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tresset_1 [31]
3 years ago
10

A= x^2 -30x+225 The polynomial represents the area (in square feet) of the square playground. a. Write a polynomial that represe

nts the side length of the playground. b. Write an expression for the perimeter of the playground. (I factored it and got (x-15)^2)
Mathematics
1 answer:
Setler [38]3 years ago
7 0

Answer:

a) The polynomial is equivalent to A = (x-15)^{2}, and the side length of the playground is x -15 feet.

b) The perimeter of the playground, measured in feet, is p = 4\cdot (x-15).

Step-by-step explanation:

a) By Geometry, we know that area of the square is equal to the square of the side length. If A = x^{2}-30\cdot x +225 is the area of the square, then must be a perfect square trinomial, that is:

(x-a)^{2} = x^{2}-2\cdot a\cdot x + a^{2} (1)

Then, we must observe the following properties:

-2\cdot a = -30 (2)

a = 15

a^{2} = 225 (3)

a = 15

Therefore, the polynomial is equivalent to A = (x-15)^{2}, and the side length of the playground is x -15 feet.

b) The perimeter of a square equals four times the side length. That is:

p = 4\cdot (x-15)

The perimeter of the playground, measured in feet, is p = 4\cdot (x-15).

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x=\frac{12}{7} \\y=\frac{12}{5} \\z=-12

Step-by-step explanation:

Let's re-write the equations in order to get the variables as separated in independent terms as possible \:

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\frac{xy}{x+y} =1\\xy=x+y\\1=\frac{x+y}{xy} \\1=\frac{1}{y} +\frac{1}{x}

Second equation:

\frac{xz}{x+z} =2\\xz=2\,(x+z)\\\frac{1}{2} =\frac{x+z}{xz} \\\frac{1}{2} =\frac{1}{z} +\frac{1}{x}

Third equation:

\frac{yz}{y+z} =3\\yz=3\,(y+z)\\\frac{1}{3} =\frac{y+z}{yz} \\\frac{1}{3}=\frac{1}{z} +\frac{1}{y}

Now let's subtract term by term the reduced equation 3 from the reduced equation 1 in order to eliminate the term that contains "y":

1=\frac{1}{y} +\frac{1}{x} \\-\\\frac{1}{3} =\frac{1}{z} +\frac{1}{y}\\\frac{2}{3} =\frac{1}{x} -\frac{1}{z}

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Now we use this value for "x" back in equation 1 to solve for "y":

1=\frac{1}{y} +\frac{1}{x} \\1=\frac{1}{y} +\frac{7}{12}\\1-\frac{7}{12}=\frac{1}{y} \\ \\\frac{1}{y} =\frac{5}{12} \\y=\frac{12}{5}

And finally we solve for the third unknown "z":

\frac{1}{2} =\frac{1}{z} +\frac{1}{x} \\\\\frac{1}{2} =\frac{1}{z} +\frac{7}{12} \\\\\frac{1}{z} =\frac{1}{2}-\frac{7}{12} \\\\\frac{1}{z} =-\frac{1}{12}\\z=-12

8 0
3 years ago
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