A 5.8-ft-tall person walks away from a 9-ft lamppost at a constant rate of 3.4 ft/sec. What is the rate that the tip of the person's shadow moves away from the lamppost when the person is 11 ft away from the lampost

9.56 ft/sec

Step-by-step explanation:

We are told that a 5.8-ft-tall person walks away from a 9-ft lamppost at a constant rate of 3.4 ft/sec.

I've attached an image showing  triangle that depicts this;

Thus; dx/dt = 3.4 ft/sec

From the attached image and using principle of similar triangles, we can say that;

9/y = 5.8/(y - x)

9(y - x) = 5.8y

9y - 9x = 5.8y

9y - 5.8y = 9x

3.2y = 9x

y = 9x/3.2

dy/dx = 9/3.2

Now, to find how fast the tip of the shadow is moving away from the lamp post, it is;

dy/dt = dy/dx × dx/dt

dy/dt = (9/3.2) × 3.4

dy/dt = 9.5625 ft/s ≈ 9.56 ft/sec

step-by-step explanation:

well if you round, things become much more simple. round 5.95 to 6, 4.25 to 4, and 1.05 to 1. now, add 6+4+1 and that would equal 11. so the items together would equal to about $11.