SOLUTION: Charles Lindbergh and Don Hall calculated that at the begining of the transatlantic flight, the most economical air speed would be 98 mph which would get 1.2 miles per pound of fue

Algebra ->  Linear-equations -> SOLUTION: Charles Lindbergh and Don Hall calculated that at the begining of the transatlantic flight, the most economical air speed would be 98 mph which would get 1.2 miles per pound of fue      Log On


   

Question 366067: Charles Lindbergh and Don Hall calculated that at the begining of the transatlantic flight, the most economical air speed would be 98 mph which would get 1.2 miles per pound of fuel. At the end of the flight, with the plane almost out of fuel, the most economic air speed would be 62 mph, which he would get 2.3 miles per pound of fuel.
Algebraically, determine the linear function for this data.
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
Charles Lindbergh and Don Hall calculated that at the begining of the transatlantic flight, the most economical air speed would be 98 mph which would get 1.2 miles per pound of fuel. At the end of the flight, with the plane almost out of fuel, the most economic air speed would be 62 mph, which he would get 2.3 miles per pound of fuel.
Algebraically, determine the linear function for this data.
This problem does not state whether the linear function you are to find
is to be for 

1. the most economical airspeed as a function of miles per pound of fuel,

or whether the linear function you are to find is to be for 

2. miles per pound of fuel as a function of the most economical airspeed.

You'll have to ask your teacher which they want.  I'll do both


When his speed was 98, he would get 1.2 miles per pound of fuel.
When his speed was 62, he would get 2.3 miles per pound of fuel.

1. the most economical airspeed as a function of miles per pound of fuel,

This amounts to the problem:

Find the equation of a line through the points (1.2,98) and (2.3, 62)

m=%28y%5B2%5D-y%5B1%5D%29%2F%28x%5B2%5D-x%5B1%5D%29

m=%2862-98%29%2F%282.3-1.2%29+=+%28-36%29%2F%281.1%29+=+-360%2F11

y-y%5B1%5D=m%28x-x%5B1%5D%29

y-98=expr%28-360%2F11%29%28x-1.2%29

y-98=expr%28-360%2F11%29x%2B432%2F11%29

y=expr%28-360%2F11%29x%2B432%2F11%2B98%29

y=expr%28-360%2F11%29x%2B432%2F11%2B1078%2F11%29

y=expr%28-360%2F11%29x%2B1510%2F11%29

I did it in exact fractions, but you can do it in approximate
decimals, if you like.

In decimals rounded to hundredths:

y=-32.73x%2B137.27

Or substituting m miles per pound of fuel and S(m) for the most economical airspeed 

%22S%28m%29%22=expr%28-360%2F11%29m%2B1510%2F11%29

or

%22S%28m%29%22=-32.73m%2B137.27

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2. Miles per pound of fuel as a function of the most economical airspeed.


This amounts to the problem:

Find the equation of a line through the points (98,1.2) and (62,2.3)

m=%28y%5B2%5D-y%5B1%5D%29%2F%28x%5B2%5D-x%5B1%5D%29

m=%282.3-1.2%29%2F%2862-98%29

m+=+%281.1%29%2F%28-36%29+=+-ll%2F360

y-y%5B1%5D=m%28x-x%5B1%5D%29

y-1.2=expr%28-11%2F360%29%28x-98%29

y-1.2=expr%28-11%2F360%29x%2B539%2F180%29

y=expr%28-11%2F360%29x%2B539%2F180%2B1.2%29

y=expr%28-11%2F360%29x%2B539%2F180%2B12%2F10%29

y=expr%28-11%2F360%29x%2B539%2F180%2A72%2B12%2A18%2F180%29

y=expr%28-11%2F360%29x%2B539%2F180%2B216%2F180%29

y=expr%28-11%2F360%29x%2B755%2F1800

y=expr%28-11%2F360%29x%2B151%2F36

Again, I did it in exact fractions, but you can do it in 
approximate decimals also, if you like.

In decimals rounded to hundredths:



In decimals rounded to hundredths:

%22M%28s%29%22=-.03s%2B4.19

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Be sure to find out whether you wanted 1 or 2.

Edwin