distributive (20x5)+(4x5) = (20+4)x5 = 24x5
yes, adding c to the function shifts the graph vertically by that amount.
Calculus kinda with the area of two curves
the two curves intersect where x^2-c^2 = c^2-x^2 that is, where x=±c The area is thus ∫[-c,c] (c^2-x^2)-(x^2-c^2) dx = 4∫[0,c] c^2-x^2 dx = 4(c^2 x - 1/3 x^3)[0,c] = 4(c^3 - 1/3 c^2) = 8/3 c^3 8/3 c^3 = 270 8c^3 = 30*3^3 c = 3/2 ∛30
f = k(x-a) 1/f = 1/k(x-a) so the vertical asymptote is x=a similarly for the quadratic. the asymptotes are at x=a and x=b, where the denominator is zero.
if the buildings are of height k and p, k/46 = tan40° p/39 = tan50° now just evaluate k and p, take the difference
Since the semi-diagonals form an isosceles triangle with the sides of the rectangle, the diagonals make an angle of 60° with the short side, and 30° with the long side. length = 24cos30° height = 24sin30° p = 2(length+width)
#1. ok #2. I get 160 -- better check again #3. solve for t when h=0 #4. 180 + 20t - 16t^2
12 = mu * 24g
assuming a constant volume v, we have v = pi r^2 h so, h = v/(pi r^2) the surface area is a = 2pi r(r+h) = 2pi r(r + v/(pi r^2)) = 2 pi r^2 + 2v/r da/dr = 4pi r - 2v/r^2 = 2(2pi r^3 - v)/r^2 we want da/dr=0 for max/min area, so r = ∛(v/(2pi)) h = v/(pi r^2) = ∛(4v/...