SOLUTION: I have the correct answer of 13x+210y- 273=0,17.8 I just don't know how to achieve this answer, please help... Its a long one! An airliner began its final descent from an altitu

Question 309286: I have the correct answer of 13x+210y- 273=0,17.8
I just don't know how to achieve this answer, please help...
Its a long one!
An airliner began its final descent from an altitude of 1.3 miles six minutes before landing. During the descent its speed with respect to the ground averaged 210 miles per hour.
a) What is the equation for the altitude during descent relative to the airport? Assume the relationship is linear. Place your orgin directly under the plane on the ground at the point where it begins its descent. Use y for altitude and x for distance from the airport. Express the expression in standard form with integer coefficients simplified as much as possible.
b) How many miles was it from the airport when the altimeter read 1.1 miles? Enter answer to nearest 10th of a mile. Thank you so much!!
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
Plane is traveling at 210 miles per hour.
Divide 210 by 60 and you get 3.5 miles per minute.
The plane takes 6 minutes to land.
This means that in 6 minutes, the plane will drop from an altitude of 1.3 miles to an altitude of 0.
This means that the plane is dropping (1.3/6) miles per minute.

Since the plane's speed is 3.5 miles per minute, this means that the plane is dropping (1.3/(6*3.5)) miles of altitude per miles of distance from the airport.

The equation for the rise in altitude is equal to:

y = ((1.3)/(6*3.5))* x

A graph of this equation looks like this:

I needed to use a rise in altitude rather than a drop in altitude because you wanted the y-axis to reflect the altitude and the x-axis to reflect the distance from the airport.

You read the graph as follows:

y is the altitude of the plane.
x is the distance from the airport.

When x = 0, the altitude is 0 because the plane has landed.

When x = 21 miles form the airport, the altitude is 1.3 miles high.

You can see that by tracing a vertical line from the intersection of the equation of the plane's altitude with the line at y = 1.3 miles to see that the intersection occurs at about x = 21.

To determine the distance from the airport when the altitude is 1.1 miles, you would calculate the equation when y = 1.1

The equation is y = ((1.3)/(6*3.5))*x

Set y equal to 1.1 to get:
1.1 = ((1.3)/(6*3.5))*x
Solve for x to get:

x = (1.1*6*3.5)/(1.3) which becomes:

x = 17.769230977

The altitude of the plane is 1.1 miles when the plane is 17.769230977 miles from the airport.

The graph is shown again below with a horizontal line at y = 1.1.

Trace a vertical line from the intersection of this line with the graph of the equation of the plane's altitude to see that the value of x is somewhere around 17.8 when this occurs, as it should.

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