1+4.9T^2=6T What Is T? Also, 1-4.9T^2=6T , T=??
595518
1 + 4.9t^2 = 6t\n4.9t^2 - 6t + 1 = 0\n\nusing the quadratic formula\n\nt = (-b ± sqrt(b^2 - 4ac))/2a\nt = (-(-6) ± sqrt((-6)^2 - 4(4.9)(1)))/2(4.9)\nt = (6 ± sqrt(36 - 19.6))/9.8\nt = (6 ± sqrt(16.4))/9.8\nt = (6 ± sqrt(4 * 4.1))/9.8\nt = (6 ± 2sqrt(4.1))/9.8\nt = (3 ± sqrt(4.1))/4.9\nt = about 1.02548 or .199011\n\n-----------------------------------\n\n1 - 4.9t^2 = 6t\n4.9t^2 + 6t - 1 = 0\n\nt = (-6 ± sqrt(36 + 4(4.9)))/9.8\nt = (-6 ± sqrt(36 + 16.4))/9.8\nt = (-6 ± sqrt(52.4))/9.8\nt = (-6 ± sqrt(4 * 13.1))/9.8\nt = (-6 ± 2sqrt(13.1))/9.8\nt = (-3 ± sqrt(13.1))/4.9\nt = about -1.37312 or .148627
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!
954091
-4t^2+36-72\nFactor-2t+12\n +2t-4t^2+24\n-6+12-72\nIt is (2t-6)(-2t+12)\n\nFirst you put the first trinomial in the upper left corner of the box (-4t^2).\nThen you put the last trinomial in the bottom right corner of the box(-72).\nYou multiple (-4t^2 & -72) then you make a list of factors that will equal that number. In this case they are 12 & 24.\nYou know that one number must be negative because -4t^2. Fill in the info in the box and multiple them out. \nIf you do this you can see that it equals the trinomial.\nAdd down the left side of the box to get the first set of numbers.\nThen add across the box to get the second of numbers.\nYou can see how I did this.\n\nYou can email me if this doesn't make sense.
704730
x(6)^2/4=117\n=>36x/4=117\n=>9x=117\n=>x=117/9=13
812111
x^2 = 96\n\nx= +/- sqrt(96) = +/- 4 sqrt(6)
122527
1 - 64x^2 = (1 + 8x)(1 - 8x)\n\ny^2 - 12y + 36 = (y - 6)(y - 6)  or (y - 6)^2
957344
t=2.512 seconds\n\n10.3=0.5(0+8.2)t\nTherefore \n\n1 = 4.1t/10.3\n\n1/t = 4.1/10.3\n\nt = 10.3/4.1 = 2.512 seconds
798514
do you mean 6x([1/2]x^2)?\nif so 3x^3\nor 6x(1/(2x^2))?\nthen its 3x^-1
702018
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
8794
5t+1>3t-13\n-> 5t-3t>-13-1\n-> 2t>-14\n-> t>-14/2\n-> t>-7\nKeep Asking
599170
For the value of t for which the height is 12, substitute 12 in for h, and solve for t.\n\nh = -16t^2 + 32t\n12 = -16t^2 + 32t\n16t^2 - 32t + 12 = 0\n4t^2 - 8t + 3 = 0\n(2t-3)(2t-1) = 0\nt=1.5s or t=0.5s\n\nThere are two answers: one is on the way up, and the other is on the way down.\n\nFor the second question, how long was he above 14 feet, you can do the same thing.\n\n14 = -16t^2 + 32t\n16t^2 - 32t + 14 = 0\n8t^2 - 16t + 7 = 0\n\nThis cannot be factored, but can be solved using the quadratic formula, giving the solutions:\nt = 1 + (sqrt2)/4\nt = 1 - (sqrt2)/4\n\nThese solutions again represent a value for on the way up and on the way down.  The difference between the two values is:\n\n2*(sqrt2)/4 = (sqrt2)/2 which is approx 0.707 s
745510
5x + 3t = 19\n-(5x -3t = 1)\n------------------\n6t = 18\nt = 3\n\n5x + 3(3) = 19\n5x + 9 = 19\n5x = 10\nx = 2\n\n5(2) - 3(3) = 1\n10 -9 = 1
786382
x^2-6x+9=3+9\n(x-3)^2=+/-rt12\nx-3=+/-2rt3\nx=3+/-2rt3\n\n2.2(x^2+5x)=-11\n2(x^2+5x+25/4-25/4)=-11\n2(x+5/2)^2-25/2=-22/2\n2(x+5/2)^2=3/2\n(x+5/2)^2=3/4\nx+5/2=+/-(rt3)/2\nx=-5/2+/-(rt3)/2\n=1/2[-5+/-rt3]
156051
first let T =X^2\nthan we get a second degree equation\n2T^2-7T-4=0\nso you can solv it easyer
221833
1) 1/3(T-1)=2(T-1/3)\n2) 1/3T- 1/3= 2T-2/3\n3) 2 1/3T-1/3= 2/3\n4) 2 1/3T= 1\n5) T= 0.375\n\nThere you go, man.
347487
-6-9x=-9x-4\n-6=-4.....\nno solution?? whats so hard to that?
839617
i think you meant F(0)=0\nand f(t)=6sec^2(t)-9t^3\nso\nF(t)=6tan(t)-(9/4) t^4+c\nF(0)=c=0\nF(t)=6tan(t)-(9/4)t^4