a commercial jet airplane climbs at takeoff with slope m=4/9. How far in the horizontal direction will the airplane fly to reach am altitude of 20,000 ft above the takeoff point?

[tex]\bf slope = {{ m}}= \cfrac{rise}{run}\implies \cfrac{\textit{vertical direction}}{\textit{horizontal direction}}\implies \cfrac{4}{9}=\cfrac{20000}{h}[/tex]

solve for "h".

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jainveenamrata jainveenamrata · Answer 2 · 2022-08-17T08:15:57+02:00

The horizontal direction in will the airplane flies to reach an altitude of 20,000 ft above the takeoff point will be 45,000 feet.

What is a linear equation?

A relationship between two or more parameters that, when shown on a graph, produces a linear model. The degree of the variable will be one.

The linear equation is given as,

y = mx + c

Where m is the slope of the line and c is the y-intercept of the line.

A commercial jet airplane climbs at takeoff with slope m = 4/9.

Then the equation of the line will be given as,

y = (4/9)x

The y-intercept is zero.

The horizontal direction will the airplane fly to reach an altitude of 20,000 ft above the takeoff point will be

20,000 = (4/9)x

180,000 = 4x

x = 180,000 / 4

x = 45,000 feet

The horizontal direction in will the airplane flies to reach an altitude of 20,000 ft above the takeoff point will be 45,000 feet.

More about the linear equation link is given below.

https://brainly.com/question/11897796

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