A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from
<span>the pole with a speed of 5 ft/s along a straight path. How fast is the tip of his shadow moving when </span>
<span>he is 40 ft from the pole? </span>
<span>Solution: USE SIMILAR TRIANGLES. Let d be the distance between the man and the pole and let x be the length of the man's shadow. Then x=(d + x) = 6=15, so 15x = 6x + 6d and 9x = 6d. Then </span>
<span>x = (2=3)d. The problem wants us to find d(d + x)=dt = d(d)=dx + dx=dt, since this is the speed </span>
<span>of the tip of his shadow. We know dx=dt = (2=3)d(d)=dt and we know d(d)=dt = 5 ft/s, from the </span>
<span>problem. Then d(d + x)=dt = (5=3)d(d)=dt = 25=3 ft/s.</span>
Answer:
ooof some ppl are salty man idek
Step-by-step explanation:
Answer:
120 unit^2.
Step-by-step explanation:
The lateral area consists of triangles so the required area is
1/2 * perimeter * slant height.
= 1/2 * 24 * 10
= 120 unit^2.
