One airplane is at 7,000 feet and is descending 50 feet every second. Another plane is at 1,000 feet and is climbing at a rate of 70 feet per second, How long will it take for the
planes to be at the same heigh

Given:

One airplane is at 7,000 feet and is descending 50 feet every second.

Another plane is at  1,000 feet and is climbing at a rate of 70 feet per second.

To find:

The time taken by planes to reach at the same height.

Solution:

Let x be the number of seconds after which the planes be at the same height.

One airplane is at 7,000 feet and is descending 50 feet every second.

Change in 1 second = -50 feet   (Here negative sign means decrease)

Change in x second = -50x feet

Height of first airplane after x seconds is

          ...(i)

Another plane is at  1,000 feet and is climbing at a rate of 70 feet per second.

Change in 1 second = 70 feet

Change in x second = 70x feet

Height of first airplane after x seconds is

    ...(ii)

Equate (i) and (ii), to find the time after which both planes are on the same height.

Divide both sides by 120.

Therefore, both planes are at the same height after 50 seconds.

a) 32pie square meters

step-by-step explanation:

answer is 8x+24

area of vanessa's patio = length × width

given, length = (3x² + 5x + 10)

width = (x² - 3x - 1)

area of vanessa's patio = (3x² + 5x + 10)(x² - 3x - 1)

= 3x²(x² - 3x - 1) + 5x(x² - 3x - 1) + 10(x² - 3x - 1)

= 3x⁴ - 9x³ - 3x² + 5x³ - 15x² - 5x + 10x² - 30x - 10

= 3x⁴ - 4x³ - 8x² - 35x - 10

hence, expression of vanessa's patio = (3x⁴ - 4x³ - 8x² - 35x - 10)