A balloon is 200 ft. off the ground and rising vertically at the constant rate of 15 ft/sec. An automobile passes beneath it traveling along a straight road at the constant rate of 45 mph or 66 ft/sec. How fast is the distance between them changing 1 sec. later?

After 1 second has passed, the balloon will be 215 ft above the ground, and the automobile will be 66 ft away from its initial position directly below the balloon, forming a right triangle whose hypotenuse has length ft.

The distance between the balloon and automobile is related to the height of the balloon and the distance the automobile has moved away from its starting position by the Pythagorean formula,

Differentiate both sides with respect to time :

Plug in everything you know and solve for the rate you want:

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

17   x 256=4352

sorry if this is wrong, i'm new.

step-by-step explanation: