HELP PLS “geometry”
1 answer:
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Answer:
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
