please help me to work this formula out - find v, if u = 6 ad t = 3 v=u-t over 3?
928623
Hi Kenima ;\n\n-if u = 6   & t = 3  & v = (u -t )/ 3 \n-just fill the value of " u " & " t " in to Formula of v \n\nv = ( 6 - 3 ) / 3 \nv = 3 / 3\nv =1 \n\nGood Luck ♣
586962
(6y^2-3y)(y+7)-->(2y)(y-1)(y+7)\nif that helps
956088
x = -11 + 12t + 3t² + 4t³ - t^4\nv = 12 + 6t + 12t² - 4t³\na = 6 + 24t - 12t²\nfor max or min a, \n24 - 24t = 0\nt = 1 (and this yields a miximum)\na = 6 + 24 - 12 = 18 units/sec²\nv = 12 + 6 + 12 - 4 = 26 units/sec\nx = -11 + 12 + 3 + 4 - 1 = 7 units
370398
O.K.\n\nUse the substitution:\n\ny = uv , x = u/v\n==>\n(1+u^2)(u/v)(udv + vdu) + (1-u^2)uv(du/v - udv/v^2) = 0\n==>\n(1+u^2)(udv + vdu) + (1-u^2)(vdu-udv) = 0\n==>\n2vdu + u^2(2udv) = 0\n==>\ndu/u^3 + dv/v = 0\n\nIntegrating gives:\n\nv = (constant) e^(1/(2u^2))\n==>\nv^2 = A e^(1/u^2)\n\nReverse substituting x,y gives the answer as an implicit function:\n\ny/x = A exp(1/xy)
221582
T = 6a^2\nV = a^3\n\nV = a^3\n108 = a^3\na = 4.76cm\n\nT = 6(4.76)^2\nT = 6(22.68)\nT = about 136cm^2
747340
-2x-x > -6 + 4\n-3x > -2 \nx < 2/3
709461
2x^2-6x+3\nuse this formula\n-b+or-root b^2-4ac/2a\nIn this case \na=2   b=-6    c=3\njust substitute this value in above formula\nYour answer must be\n3+root3/2     3-root3/2
929521
V=(4/3)pir^3\ndV/dt=4pir^2*dr/dt\n10=4pir^2*dr/dt\nA=4pir^2\ndA/dt=8pir*dr/dt\n=4pir^2*dr/dt*/r\n=10/r\nwhen r=8 dA/dt\n=10/8cm^2/sec
823232
y = (3x - 2)/(5x² + 4)\n\nW.w.t.f dy/dx.  You can use the formula:\nd/dx (u/v) = (v du/dx - u dv/dx)/v²\n\nNow, we have u = 3x - 2, v = 5x² + 4.  du/dx = 3 and dv/dx = 10x. We can now substitute them to the formula:\ndy/dx = [(5x² + 4)(3) - (3x - 2)(10x)]/(5x² + 4)²\n\nNow,\ndy/dx = (15x² + 12 - 30x² + 20x)/(5x² + 4)²\n\nThus,\ndy/dx = (-15x² + 20x + 12)/(5x² + 4)²\n\ntherefore,\ndy/dx = -(15x² - 20x - 12)/(5x² + 4)²\n\n^_^
998009
You need to use the distributive property where:\n8(4w+x-6y)=\n8(4w)+8(x)-8(6y)=       \n32w+8x-48y
730085
If x is the angle between two vectors u and v, then:\n\ncos x = (u * v) / ||u|| ||v|| \n\nIf u = (u1, u2) and v = (v1, v2), u * v is the dot product of u and v, equal to u1 v1 + u2 v2, and ||u|| and ||v|| denote the magnitudes of each vector [ ||u|| = sqrt(u1^2 + u2^2), similarly for ||v|| ].
24383
here are some simple mechanics formulas dealing with velocity:\n\nv=d/t\na=v/t\nvf^2=vi^2+2ad\nd=vit + 1/2at^2\n\nwhere v is velocity, vi is initial velocity, vf is final velocity, a is acceleration, and t is time.  also the i and the f after the v are just subscripts, and ^2 just means sqared.
1031456
g(1-t) = 3(1-t)^2 - 2(1-t) + 1\nFirst expand (1-t)^2\n(1-t)^2 = (1-t)(1-t) = 1 - t - t + t^2\n= 1 - 2t + t^2 \nnow multiply by 3\n3(1-t)^2 = 3(1 - 2t + t^2 )\n= 3 - 6t + 3t^2\n\nNow move on to the next term:\n-2(1-t) = -2 + 2t\n\nPut everything together and combine like terms.\ng(1-t) = 3(1-t)^2 - 2(1-t) + 1\n= 3 - 6t + 3t^2 -2 + 2t + 1\n= 3t^2 - 4t +2
929743
∫t²cos(1 - t³) dt\n\nI note that d(1 - t)/dt = -3t² \nThis is set for substitution of 1 - t³\n\nLet u = 1 - t Thus du = -3t²dt ie  -1/3 . du = t²dt\n\nSo ∫t²cos(1 - t³) dt\n= ∫cos(1 - t³) t²dt\n= ∫cosu (-1/3) . du\n= -1/3 sinu + c\n= -1/3 sin (1 - t³) + c
246336
r = 20+t\nf = 6+y\ns = 15-g+u\ng = 15-s+u\nu = s+g-15
443446
S (x^2-4x) / (x^3-6x^2) dx\n\nlet u = x^3-6x^2\nthen du = 3x^2-12x\n\nS (x^2-4x) / (x^3-6x^2) dx = (1/3) du/u\n= (1/3) ln u + C\n= (1/3) ln (x^3-6x^2) + C\n\nAnswer B.
878312
V=4/3pier^3\nA=4pier^2