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Thread: HELP with INEQUALITIES

  1. #1
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    HELP with INEQUALITIES

    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 <
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  2. #2
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    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: .$\displaystyle P \:=\:2L + 2W$

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

    The perimeter is between 170 and 230 inches.

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

    Therefore, the length $\displaystyle L$ 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 $\displaystyle x$ = side of the square.
    The perimeter of a square is: .$\displaystyle 4x$

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

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

    The length of the side is between 11 and 22 meters.


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

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


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

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

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