SOLUTION: The width of a rectangle is 53 yd. Find all possible lengths such that the perimeter of the rectangle will be atleast 486yd.

Question 679931: The width of a rectangle is 53 yd. Find all possible lengths such that the perimeter of the rectangle will be atleast 486yd.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The width of a rectangle is 53 yd. Find all possible lengths such that the perimeter of the rectangle will be atleast 486yd.
---------
Perimeter = 2(length + width)
486 <= 2(x + 53)
x+53 >= 243
----
length >= 190
---------
Cheers,
Stan H.
=========
RELATED QUESTIONS
Please help me on my problem: The width of a rectangular ballfield is 96 yd. Find all... (answered by longjonsilver)

I would appreciate some help with my homework problem # 33. Solve. The width of a... (answered by Cintchr)

If a rectangle's length is 4 meters more than its width, find all possible lengths so... (answered by josgarithmetic)

the perimeter of a rectangle is 36 yd. The width is 18 yd less than twice the length.... (answered by mandi08)

the perimeter of a rectangle is 24 units. Find all possible integet lengths and widths (answered by Jc0110,josgarithmetic)

The area of a certain rectangle is 288 yd the perimeter is 68 yd if u double the width... (answered by josgarithmetic)

The length of a rectangle is 6 yd longer than its width. If the perimeter of the... (answered by CharlesG2)

The length of a rectangle is 3 yd longer than its width. If the perimeter of the... (answered by rapaljer)

The length of a rectangle is 2 yd longer than its width. If the perimeter of... (answered by IBMath)