A pilot flew his single-engine airplane 6060 miles with the wind from City A to above City B. He then turned around and flew back to City A against the wind. If the wind was a constant 3030 miles per hour, and the total time going and returning was 1.31.3 hours, find the speed of the plane in still air.

the speed of the plane in still air is 101.2 mph

Step-by-step explanation:

Let x represent the speed of the plane in still air.

The pilot flew his single-engine airplane 60 miles with the wind from City A to above City B. If the wind was a constant 30 miles per hour, it means that the total speed at which he flew the plane while going is (x + 30) mph.

Time = distance/speed

Time spent while going is

60/(x + 30)

He then turned around and flew back to City A against the wind. it means that the total speed at which he flew the plane while returning is (x - 30) mph.

Time spent while returning is

60/(x - 30)

If the total time going and returning was 1.31.3 hours, it means that

60/(x + 30) + 60/(x - 30) = 1.3

Cross multiplying, it becomes

60(x - 30) + 60(x + 30) = 1.3(x - 30)(x + 30)

60x - 1800 + 60x + 1800 = 1.3(x² + 30x - 30x - 900)

120x = 1.3x² - 1170

1.3x² - 120x - 1170 = 0

The general formula for solving quadratic equations is expressed as

x = [- b ± √(b² - 4ac)]/2a

From the equation given,

a = 1.3

b = - 120

c = - 1170

Therefore,

x = [- - 120 ± √(- 120² - 4 × 1.3 × - 1170)]/2 × 1.3

x = [120 ± √(14400 + 6080)]/2.6

x = [120 ± √20480]/2.6

x = (120 + 143.1)/2.6 or x = (120 - 143.1)/2.6

x = 101.2 or x = - 8.9

Since the speed cannot be negative, then x = 101.2 mph

the final function for the graph becomes x-2=-1. before graphing you solve the equation, where x=1. then graph accordingly.   ( on the x-axis, 1 on the line will possess a vertical line.)

answer: 2 and 1

step-by-step explanation: got this question