how do you solve for t in 6*2^t=3^t-1?
487376
Well, it's hard to tell from the way you wrote\nthis if the right hand side is:\n\n3^T - 1\n\nor if it is:\n\n3^(T-1)\n\nIf it's the first, then I'm stuck too, although,\nnumerically, the answer would be approximately 4.437915.\n\nHowever, if the equation is really:\n\n6*2^T = 3^(T-1)\n\nthen you can solve it using logs.\n\n\nTake the log of both sides:\n\nlog(6*2^t) = log(3^(T-1))\n\nlog(6) + T*log(2) = (T-1) * log(3)\n\nDistribute the (T-1):\n\nlog(6) + T*log(2) = T*log(3) - log(3)\n\nMove the T's to one side and constants to the other:\n\nlog(6)+log(3) = T [ log(3)-log(2) ]\n\nDivide the factors out to get T by itself:\n\n[log(6)+log(3) ] / [ log(3)-log(2) ] = T\n\nPlug this into a calculator, and it appears that\nT is approximately equal to 7.1285.
901005
Use the following equation and solve:\n\nP1V1/T1=P2V2/T2\n\nT2=P2V2T1/P1V1\n\nalso note the significant figures!!!!
268383
2x^2+6x-1x-3\n2x(1x+3)-1(1x+3)=0\n(2x-1)(x+3)=0\nx=1/2 or -3
948100
t-(4/t)=3\ncan be rewritten as\nt^2 - 3t - 4 = 0\n\nthis has 2 possible solutions:\nt=-1\nor\nt=4
1160074
18r^4s^5t^6\n---------------- =\n-3r^2st^3\n\n-6r^(4-2)s^(5-1)t^(6-3)= -6r^2s^4t^3\n\nhope it's not too late.\n\nGood luck!
918458
Look at it this way:\n(-1)*(6x + 6x) = (-1)*(12x) = -12x
344350
=f^3(6f^2+6f+4-9f) \n=f^3(6f^2-3f+4)\n\nsolve 6f^2-3f+4 using f=[3 +/- sqrt(9-4(6)(4))]/12\n\nNote that the solution (other than f=0) is imaginary since the term under the sqrt sign is negative
791030
3x – 2y = 6\n3x = 6 + 2y\nx = 2 + 2/3y
107615
t=theta\n^=to the power of...\n\ncos (2t) + 2 sin ^2 (t) = cos (2t) + 2 - 2 cos ^2 (t)
640891
I'm guessing you meant 17x^3 - 10x^2 - 6x +4 and 2x^2- 6x +2\nIf that's right, you just add like powers of x and their coefficients to get 17x^3 - 8x^2 -12x +6.
487411
e^3x+1=e^-2x-6\nor 3x+1=-2x-6\nor 3x+2x=-6+1\nor e=-5/5=-1
652953
I got:\n1. 6x^2 - 36x +48\n2. 6x + 22
723707
s'(t) =(t-3)(-2t^{-3}) + t^{-2}\n=( -2t+6 +t)/ t^{-3}\n=(6-t) / t^{-3}\nso the velocity is 0 when the top =0\ni.e 6-t-0\nt=6\ns
848441
If you are solving for t and if the problem is supposed to be\n 0=16*(t^2)+80t, then t = 0 and/or -5. If the problem is supposed to be 0=(16t)^2+80t, then t = 0 and/or -5/16. To solve the problem, you just factor out a t, which implies that t is equal to zero, then set the rest of it equal to zero and solve for t.
156051
first let T =X^2\nthan we get a second degree equation\n2T^2-7T-4=0\nso you can solv it easyer
741312
If f(t)=(t-a)*(t-b) = t^2-(a+b)*t+ab; then \nI=(1/3)*t^3 - (a/2+b/2)*t^2 + a*b*t+C
33091
3m(-6m-2m^2 p - 10p) = 3m(-6m) - (3m)(2m^2 p)- 3m(10p)\n= -18m^2- 6m^2 p - 30mp