A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 7 ft/s along a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole?

Dy/dt = 11.67 ft/s

Step-by-step explanation: See Annex

In Annex, Triangles ACE and BCD are similar then, we can write

( y- x ) / y  =  6 / 15

15*y -15*x  =  6*y

15*y - 6*y = 15*x

9*y = 15*x

y = (15/9)*x

Differentiating with respect time on both sides of the equation we get

Dy/dt = (15/9) Dx/dt  (1)

Where we know Dx/dt = 7 ft/s, and according to (1)  Dy/dt does not depend on x (distance between man and the pole, only depend on the speed f the man

Dy/dt = (15/9) * 7

Dy/dt = 11.67 ft/s

the answer is b i just did that

step-by-step explanation:

hello from mrbilldoesmath!

answer:

x - 3

discussion:

x^2 - 9 = (x-3) (x+3)   so

(x^2-9)/ (x + 3) = x-3

you,

mrb