S=1000(1.02)to the power of 4x\n\nWhat amount will result in 8 years?
133670
A = P(1+i)^n\nP = 1000\ni = .08/4 = .02\nn = 8*4 = 32\n\nA = 1000(1+.02)^32\n= 1884.54
487506
.08y + .12(1000-y) = 960\nFirst distribute the .12\n.08y + 120 - .12y = 960\nCombine .08y and -.12y\n-.04y + 120 = 960\nSubtract 120 from each side\n-.04y = 840\nDivide both sides by -.04\ny = 840/-.04\nAnswer: y = -21000
342825
this is a mixture of product power and chain ..simple but takes time\n((x^2+1)^9*(36x^4+77x^2+1).....\ne^(9x^2))/(2sqrtx)\nsry it cut my answer in half\ni didnt have to use ln\nmine is right i checked it on my calculator
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
378986
find the stoichiometric yield.  then multiply by .8962 voila.
621470
T=.00139x^2+.0304x=1.25 \n\nmaybe you meant\n\nT=.00139x^2+.0304x-1.25 \nor\nT=.00139x^2+.0304x+1.25 \n\n?????
178322
let x be the number from A\nlet y be the number from B\nlet z be the number from C\nthey have 500 pounds\nso\n2x+2y+3z=500\nand we have 1000 hours to produce \nso \n4x+5y+5z=1000\nsolve this system with respect to x u will get the following\n-2(2x+2y+3z=500)\n   4x+5y+5z=1000\nadd them u get \ny-z=0\nthen y=z\n5(2x+2y+3z=500)\n-2(4x+5y+5z=1000)\nu add and get\n2x+5z=500\nthen z=(500-2x)/5=100-2x/5\nz=y=100-2x/5\nlet f(x)=500x+600y+1000z=500x+y(1000+600) since y=z\n        =500x+1600y\nsubstitute y and z\nu will get\nf(x)=500x+1600(100-2x/5)=500x+160000-640x\n    =160000-140x\ndifferentiate f(x)\nu get\nf'(x)=-140\nso it is maximal when x=0\nthen y=z=100\nand profit =f(0)=160000$\nf(x)=
287771
I figured it out... the answer is 1/(13*(2-3x))\nIf you need an explanation tell me
59978
It's multiplication.\n\n(64500)(0.02) = 1290\n\nScientific notation is 1.29 * 10^3
55745
6250\n\nX    2    10    50    250   1250    6250\n\nEach term is 5x the previous.
429077
ok 1.plz spess equation right\nnxt\n2. 6284000 X .01= 62840\n\n3. it would end up being well over 10 mil.
344540
1.68 * 10^14 = 5.8*e^(.0114t)\n\n(1.68*10^14)/5.8 = e^(.0114t)\n\nln[(1.68*10^14)/5.8] = ln(e^(.0114t)\n\nln[(1.68*10^14)/5.8] = .0114t*ln(e)\n\nln[(1.68*10^14)/5.8] = .0114t\n\n\n{ln[1.68*10^14)/5.8]}/.0114 = t\n\nt = 2719.046 years
352821
10.3 = 0.5 x (0 + 8.2) x t\n\n10.3 = 4.1 x t\n\nt = 10.3 / 4.1 = 2.512 s
508169
800*0.15+x=(x+800)*0.32\n120+x=0.32x+256\nx-0.32x=256-120\n0.68x=136\nx=136/0.68\n=200\nhe has to add 200 ml
54208
2x is the largest possible factor, leaving:\n2x(4x^2-5x-7)
619556
888+ 88+ 8+ 8+ 8 = 1000\nComplete answer-\n1) [(8+8)/8] * [(8* 8* 8)- 8] - 8 = 1000\n\n2) (8888-888) / 8 = 1000 \n\n3) 8 [(8*8) + (8*8)] -8 -8 -8
158984
(.8+.9)/(1+.8*.9)=.98837c