SOLUTION: Can I Ask More Than One??
5.The Length Of A Rectangle Is Twice The Width. The Perimeter Is 48 Inches.Find The Dimension Of The Rectangle.
6.At 10:00 A.M., A Car Leaves House At
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Question 464492: Can I Ask More Than One??
5.The Length Of A Rectangle Is Twice The Width. The Perimeter Is 48 Inches.Find The Dimension Of The Rectangle.
6.At 10:00 A.M., A Car Leaves House At A Rate Of 60 Mi/h. At The Same Time,Another Car Leaves The Same House At A Rate Of 50 Mi/h In The Opposite Direction.At What Time Will The Cars Be 330 Miles Apart?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
It would be better to post the problems separately.
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5.The Length Of A Rectangle Is Twice The Width. The Perimeter Is 48 Inches.Find The Dimension Of The Rectangle.
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Equations:
L = 2W
48 = 2(L+W)
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Substitute for "L" and solve for "W":
48 = 2(2W+W)
48 = 2(3W)
48 = 6W
Width = 8 inches
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Solve for "L":
L = 2W
L = 2*8 = 16 inches
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6.At 10:00 A.M., A Car Leaves House At A Rate Of 60 Mi/h. At The Same Time,Another Car Leaves The Same House At A Rate Of 50 Mi/h In The Opposite Direction.At What Time Will The Cars Be 330 Miles Apart?
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The two cars are separating at a rate of (60+50) = 110 mph
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Distance = rate*time
330 = 110*t
t = 330/110 = 3 hrs
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The cars will be 330 miles apart at 10:00AM + 3 hrs = 1:00PM
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Cheers,
Stan H.
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