What are the answers to these math problems?
827033
36=36 (6^2)\n100,000=100,000 (10^4)\n196=196 ( 14^2)\n512=512 (2^9)\n81=81 (9^2)
222592
1. p'q\n\n2. p'\n\n3. pq' + p'\n\nI think i got the correct ones !
775434
1) x= 2\n2) correct\n3)correct\n4)correct, u got the rest right!
706334
To solve #1, you  have to use partial fractions.  \n\nYou have to convert \n\n** x / [(3x-1)(x+2)] = A/(3x-1) + B/(x+2) for some constants A,B.\n\nTaking the integral of x / [(3x-1)(x+2)] would be the same as taking the integral of A/(3x-1) + B/(x+2), so let's do that.\n\nIntegral (A/(3x-1) + B/(x+2))dx = \nIntegral (A/(3x-1))dx + Integral (B/(x+2))dx\n\nAlways pull out constants.\n\nA*Integral(1/(3x-1))dx + B*Integral(1/(x+2))dx\nA * 1/3 * ln |3x - 1| + B * ln|x+2| + C\n\nNow, solve for A and B using the equation marked as **.\n\nx / [(3x-1)(x+2)] = A/(3x-1) + B/(x+2)\n\nFirst, multiply both sides by (3x-1)(x+2) to get\n\nx = A(x+2) + B(3x -1)\n\nRemember that the above is a statement that is __true for all x___, so in order to solve for A and B, we can let x be ANY value.\n\nFor simplicity purposes, we're going to let x=-2 and then x=1/3.\n\nx = -2:  \n\n-2 = A(-2 + 2) + B(3(-2) - 1)\n-2 = B(-7)\nB = 2/7\n\nx = 1/3:\n\n1/3 = A(1/3 + 2) + B(3(1/3) -1)\n1/3 = A(7/3)\nA = 1/7\n\nTherefore, substitute A=1/7 and B=2/7 back into here:\n\nA * 1/3 * ln |3x - 1| + B * ln|x+2| + C\n\nto get\n\n1/7 * 1/3 * ln |3x - 1| + 2/7 * ln|x+2| + C\n7/3 * ln |3x - 1| + 2/7 * ln|x+2| + C\n\nTo solve #2:\n\nUse integration by parts.\n\nlet u = arcsin(3x)                  dv = dx\ndu = 1/sqrt(1-(3x)^2) * 3         v = x\n\nSo our new integral becomes\n\nx*arcsin(3x) - Integral [ (3x)/sqrt(1 - (3x)^2) ] dx\n\nAfter that it becomes substitution.
133517
the solutions is 7,8,3,2 or 7,6,5,2. The problem works with both sets of numbers.
970314
No you can't have the answers.  This is HOMEWORK HELP not DO MY HOMEWORK FOR ME.  You will have to open your text or do you own online search.
843058
1) 2.01 * 10^2\n\n2) 1.995 * 10^23\n\n3) 5.35 * 10^-4\n\n4) 3.8418 * 10^-7
752959
Problem 1: Find an exponential function of the form f(x)=ba^x+c with y-intercept 2, horizontal asymptote y=-2, that passes through the point P(1,4). \n\nYou find the y-intercept by replacing x with zero, which gives f(x) [which is y on the graph] = 2, so b + c = 2 (since a^0 = 1)\n\nThe horizontal asymptote is y = c, so c = -2.  That means b must be [you figure it out]\nNow it is y = 4a^x - 2.  This must be true when x=1 and y=4.\nSo 4 = [you figure it out; answer deleted due to complaint from below]\n\n\nProb. 3.  What is the amount in the account if $500 is invested for 10 years at the annual rate of 5% compounded continuously? \n  Just plug in the values.\nA = 500 e^(.05 x 10)\n\n[answer deleted]
909535
1) \n76>40+4c\n36>4c\nc<9\n\n2) 1500>100+17.5p\n1400>17.5p\np<80
761571
1.(4x+1)(16x^2-4x+1)\n\n2.(1-5x)(1+5x+25x^2)\n\n3.(2xy^2+3)(4x^2y^4-6xy^2+9)\n\n4.b^3(a^3-b^3)\nb^3(a-b)(a^2+ab+b^2)\n\n5.(3x-2y)(9x^2+6xy+4y^2)
805274
1) 3 * sqrt(-8)\n= 3 * sqrt(4) * sqrt(-2)\n= 6 * sqrt(2) * i\n\n2) sqrt(1/6)\n= sqrt(6/36)\n= sqrt(6) / sqrt(36)\n= sqrt(6) / 6
949969
Problem #1:\n2^(3x) = 4\n\nSince you have powers of 2, I would take the log base 2 of both sides:\nlog[2](2^(3x)) = log[2](2²)\n3x = log[2](2²)\n3x = 2\nx = 2/3\n\nProblem #2:\n5/6x = 10\n\nMultiply both sides by x\n5/6 = 10x\n\nDivide both sides by 10:\n5/60 = x\n\nReduce:\nx = 1/12\n\nYour answers are:\n2/3 and 1/12
464335
I normally hate referring people to wikipedia articles, since it is so often inaccurate, but this one is quite good (link is below).  If you scroll down, there is a table that gives just what you're looking for.\n\n2x2 grid: 288 possible solutions\n3x3 grid: 6.7*10^21 possible solutions!
221530
Add them together and divide by 5.\n\nWow. I could seriously have you people do my homework for me. Every night. I'll type out the problems and see how you all do. :P JP
778763
1) (a - b)(x - y)\n2) (x^9 + y^3)(x^9 - y^3)\n3) (6x + 5)(x + 8)\n4) (x - 9)(x - 7)
884068
2|1 1 -2 -3 -1\n..... 2 6 ..8 10\n----------------\n...1 3 4 ..5 9\n\nSo the quotient is x³ + 3x² + 4x + 5, and the remainder is 9
937250
PROBLEM 1:\nBobby is 3 years old.  His sister Nancy is four more than six times as old as Bobby.  How old is Nancy?\n\nPROBLEM 2:\nThe pizza parlor cuts large pizzas into 6 slices and small pizzas into 4 slices.  If Ryan buys 3 large pizzas and a small pizza, how many total slices will he have?\n\nPROBLEM 3:\nThere are 3 hexagons and a square drawn on a piece of paper.  How many total sides are there?\n\nPROBLEM 4:\nWhat is the base-10 equivalent of 34 base-6?\n\nPROBLEM 5:\nAdult tickets are $6 each and child tickets are $4 each.  How much will it cost for a family of 3 adults and 1 child?