1-3x+x^2 on top and 0 on the bottom of the function\n\n                   1\n               _________                 dt\n                  T-1               \nPlease show the steps so I can understand how to do these. Thanks for your help.
203732
Okay, this is how I read this problem:\n\nThe integral of 1/(t-1) dt on the interval from 0 to (1-3x+x^2).\n\nFirst look at what you're integrating 1/(t-1) dt.  This can be rewritten (t-1)^-1 dt.  If you substitute u for t-1 and du for dt (the derivative of t-1 = 1) then you have the integral of u^-1 du.  This is ln |u| (ln is the natural log).  Sub t-1 for u and you get ln |t-1|.\n\nNow you use your interval and you get:\n\nln |(1-3x+x^2) - 1| - ln |0|\n\nSimplify and you get:\n\nln |x^2 - 3x| - ln |0|
353839
x/(x+3) - 2(x-1)/(x+3)(x+2)\n\nx(x+2) - 2 (x-1)\n--------------------\n   (x+3) (x+2)\n\n\nx^2 +2\n----------\n(x+3)(x+2)
447929
divide through\n\ny = [x^(3/2)]/3 - 2x^(1/2) -x^(-1/2)\n\nthen take the derivative\n\ndy/dx = [x^(1/2)]/2 - x^(-1/2) + [x^(-3/2)]/2\n\nthen put x^(3/2) in the denominator\n\ndy/dx = [1/x^(3/2)]*{ [x^2]/2 - x + 1/2 }\n\nwe take out the one half to find\n\ndy/dx = [1/2x^(3/2)] * [ x^2 - 2x + 1 ]\n\nwhich is the same as writing \n\ndy/dx = [1/2x^(3/2)] * (x - 1)^2\n\nor a = -1 and b = 2
505392
Factor out 3.\n3(x^2+2x+1)=0\n3(x+1)^2=0\nThe only answer then is x=-1 to make the expression equal 0.
485338
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
512937
∫ax(2 -x)dx\n= ∫(2ax - ax²)dx\n= 2a * ½x² - a * ⅓x³ + c\n= ax² - ⅓ax³ + c
401666
1/x + y\n______\nx + y\n\nMultiply the top and bottom by x\n= \n1 + xy\n______\nx^2 + xy\n\n\n(x-1)/x divided by ((x^2 -1)/(x+4)\n= (x-1)/x times (x+4)/(x^2 -1)\nFactor x^2 -1 = (x-1)(x+1)\nYou can cancel the (x-1)'s leaving you with\n(x+4)/[x(x+1)]\n\nNote:  For the first problem, if you meant\n1/(x+y)\n______\nx + y\nthen the answer is 1/(x+y)^2
31021
if 2x-1=T so x=(T+1)/2 \nf(T)=4((T+1/2)power to 2)-1 \nf(T)=4((T+1)power to 2)/4)-1\nf(T)=((T+1)power to 2 )-1\nf(T)= T power to 2+ 2T+1-1\nf(T)= T power to 2 + 2T\nwe can replace T with everything like x\n\nf(x)= x power to 2 + 2x
217152
Factor each one.\n(x^2-3x-18) / (x^2-4x-32) divided by (x^2-2x-24) / (x^2-x-20)\n= (x^2-3x-18) / (x^2-4x-32) times (x^2-x-20)/(x^2-2x-24)\n= [(x-6)(x+3)]/[(x-8)(x+4)] times [(x-5)(x+4)/(x-6)(x+4)]\n= [(x-6)(x+3)(x-5)(x+4)] / [(x-8)(x+4)(x-6)(x+4)]\nYou can now cancel anything on top with anything on bottom\nx-6 and x+4 cancels\n= [(x+3)(x-5)]/[(x-8)(x+4)]
353301
1.Xc=1/2pifc\nc=1/2pifXc\n\n2.ib=ibq+rt(2is)\nib(1-q)=rt(2is)\nsquaring\n[ib(1-q)]^2=2is\ns=[ib(1-q)]^2/2i\n\n3.Q=Xl/R\nXl=Q/R\n\n4.Lt=L1+L2+2Lm\n2Lm=Lt-L1-L2\nLm=[Lt-L1-L2]/2\n\n5.Es=E1+E2+E3\nE2=[Es-E1-E3]\n\n6.R1=i2R^2/i2\ni2R^2=i2R1/i2\nR=rt[i2R1/i2]\n\n7.R1/Rx=R2/R3\nRx=R1R3/R2\n\n8.y=mx+b\nb=y-mx\n\n9.i2=i1R1/i1
470893
1.  (x + 9)(x - 9)\n2.  (t + 2)(t - 2)\n3.  (b + 12)(b - 12)
400274
Step 1: find y '\nStep 2: fill in x=1\nThat gives the right answer 2\n\nTh
522182
The first step in these type of questions is to get the expression into index form.\n\nThis becomes x - 3 x^1/2.  Also, remember x is the same as x^1\n\nSo dy/dx =  1*x^(1-1) - 3*1/2 * x^(1/2 -1)\n\n               =  1 * x^0 - 3/2 * x^(-1/2)\n\n               =  1 - 3/2* x1/sq.root x     remember x^0 =1\n\n                = 1 - 3/(2sq.root x)
364846
put sinx=t\ncosxdx=dt\nI=fcos(pi t)dt with limits -1,+1\nI=(sin(pi t))/pi with limits +1,-1\nI=(sin(pi)-sin(-pi))/pi=0\nthanks & good luck
372657
1.f(x)=x^2+4x-7\nx=[-4+/-rt(16+28)]/2\n=[-4+/-2rt(11)]/2\n=-2+/-rt11\n\n2.3x^2-11x-20=0\n=3x^2-15x+4x-20=0\n=(3x+4)(x-5)=0\nx=-4/3 or 5
394434
x² / √(1-x) \n= x² √(1-x)/(1 - x)\n= (-1 + 1 - x²)√(1-x)/(1 - x)\n= (1 - x²)√(1-x)/(1 - x) - √(1-x)/(1 - x)\n= (1 + x)√(1-x) - √(1-x)/(1 - x)\n= x√(1-x) + √(1-x) - 1/√(1-x)\n\nSo ∫x² / √(1-x) . dx\n= ∫[x√(1-x) + √(1-x) - 1/√(1-x)] . dx\n\nNow ∫x√(1-x)dx \n= x * -1 * 2/3 (√(1-x)³ - ∫[-1 * 2/3 (√(1-x)³ . 1)dx\n= -2x/3 * √(1-x)³ + 2/3∫(√(1-x)³ dx\n= -2x/3 * √(1-x)³ + 2/3*2/5 * -1 * √((1-x)^5)\n= -2x/3 * √(1-x)³ - 4/15 * √((1-x)^5)\n\n∫√(1-x) dx = \n\n∫1/√(1-x) dx = -2 *√(1-x)\n\nSo ∫x² / √(1-x) . dx\n= ∫[x√(1-x) + √(1-x) - 1/√(1-x)] . dx\n=  -2x/3 * √(1-x)³ - 4/15 * √((1-x)^5) - 2/3 * √(1-x)³ + 2 *√(1-x) + c\n= - 4/15 * √((1-x)^5) -2/3 (x + 1)√(1-x)³ + 2 *√(1-x)  + c
455733
x = [-b +/- sqrt(b^2 - 4ac)] / (2a)\nf(x): a = 1, b = 0, c = -4\ng(x): a = 1, b = -1, c = -12