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Pachacha [2.7K]
3 years ago
11

The rectangle below has an area of x^2+8x+15x

Mathematics
1 answer:
Darina [25.2K]3 years ago
7 0

Answer:

The length of the rectangle is x+5.

Step-by-step explanation:

Given : The rectangle below has an area of x^2+8x+15 square meter and a width of  x+3 meter.

To find : What expression represents the length of the rectangle?

Solution :

The area of the rectangle is given by,

\text{Area}=\text{Length}\times \text{Breadth}

\text{Area}=x^2+8x+15

\text{Breadth}=x+3

x^2+8x+15=\text{Length}\times(x+3)

\text{Length}=\frac{x^2+8x+15}{x+3}

\text{Length}=\frac{(x+5)(x+3)}{x+3}

\text{Length}=x+5

Therefore, the length of the rectangle is x+5.

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Annie is framing a photo with a length of 6 inches and a width of 4 inches. The distance from the edge of the photo to the edge
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Answer:

Part a) The quadratic function is 4x^{2} +20x-39=0

Part b) The value of x is 1.5\ in

Part c) The photo and frame together are 7\ in wide

Step-by-step explanation:

Part a) Write a quadratic function to find the distance from the edge of the photo to the edge of the frame

Let

x----> the distance from the edge of the photo to the edge of the frame

we know that

(6+2x)(4+2x)=63\\24+12x+8x+4x^{2}=63\\ 4x^{2} +20x+24-63=0\\4x^{2} +20x-39=0

Part b) What is the value of x?

Solve the quadratic equation 4x^{2} +20x-39=0

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem

we have

4x^{2} +20x-39=0

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a=4\\b=20\\c=-39

substitute in the formula

x=\frac{-20(+/-)\sqrt{20^{2}-4(4)(-39)}} {2(4)}

x=\frac{-20(+/-)\sqrt{1,024}} {8}

x=\frac{-20(+/-)32} {8}

x=\frac{-20(+)32} {8}=1.5\ in  -----> the solution

x=\frac{-20(-)32} {8}=-6.5\ in

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(4+2x)=4+2(1.5)=7\ in

5 0
3 years ago
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