HELP with INEQUALITIES

**epetrik** · October 7th 2008, 12:29 PM

I'm having issues figuring out an equation to figure this out...any help would be greatly appreciated!

Applications The perimeter of a rectangle is to be between 170 and 230 inches. Find the range of values for its length when its width is 45 inches.

< length <

Applications The perimeter of a square is to be from 45 meters to 90 meters. Find the range of values for its area.

< area <

**Soroban** · October 7th 2008, 02:36 PM

Hello, epetrik!

The perimeter of a rectangle is to be between 170 and 230 inches.
Find the range of values for its length when its width is 45 inches.

We know that: . $P \:=\:2L + 2W$

Since $W = 45$ , we have: . $P \:=\:2L + 90$

The perimeter is between 170 and 230 inches.

. . $\begin{array}{cccccc}<br /> \text{So we have:} & 170 & < & 2L + 90 & <& 230 \\ \\[-3mm]<br /> \text{Subtract 90:} & 80 &<& 2L &<& 140 \\ \\[-3mm]<br /> \text{Divide by 2:} & 40 &<& L &<& 70 \end{array}$

Therefore, the length $L$ must be between 40 and 70 inches.

The perimeter of a square is to be from 45 meters to 90 meters.
Find the range of values for its area.

Let $x$ = side of the square.
The perimeter of a square is: . $4x$

The perimeter is between 45 and 90: . $45 \;<\;4x \;<\;90$

Divide by 4: . $\frac{45}{4} \;<\:x\;<\;\frac{45}{2}$

The length of the side is between 11¼ and 22½ meters.

The area of the square is: . $x^2$

So we have: . $\left(\frac{45}{4}\right)^2 \;<\;x^2 \;<\:\left(\frac{45}{2}\right)^2$

Therefore: . $\frac{2025}{16} \;<\;\text{Area} \;<\;\frac{2025}{4}$

The area is between: . $126\frac{9}{16}\,\text{ and }\,506\frac{1}{4}\text{ square meters.}$

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