Friday

October 24, 2014

October 24, 2014

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

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.

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

I hope this helps. Thanks for asking.

**Answer this Question**

**Related Questions**

math,correction please.. - (x)/(x-2) - (x+1)/(x) = (8)/(x^2-2x) I keep getting x...

Algebra 1 - Solve the equation by multiplying each side by the least common ...

2nd grade math - If you have a rectangle with 24 square units 3 lines down 8 ...

Ratios and proportions - RATIOS AND PROPORTIONS x 3 - = - 4 2 The boards don't ...

algebra - Solve for x x/6-x/8 equals 1 Do fraction busters Find the least common...

math - The rectangle given has a perimeter of 14 units. Find the value(s) of x ...

Pre-Algebra - If the perimeter of a rectangle is 38 ft and the width is 9 ft, ...

College Algrebra Math 133 unti 1 IP 1 - Suppose the width of a rectangle is 5 ...

Algebra - 2 more questions... ^^ - Thank you for helping on the other problem, ...

Calculus - 1/a+1 + 1/a-1 I solved this out by multiplying the (a+1) by (a-1) and...