Posted by **jasort20** on Sunday, March 18, 2007 at 7:39pm.

help

Probblem #6

solve by multiplying through with the common denominator

(11)/(3)x + (1)/(6) = (8)/(3)x + (19)/(6)

Problem #20

the perimeter of a rectangle is to be no greater than 300in. and the length must be 125in. write an inequality representing the maximum perimeter.

i know that the perimeter is length times width...

#6. First, find the common denominator.

(22/6)X + 1/6 = (16/6)X + 19/6

(22/6)X - (16/6)X = 19/6 - 1/6

You should be able to take it from here.

#20.

*Area* — NOT perimeter — is length times width. Perimeter is obtained by adding the length of all the sides.

300 > 2(125) + 2X, where X = width

I hope this helps. Thanks for asking.

## Answer This Question

## Related Questions

- MATH-HELP - 1.THE PERIMETER OF A RECTANGLE IS TO BE NO GREATER THAN 300IN.,AN ...
- maths - the length of a rectangle is 3 less than 5 times its width. Write a ...
- Algebra - Can you please review to see if I am on the right track with this ...
- Math - The length of a rectangle is 5 cm greater than it width. The perimeter is...
- math,correction please.. - (x)/(x-2) - (x+1)/(x) = (8)/(x^2-2x) I keep getting x...
- math/algebra - 1) The length of one of the equal legs of an isisceles triangle ...
- basic geometric - 10. a rectangle has width the same as a side of a square whose...
- Math 116 - The length of a rectangle is fixed at 23cm What lengths will make the...
- math - Can someone check over these problems for me? 1. The width of a rectangle...
- College Algrebra Math 133 unti 1 IP 1 - Suppose the width of a rectangle is 5 ...

More Related Questions