(3) Two people are at an elevator. At the same time one person starts to walk away from the elevator at a rate of 2 ft/sec and the other person starts going up in the elevator at a rate of 7 ft/sec. What rate is the distance between the two people changing 15 seconds later

The rate at which the distance between the two people is changing 15 seconds later is 7.28 ft/sec.

Step-by-step explanation:

Here, we note the following

Let the person walking way fom the elevator be X

Let the other person going up in the elevator be Y

Therefore after 15 seconds, their positions will be;

For X,  2 ft/sec × 15 s = 30 ft away from the elevator

For Y,  7 ft/sec × 15 s = 105 ft up in the elevator

At that instant, the distance between them is given as

d² = x² + y²

= 30² +105² = 11925 ft²

d = √11925 ft²  = 109.202 ft

The rate of change of the distance between the two peopl, X and Y is given as

Therefore,   or

Since is given as 2 ft/sec and = 7 ft/sec

Then  

50-60=10

step-by-step explanation:

answer: it is 8

step-by-step explanation:

you know why its 8