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Masja [62]
2 years ago
8

The Johnsons’ new swimming pool measures 3x^2-2x+1 ft in length and x+2 ft in width. What is the area of the new swimming pool,

in square feet?
Mathematics
1 answer:
Usimov [2.4K]2 years ago
8 0

Answer:

3x^3 + 4x^2 -3x + 2

Step-by-step explanation:

The area of the swimming pool will be equal to length * width.

We know that the length is 3x^2-2x+1 and the width is x+2.

Therefore we can substitute those into the equation:

(3x^2-2x+1)(x+2)\\

Then, we expand the brackets using a grid (attached).

After this, we collect like terms to find the final answer:

3x^3+6x^2-2x^2-4x+x+2\\3x^3 + 4x^2 -3x + 2

Our final answer is: 3x^3 + 4x^2 -3x + 2

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X2 - 6x + 12 and y = 2x - 4, algebraically are
olga_2 [115]

<em><u>The solution is (4, 4)</u></em>

<em><u>Solution:</u></em>

<em><u>Given system of equations are:</u></em>

y = x^2 - 6x + 12 ------ eqn\ 1\\\\y = 2x - 4 ---------- eqn\ 2

<em><u>Substitute eqn 2 in eqn 1</u></em>

x^2 - 6x + 12 = 2x - 4

Make the right side of equation 0

x^2 - 6x + 12 - 2x + 4 = 0\\\\x^2 -8x + 16 = 0

<em><u>Solve by quadratic equation</u></em>

\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=1,\:b=-8,\:c=16:\\\\x=\frac{-\left(-8\right)\pm \sqrt{\left(-8\right)^2-4\cdot \:1\cdot \:16}}{2\cdot \:1}\\\\x=\frac{-\left(-8\right)\pm \sqrt{0}}{2\cdot \:1}\\\\x = \frac{8}{2}\\\\x = 4

<em><u>Substitute x = 4 in eqn 2</u></em>

y = 2(4) - 4

y = 8 - 4

y = 4

Thus solution is (4, 4)

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What is 4.39 mutiply 10 exponents 2
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The sum of the ages of Berma, her mother Rinna and her father Erwin is 80. Two years from now, Rinna’s age will be 13 less than
jekas [21]

Answer: Berma is 5 years old

Rinna is 38 years old

Erwin is 37 years old

Step-by-step explanation:

Let x represent Berma's age

Let y represent Rinna's age

Let z represent Erwin's age

Since the sum of their ages is 80,

x + y + z = 80 - - - - - - -1

Two years from now, Rinna’s age will be 13 less than the sum of Erwin’s age and twice Berma’s age. This means that

y +2 = [ (z+2) + 2(x+2) ] - 13

y +2 = z + 2 + 2x + 4 - 13

2x - y + z = 13 + 2 - 4 -2

2x - y + z = 9 - - - - - - -2

Three years ago, 15 times Berma’s age was 5 less than the age of Rinna. It means that

15(x - 3) = (y - 3) - 5

15x - 45 = y - 3 - 5

15x - y = - 8 + 45

15x - y = 37 - - - - - - - -3

From equation 3, y = 15x - 37

Substituting y = 15x - 37 into equation 1 and equation 2, it becomes

x + 15x - 37 + z = 80

16x + z = 80 + 37 = 117 - - - - - - 4

2x - 15x + 37 + z = 9

-13x + 2 = -28 - - - - - - - - -5

subtracting equation 5 from equation 4,

29x = 145

x = 145/29 = 5

y = 15x - 37

y = 15×5 -37

y = 38

Substituting x= 5 and y = 38 into equation 1, it becomes

5 + 38 + z = 80

z = 80 - 43

z = 37

6 0
3 years ago
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