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

Algebra ->  Customizable Word Problem Solvers  -> Travel -> 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      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



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) About Me  (Show Source):
You can put this solution on YOUR website!
It would be better to post the problems separately.
---------------
5.The Length Of A Rectangle Is Twice The Width. The Perimeter Is 48 Inches.Find The Dimension Of The Rectangle.
---
Equations:
L = 2W
48 = 2(L+W)
----
Substitute for "L" and solve for "W":
48 = 2(2W+W)
48 = 2(3W)
48 = 6W
Width = 8 inches
---
Solve for "L":
L = 2W
L = 2*8 = 16 inches
-------------------------
===============

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?
---
The two cars are separating at a rate of (60+50) = 110 mph
---
Distance = rate*time
330 = 110*t
t = 330/110 = 3 hrs
---
The cars will be 330 miles apart at 10:00AM + 3 hrs = 1:00PM
====================
Cheers,
Stan H.
============