5(r+5) is.
13
a(r) = 2pi*r^2 and d = 2r 5 = 2r => r = 5/2 a(r) = 2pi*(5/2)^2 a(r) = 2pi*25/4 a(r) = 25pi/2 or, in decimal form, 39.2699 millimetres2
(r + 5)(r + 5)
one fifth of r
5(r+5) is.
13
r2 + r - 20 = 0(r + 5)(r - 4) = 0r + 5 = 0 or r - 4 = 0r = -5 or r = 4
6
Moment of inertia of hollowsphere = 2/5 M(R^5-r^5)/(R^3-r^3)
No, they are not.
if: r = 5z 15z = 3y then: z = y/5 r = 5(y/5) r = y
factor out the common monomial, 5. 5(r^3-1) Factor as a difference of cubes. 5(r-1)(r^2+r+1)
5*[sqrt(r)]5 = 27 => r = 1.9632, approx.
a(r) = 2pi*r^2 and d = 2r 5 = 2r => r = 5/2 a(r) = 2pi*(5/2)^2 a(r) = 2pi*25/4 a(r) = 25pi/2 or, in decimal form, 39.2699 millimetres2
(r + 5)/b
4 -5 on the r ax = -1