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A=LW, and we are told that L=W+10

Using L=W+10 in A=LW you have:

A=(W+10)W

A=W^2+10W

W^2+10W=880, we are told that A=880 so

W^2+10W=880  now we can complete the square...

W^2+10W+25=905

(W+5)^2=905

W+5=±√905

W=-5±√905, since W>0

W=-5+√905  and since L=W+10

L=5+√905

W=-5+√905 ft and L=5+√905 ft

If we want an approximation:

W≈25.08 ft and L≈35.08 ft  (both dimensions to the nearest hundredth)
Lanuel

a. The length of the rectangular pool is equal to 35.08 feet

b. The width of the rectangular pool is equal to 25.08 feet.

  • Let the length of the rectangular pool be L.
  • Let the width of the rectangular pool be W.

Given the following data:

  • Area of rectangular pool = 880 square feet.

To find the dimensions of the rectangular pool:

Translating the word problem into an algebraic equation, we have;

[tex]L = 10+W[/tex]

Mathematically, the area of a rectangle is given by the formula;

[tex]A = LW[/tex]

Substituting the values into the formula, we have;

[tex]880 = (10+W)W\\\\880=10W+W^2\\\\W^2+10W-880=0[/tex]

Solving the quadratic equation by using the quadratic formula, we have:

[tex]W = \frac{-b\; \pm \;\sqrt{b^2 - 4ac}}{2a}[/tex]

W = 25.08 or -35.08 feet

For the value of L:

[tex]L = 10+W\\\\L=10+25.08[/tex]

L = 35.08 feet

Find more information: brainly.com/question/897975

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