SOLUTION: an insurance representative traveled 735 mi by commercial jet and then an additional 105 mi by helicopter. The rate of the jet was four times the rate of the helicopter. the entire
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Question 78653This question is from textbook intermediate algebra w/appl
: an insurance representative traveled 735 mi by commercial jet and then an additional 105 mi by helicopter. The rate of the jet was four times the rate of the helicopter. the entire trip took 2.2 hours. Find the rate of the jet. This question is from textbook intermediate algebra w/appl
You can put this solution on YOUR website! an insurance representative traveled 735 mi by commercial jet and then an additional 105 mi by helicopter. The rate of the jet was four times the rate of the helicopter. the entire trip took 2.2 hours. Find the rate of the jet.
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Jet DATA:
distance=735 mi ; rate = 4x mi/hr ; time=d/r = 735/4x hrs
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Helicopter DATA:
distance= 105 mi ; rate = x mi/hr ; time=d/r= 105/x hrs
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EQUATION:
time + time = 2.2 hrs
Note: 2.2(10/10)=22/10
735/4x + 105/x = 22/10
Note: the least common multiple of 4x, x, and 10 is 20x.
Multiply thru by 20x to get:
5*735 + 2100 = 44x
x=131.25 mph (speed of helicopter)
4x=525 mph (speed of jet)
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Cheers,
Stan H.