My Notes 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 4 ft/s along a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole?

6.667 ft/s

Step-by-step explanation:

Given

Height of street light = 15ft

Height of man = 6ft

Speed away from pole = 4ft/s

Let x represents the distance between the man and the pole, and Let y represents the distance between the tip of the man's shadow to the pole.

This forms a similar triangle (see attachment below).

From similar triangles, we have

(y - x)/6 = y/15 ----- Solve equation

15(y - x) = 6 * y

15y - 15x = 6y ---- Collect Like Terms

15y - 6y = 15x

9y = 15x --- divide through by 9

y = 15x/9

y = 5x/3 ---- differentiate with respect to time, t

dy/dt = 5/3 dx/dt

dx/dt represents rate of change of distance per time = 4ft/s

while dy/dt represents rate of movement of his shadow tips

dy/dt = 5/3 * 4

dy/dt = 20/3 = 6.667 ft/s

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!

the answer will be b x² +x-1

step-by-step explanation: