A plane is flying at 32, 460 feet. To avoid some turbulence, the plane ascends 1645 feet. It then descends 3/5 of its new height to prepare for landing. What are the heights that the plane will be at after it ascends to avoid turbulence and after it descends to prepare for landing? Select all that apply

A) 20, 463

B) 12, 326

C) 34, 105

D) 30, 814

E) 13, 642

F) 18, 489

Question

Mathematics

Piper Farrell

1 February, 11:58

A plane is flying at 32, 460 feet. To avoid some turbulence, the plane ascends 1645 feet. It then descends 3/5 of its new height to prepare for landing. What are the heights that the plane will be at after it ascends to avoid turbulence and after it descends to prepare for landing? Select all that apply

A) 20, 463

B) 12, 326

C) 34, 105

D) 30, 814

E) 13, 642

F) 18, 489

+5

Answers (1)

Know the Answer?

Peter Little · Answer 1

Answer: C and E

Step-by-step explanation:

The plane starts at 32,460 feet.

It then ascends 1645 feet. Ascending means it's going up higher. So we add.

32,460 + 1645 = 34,105 (C)

From here (34,105 feet) the plane descends by 3/5 of it's height. Descending means it's lowering. So we subtract.

But first we have to figure out how much we are subtracting.

3/5 is equal to. 6

Multiply to find 3/5 of current height.

34,105 *.6 = 20,463

So the plane descends by 20,463 feet.

34,105 - 20,463 = 13,642 feet (E)