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Sagot :
The given problem is about rectangle areas.
The area of a rectangle is defined as
[tex]A=w\cdot l[/tex]Where w is width and l is the length.
In this case, we know that the length is 5 yards more than twice the width, where the last one is expressed as x.
[tex]l=2x+5,w=x[/tex]We also know by given that
[tex]A=462yd^2[/tex]Using all the given information, we have
[tex]462=x(2x+5)[/tex]Where we solve for x, first, we apply the distributive property.
[tex]462=2x^2+5x[/tex]Now, we move all terms to one side only
[tex]2x^2+5x-462=0[/tex]Using a calculator, we have
[tex]x_1=14,x_2=-\frac{33}{2}[/tex]Where 14 represents the width because distances cannot be expressed by negative numbers.
[tex]l=2(14)+5=28+5=33[/tex]Therefore, the width of the rectangle is 14 yards, and its length is 33 yards.
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