RudyLopez
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☻-------SOLVE------☺.The area of a playground is 160 yd^2. The width of the playground is 6 yd longer than its length. Find the length and width of the playground.

A.) length = 22 yd, width = 16 yd

B.) length = 16 yd, width = 10 yd

C.) length = 16 yd, width = 22 yd

D.) length = 10 yd, width = 16 yd

Respuesta :

i think the answer is D
calculista

Answer:

Option D.) [tex] length=10\ yd[/tex], [tex] width=16\ yd[/tex]

Step-by-step explanation:

we know that

The area of a rectangle (playground) is equal to

[tex]A=LW[/tex]

we have

[tex]A=160\ yd^{2}[/tex]

so

[tex]160=LW[/tex] -----> equation A

[tex]W=L+6[/tex] -----> equation B

substitute equation B in equation A and solve for L

[tex]160=L(L+6)[/tex]

[tex]L^{2}+6L-160=0[/tex]

using a graphing tool

solve the quadratic equation

see the attached figure

The solution is

[tex]L=10\ yd[/tex]

Find the value of W

[tex]W=L+6[/tex] -----> [tex]W=10+6=16\ yd[/tex]

Ver imagen calculista

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