(5 points) A street light is at the top of a 18 ft pole. A 5 ft tall girl walks along a straight path away from the pole with a speed of 6 ft/sec.
At what rate is the tip of her shadow moving away from the light (ie, away from the top of the
pole) when the girl is 22 ft away from the pole?

2.31 ft/sec

Step-by-step explanation:

In the picture attached, a representation of the problem is shown, where w(t) is the walking distance (as a function of time) and s(t) is the position of the tip of her shadow (as a function of time).

We want to find the derivative of s(t).

From triangles similarity:

s(t)/5 = [w(t) + s(t)]/18

18s(t) = 5w(t) + 5s(t)

13s(t) = 5w(t)

w(t) = 13/5*s(t)

We know that her speed is 6 ft/sec, that is:

d(w(t))/dt = 6 ft/sec

From the previous relationship:

d(w(t))/dt = 13/5*d(s(t))/t

Replacing:

6 = 13/5*d(s(t))/t

d(s(t))/t = 5*6/13

d(s(t))/t ≈ 2.31 ft/sec

the distance between point a and point b is 6 units

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.