A street light is at the top of a 12 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 5 ft/sec along a straight path. How fast is the tip of her shadow moving along the ground when she is 50 ft from the base of the pole?

10ft/sec

Step-by-step explanation:

In t sec the lady would be at a distance of (speed)(time) = (5 ft/sec)(t sec) = 5t ft.

As the height of the pole is double the height of the lady, from similar triangles, we can deduct the tip of the shadow would be at a distance x = 2 (5t) ft = 10t.

The distance equation is x = 10t ft.

So, the speed is x' = 10 ft/sec.

The speed will be constant like the speed of the lady.

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)  

the surface area is 124cm²

step-by-step explanation:

length: 4cm

width: 9cm

height:   2cm