Question

IF YOU ANSERW MY MATH QUESTIONS ILL GIVE ADTINAL 50) POINTS FOR each question and crown evry question.

A scientist uses a submarine to study ocean life.
*She begins at sea level, which is an elevation of 0 feet.
*She descends for 91 seconds at a speed of 2.2 feet per second.
*She then ascends for 32 seconds at a speed of 0.7 feet per second.

At this point, what is her elevation, in feet?

Answer 1

220.4 feet below sea level

Explanation:

use the following formula:

rate x time = distance

(rate going down)(time going down) = distance going down.

so if she starts at 0 feet above sea level, and descends for 91 seconds at a speed of 2.2 ft/sec. you would plug that in as:

(2.2 ft/sec)(91 seconds) = distance of 198 feet.

she then ascends for 32 seconds at a speed of 0.7 feet/second.

so, substitute that;

(0.7 ft/sec)(32sec) = distance of 22.4 feet.

distance traveled downward + distance traveled upward = total distance of location.

198 + 22.4 feet = 220.4 feet

Answer 2

-177.8 feet below sea level.

Explanation:

First, we know that descend means to lower. Anything lower than 0 has to be negative. It says that she descends for 91 seconds at a speed of 2.2 feet per second. So first to find out how many feet she descended in those 91 seconds, you need to multiply 91*-2.2= -200.2. Then, it says that she ascends for 32 seconds at the speed of 0.7 feet per second. It's the same thing, but ascend means to go up. So, we multiply 32*0.7=22.4. Now we add the two numbers together to get -177.8.