))/(3/2) + c = (x^2)/2 - 2x^(3/2) + c
78524
(1-a)+3(a-2)-(2a-3)=1
368125
if this is the problem:\n((x+3)/2)*((x^2-x)/(x+3))=21\nthen you solve by\ncancelling out (x+3)\n(1/2)*(x^2-x)=21\nmultiplying by 2\nx^2-x=42\nmaking it quadratic\nx^2-x-42=0\nand using x=(-b(+-)srt(b^2-4ac))/2a\nx=(1(+-)srt(1-4*1*(-42)))/2\nx=(1(+-)srt(169))/2\nx=(1+13)/2  or x=(1-13)/2\nx=7   or x= -6
1054627
assuming this is (3a+1)/(a^2-1) - 1/(a+1)\n\nfactorise the (a^2-1) to give (a+1)(a-1)\n\nmultiply the second fraction by (a-1)\n\n-->  (3a+1)/[(a+1)(a-1)] - 1(a-1)/[(a+1)(a-1)]\n\nfactorising this gives [(3a+1)-(a-1)]/[(a+1)(a-1)]\n\nwhich equals (2a+2)/[(a+1)(a-1)]\n\n= [2(a+1)]/[(a+1)(a-1)] \n\nthe (a+1)'s cancel out...\n\ntherefore = 2/(a-1)