Question 362946: A jet flew 1200 miles with a tailwind of 50 mph. The tailwind changed to 20 mph for the remaining 520 miles of the flight. The total time of the flight was 3 hours. Find the speed of the jet relative to the ground. Please help me out here, and if you dont mind including the step by step process that would be great!
Answer by checkley77(12844)   (Show Source):
You can put this solution on YOUR website! D=RT
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1,200=(R+5O)T
T=1,200/(R+50)
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520=(R+20)(3-T)
(3-T)=520/(R+20)
-T=520/(R+20)-3
T=-520/(R+20)+3
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1,200/(R+50)=-520/(R+20)+3
1,200/(R+50)+520/(R+20)=3
(1,200(R+20)+520(R+50)/(R+50)(R+20)=3
(1,200R+24,000+520R+26,000)/(R^2+70R+1,000)=3
3(R^2+70R+1,000)=1,720R+50,000
3R^2+210R+3,000-1,720R-50,000=0
3R^2-1,510R-47,000=0
R=(1,510+-SQRT(-1,510^2-4*3*-47,000])/2*3
R=(1,510+-SQRT[2,280,100+564,000])/6
R=(1,510+-SQRT2,844,000)/6
R=(1,510+-1,686.45)/6
R=(1,510+1,686.45)/6
R=3,196.45/6
R=532.74 MPH IS THE SPEED OF THE PLANE IN STILL AIR.
1,200/(532.74+50)+520/(532.74+20)=3
1,200/582.74+520/552.74=3
2.06+.94=3
3=3
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