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the perimeter of a rectangle can be found by using the formula P=21+2w. What is the width of the rectangle if the perimeter is 100 ft and the length is 50 ft?

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\begin{gathered} P=2l+2w \\ \text{If l=50 and P=100, then:} \\ 100=2(50)+2w\Rightarrow0=2w\text{ !!!!} \end{gathered}

Since this can't happen, let's suppose that P=150, then:


\begin{gathered} P=2l+2w \\ \Rightarrow150=2(50)+2w \\ \Rightarrow150=100+2w \end{gathered}

Now we have to move the 100 to the other side of the equation with a negative sign to get this:


\begin{gathered} 150=100+2w \\ \Rightarrow150-100=2w \\ \Rightarrow50=2w \end{gathered}

Finally, to get w, we move the 2 that's multiplying to the other side dividing the 50:


\begin{gathered} 50=2w \\ \Rightarrow(50)/(2)=w \\ \Rightarrow w=25 \end{gathered}

Therefore, the width of the rectangle would be 25 ft if the perimeter is 150ft, and we can see how the rectangle would look:

the perimeter of a rectangle can be found by using the formula P=21+2w. What is the-example-1
User Nick Retallack
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