.30*679.95 = 203.99
you don't want the minima and maxima of (x^2-4)/(1+cos(x)^2) That is not f(x). It is f'(x). So, the minima and maxima of f(x) occur when (x^2-4)/(1+cos(x)^2) = 0.
convert 600mg to moles. 2HCl + Mg(OH)2 = MgCl2 + 2H2O so, each mole of Mg(OH)2 needs 2 moles of HCl. So, now you know how many moles of HCl you need. For .5M HCl solution, you will need 2L (2000ml) of acid for each mole.
using discs, v = ∫[0,4] πr^2 dy where r = x = √(24-6y) v = π∫[0,4] (24-6y) dy = π(24y-3y^2) [0,4] = 48π using shells, v = ∫[0,√24] 2πrh dx where r=x and h=y=4-x^2/6 v = 2π∫[0,√24] x(4-x^2/6) dx = 2x^2 - x^...
(i) v(t) = ds/dt (ii) set v=0 and solve for t (iii) a(t) = dv/dt; find a(5)
just expand the power and you have ∫(4x^4 - 12x^2 + 9) dx now it's easy - just powers
Math - PLEASE HELP
f(x) = ∫[4,x] 4t^10(2t+8lnt)^4 dt Don't know what the 0 is doing in there; if I got it wrong, fix it and modify the solution. Using Leibnitz's Rule for differentiation under the integral sign, f'(x) = 4x^10 (2x+8lnx)^4 so, f'(1) = 4(2+0)^4 = 64
first, collect the stuff with d on one side, and the plain numbers on the other. Recall that a true equation remains true if you do the same thing to both sides. So, start by adding 3.4 to both sides: d/2.5 - 3.4+3.4 = 4.6+3.4 d/2.5 = 8.0 Now, multiply both sides by 2.5: d/2.5...
w(4w-7) = 15 Now, 15=3x5, so try w=3. In that case, the length is 5. Algebraically, 4w^2 - 7w - 15 = 0 (4w+5)(w-3) = 0 w = 3 or -5/4
aside from x=120, the other values are hard to find.
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