solve for R in terms of I,P and T: I=PRT?
483738
Isn't this supposed to be,\n\nI = PTR/100\n\nThen the answer is,\nR=100I/PT\n\nAnyhow, for your query its,\n\nR=I/PT
184147
5t(t^2+t-1)(t^3-t^2+2t+2)\nIts a long answer but it's CORRECT.
821342
THe VdW equation is P= nRT/(V-nb) - a(n/V)^2, just plug and chug.  But dont forget to switch your T into Kalvin
796480
use v^2 =  GM/r\n\nv = root GM/2r\n\norbit = 2 pi r\n\nfor part 2 use suvat equations, s=vu+.5at^2 etc and it should fall out
724685
r-s = t\nadd s to both sides\nr = t+s
708667
I think you mean\n\nS = A - RT/(1-R)\ntake A  to the left and negate\n\nA- S = RT/(1-R)\nmutiltply by (1-R) on both sides\n\n(A-S)(1-R) = RT\nor (A-S) -R(A-S) = RT\ntake R term to the right\n\nA- S = R(T+A-S)\n\nso R = (A-S)/(T+A-S)
642651
A=P(1+r/n)^(n*t)\nA = 20,000(1+.08/1)^(1*3)\nA = 20,000(1.08)^3\nA = 25,194.24\n\nThe more frequently you compound, the more money you will make.\n\nA = Pe^(rt)\nA = 20,000e^(.08*3)\nA = 20,000e^.24\nA = 20,000(1.271249150)\nA = 25,424.98
979217
y = sqrt(1+t^2) = (1+t^2)^(1/2)\n\nby the power and chain rule\n\ndy/dt = (1/2)(1+t^2)^(-1/2) x d/dt (1+ t^2)\n\n= 2t(1/2) /sqrt(1+t^2)\n\n= t/sqrt(1+t^2)\n\nTherefore\n\ndy = (t / sqrt(1+t^2)) dt
706256
x(rt2)x+5rt2(x^2)-25y(rty)\n=6rt2x^2-25yrty
616704
pie   p power = i current x e voltage\n      e = i x r resistance
799788
You are given P(t=0), Q(t=0); P'(t=0), Q'(t=0); and P''(t=0), Q''(t=0). Given that P'' and Q'' remain the same (that is, the motion of P and Q do not change), you need to show that there is a time T such that P(t=T) = Q(t=T).\n\nIt's late and I'm drunk, so you can do the actual working for yourself :D
915753
To help solve for t you'll need to use the quadratic formula. It's too difficult to show how I got the answer since I can't use the correct symbols, but the answer is t = 1.39.
289077
add all the equations, you get\n2(r+s+t) = 56\nso,\nr+s+t = 56/2 = 28\ndone!!
893823
Use the quadratic equation \n\nt = -b + or - sqr(b^2 - 4ac)/2a\n\nyour equation is already in the form:\n\nat^2 + bt + c = 0\n\njust plug in the variables and solve for t twice
244823
The amount (A) will be doubling, so substitute "2P" in for A. Then you can drop the "P" from each side by division and solve for t.\n\nA = Pe^(rt)\n2P = Pe^(rt)\n2 = e^(rt)\nln 2 = (rt) ln e\nln 2 = .054t (ln e = 1)\n(ln 2)/.054 = t\n(approx.) 12.836 = t\n\nDoes that sound correct? (It's been awhile since I've done these kinds of problems.)
937830
T=m1*-9.8+m1*a\nT=m2*a--m2*-9.8\n\nset them equal and solve for a\n\ntrust me, i had a test on this today
865054
Let u=ln t, e^u=t, du=dt/t, dt=t du\nThen:\n∫ln² t dt = ∫t u² du = ∫u² e^u du\nIntegrating by parts:\nu² e^u - ∫2u e^u du\nu² e^u - 2u e^u + ∫2e^u du\nu² e^u - 2u e^u + 2e^u + C\nt ln² t - 2t ln t + 2t +C\nAnd we are done.