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Ratling [72]
1 year ago
11

Suppose that a rectangle has a perimeter of 26 meters. Express the area A(x) of the rectangle in terms of the length x of one of

its sides. A(x)
Mathematics
1 answer:
ch4aika [34]1 year ago
4 0

The area A(x) of the rectangle in terms of the length x of one of its sides. A(x) = x(13-x).

<h3>What is rectangle?</h3>

A rectangle is really a closed two-dimensional geometry with four sides, four corners, and four right angles (90°). A rectangle's opposite sides are equal & parallel. Because  rectangles is a 2-D form, it has two dimensions: length and width.

Some characteristics of rectangle are-

  • The length of the rectangle is the longer side, while the width would be the shorter side.
  • Because all of the angles in a rectangle are equal, it is also known as an equiangular quadrilateral. The quadrilateral is a closed 4-sided shape.
  • Since a rectangle contains parallel sides, it is also known as a right-angled parallelogram.
  • The parallelogram would be a quadrilateral with equal and parallel opposite sides. Rectangles are a type of parallelogram.

Now, according to the question,

Let 'P' be the perimeter of the rectangle.

Perimeter = 2(Length + Breadth)

P = 2(L + B)

The perimeter is 26 meters.

26 = 2(L + B)

L + B = 13

B = 13 - L

Now, the area of the rectangle is given as;

Area = Length×Breadth

A = L×B

Substitute the value of B in area.

A = L×(13 - L)

Area in terms of length x, Put L =x

A(x) = x(13 - x)

Therefore, the area A(x) of the rectangle in terms of the length x of one of its sides. A(x) = x(13 - x).

To know more about the rectangle, here

brainly.com/question/25292087

#SPJ4

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All the interior angles of a triangle always add up to 180 degrees.

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Match each equation with its solution set. Tiles a2 − 9a + 14 = 0 a2 + 9a + 14 = 0 a2 + 3a − 10 = 0 a2 + 5a − 14 = 0 a2 − 5a − 1
sattari [20]
We have that

N 1)
a²<span> − 9a + 14 = 0 
</span>

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² − 9a)=-14

Complete the square  Remember to balance the equation by adding the same constants to each side 

(a² − 9a+20.25)=-14+20.25

Rewrite as perfect squares

(a-4.5)²=6.25--------> (a-4.5)=(+/-)√6.25

a1=4.5+√6.25-----> a1=7

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the solution problem N 1 is the pair {7, 2}


N 2) 

a²<span> + 9a + 14 = 0
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Group terms that contain the same variable, and move the constant to the opposite side of the equation

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(a² +9a+20.25)=-14+20.25

Rewrite as perfect squares

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a1=-4.5+√6.25-----> a1=-2

a2=-4.5-√6.25-----> a2=-7

the solution problem N 2 is the pair {-2,-7}

N 3) 

a² + 3a − 10 = 0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² + 3a)=10

Complete the square  Remember to balance the equation by adding the same constants to each side 

(a² + 3a+2.25)=10+2.25

Rewrite as perfect squares

(a+1.5)²=12.25------> (a+1.5)=(+/-)√12.25

a1=-1.5+√12.25-----> a1=2

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the solution problem N 3 is the pair {2, -5}


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a²<span> + 5a − 14 = 0
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Group terms that contain the same variable, and move the constant to the opposite side of the equation

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Complete the square  Remember to balance the equation by adding the same constants to each side 

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Rewrite as perfect squares

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a1=-2.5+√20.25-----> a1=2

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the solution problem N 4 is the pair {2, -7}


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a² − 5a − 14 = 0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² − 5a)=14

Complete the square  Remember to balance the equation by adding the same constants to each side 

(a² − 5a+6.25)=14+6.25

Rewrite as perfect squares

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a2=2.5-√20.25-----> a2=-2

the solution problem N 5 is the pair {7, -2}

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ANSWER

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EXPLANATION

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