40 pts! and brainliest submarine a is positioned at –1,100 feet in relation to sea level. submarine b is positioned at –800 feet in relation to sea level. each submarine needs to be 200 feet below sea level in 30 minutes. which compares the speeds at which each submarine needs to travel?

submarine a must travel at a rate of (blank) ft/min, and submarine b must travel at a rate of (blank) ft/min.

43.33

33.33

Step-by-step explanation:

Speed = distance/time

Now, submarine A is presently 1100ft away from sea level and still needs to cover a distance of 200ft below sea level. The total distance it needs to cover is 1100 + 200 = 1300ft

Submarine B is at a distance of 800ft above sea level and still needs to go 200ft below. Total distance needed to be covered is 800 + 200 = 1000ft.

The speed of submarine A = 1300/30 = 43.33ft/min

The speed of submarine B = 1000/30 = 33.33ft/min

Submarine A must travel at a rate of StartFraction negative 1100 minus (negative 200) over 30 EndFraction = negative 30 ft/min, and Submarine B must travel at a rate of StartFraction negative 800 minus (negative 200) over 30 EndFraction = negative 20 ft/min.

Step-by-step explanation:

choice letter: D

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b

29933333.33

step-by-step explanation: