Moutain climber A is at altiude of 250 feet and is descending at a rate of 30 feet per hour. Moutain climber B is at an altitude of 400 feet and is descending at a rate of 50 feet per hour. In how many hours will the altitude of moutain climber B be lower than the altitude of moutain climber A?

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ayfat23 ayfat23 · Answer 1 · 2020-12-01T15:35:37+01:00

Answer:

the altitude of mountain climber b will be lower than the altitude of mountain climber A with number of hours greater than 7.5hrs.

Step-by-step explanation:

This talks about equalities in math,

We are told about Moutain climber A is at altiude of 250 feet and is descending at a rate of 30 feet per hour, this can be interpreted as

250 - 30h

Moutain climber B is at an altitude of 400 feet and is descending at a rate of 50 feet per hour, this can be interpreted as

400 - 50h

We were now asked how many hours the altitude of mountain climber b be lower than the altitude of mountain climber A, we can interpret this as

(400 - 50h) < (250 - 30h )

We can collect the like terms here

-50h + 30h < 250 - 400

-20h < - 150

The negatives sign can cancelled out, we have

Note that multiplying or dividing by a negative THE INEQUALITY SIGN MUST BE FLIPPED

20h > 150

h > 150/20

h > 7.5

Therefore, the altitude of mountain climber b will be lower than the altitude of mountain climber A with number of hours greater than 7.5hrs.