Given the formula  s=1/2(u+v)t, express u in terms of v,s and t?
375706
on crss multiplying,we get:\n2s = ut + vt\ntherefore ut = 2s - vt\ntherefore u= 2s/t -v
590233
Hint:\nDo a simple substitution u = x-5\nWhich means x = u + 5\n\nYou should get\n\n(2/5)(x-5)^(5/2) + 2(x-5)^(3/2) + C
950045
Correct formula should be as follows:\n\nE = 1/2*I*w^2
871386
∫udv = uv - ∫vdu\n∫x^2dx/√(1-x)\nlet u = x^2, dv = dx/√(1-x)\ndu = 2xdx, v = -2√(1-x)\n∫x^2dx/√(1-x) = -2x^2√(1-x) + ∫(4x√(1-x))dx\nu = 4x, dv = √(1-x))dx\ndu = 4dx, v = -(2/3)(1-x)^(3/2)\n∫x^2dx/√(1-x) = -2x^2√(1-x) - (8x/3)(1-x)^(3/2) + ∫(8/3)(1-x)^(3/2)dx\n∫x^2dx/√(1-x) = -2x^2√(1-x) - (8x/3)(1-x)^(3/2) - (16/15)(1-x)^(5/2)\n∫x^2dx/√(1-x) = √(1-x)(2x^2 - (8x/3)(1 - x) - (16/15)(1-x)^2)\n∫x^2dx/√(1-x) = √(1-x)(2x^2 - 8x/3 - 8x^2/3 - (16/15)(1 - 2x + x^2)\n∫x^2dx/√(1-x) = (1/15)√(1-x)(30x^2 - 40x - 40x^2 - 16 + 32x + 16x^2)\n∫x^2dx/√(1-x) = -2√(1 - x)(x(3x + 4) + 8)/15
840169
u= x\ndv=cosx dx\n\ndu =dx\nv= sinx\n\n\nintegral of xcosx dx = xsinx - integral of sinx dx\n= xsinx + cosx +C\n`
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
790665
starting from the very beginning\nv ≡ ds/dt\na ≡ d²s/dt² = dv/dt\ndv = adt\nif a is constant,\n∫dv = a∫dt\nv = at + C\ncall C v0, or initial velocity:\nv = v0 + at\nv = 0 +(9.80662 m/sec^2)(2 sec)\nv = 19.61324 m/s
724685
r-s = t\nadd s to both sides\nr = t+s
754733
Let y = u/v   where u=some function and v =some funtion\n\nthen dy/dx = (v*du/dx - u*dv/dx)/ (v^2)     **Quotient Rule**\n\nIn this case u = 6x^2-1 and v = 2*sqrt(x)\n\nSo you just differentiate and plug into the equation for dy/dx
376765
Let u = 1 + x^4 and then apply the Chain Rule. This allows your expression to be written as\n\n1/sqrt(u)^(1/2)\n\nwhich is equivalent to\n\n1/u^(1/4)\n\nwhich, in turn, is equivalent to\n\nu^(-1/4).\n\nDifferentiating this expression with respect to x yields the following:\n\n(-1/4)*u^(-1/4-1)*(du/dx).\n\nNote: du/dx = 4x^3.\n\nTherefore, the desired derivative is \n\n(-1/4)*(1+x^4)^(-5/4)*(4x^3).\n\nIf we simplify this a bit we are given\n\n-x^3/(1+x^4)^(5/4).
195576
Y = U + VX^2\nVX^2 = Y - U\nX^2 = (Y-U)/V\nX = sqrt ( (Y-U)/V )
904926
5+-V92\nquadratic formula
904430
(T-12)/(T-5) - (T-11)/(5-T) = (T-12)/(T-5) + (T-11)/(T-5) = (2T-23)/(T-5)\n\nPerhaps you would consider this simplified further as:\n\n2 - 13/(T-5)
473554
You have to use integration by parts.\nLet u = ln(x) and dv = dx, then\ndu = (1/x)dx and v = x\nint(udv) = uv - int(vdu)\n= xln(x) - int(x(1/x)dx)\n= xln(x) - int(dx)\n= xln(x) - x + C, where C is a constant
492690
S(I)+3f (g)--> SF (g)\n..........2...........6
684353
P1V1T2=P2V2T1 \n\nRemember to express  temperature in Kelvin (with pressure in Pascals and volume in m^3)
22859
Subract the two terms on the left giving:\n-2t < 16\n\nAdd 2t to both sides giving\n0 < 16 + 2t\n\nSubtract 16 from both sides\n\n-16 < 2t\n\nDivide both sides by 2\n\n-8 < t