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Elena L [17]
4 years ago
11

The length of a rectangular sign is 5 times its width . If the sign's perimeter is 36 inches, what is the signs area in square i

nches
Mathematics
2 answers:
swat324 years ago
4 0

Answer: The area is 45 square inches.

Step-by-step explanation:

so if we know that the the length is 5 times the width then we could represent the length as  5x   and x for the width.  

And we  know the perimeter of the sign so we could set up equation to solve for the side lengths.

The perimeter of a rectangle is  2l + 2w  which means 2 times the length plus two times the width.

2(5x) + 2(x) = 36

10x + 2x = 36

12x = 36

x = 3

Now we know the value of x  so we will plot it into the length and with to find the actually side lengths.

length= 5(3) = 15

width=  3  

Check:  15 + 15 + 3 + 3 = 36  

Now we know the side lengths so to find the area we will have to multiply the length times the width.

3 * 15 = 45  

My name is Ann [436]4 years ago
3 0

Answer:

9 inches

Step-by-step explanation:

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The actual width of the building shown in the model below is 60 feet: What is the actual height od the building shown in the mod
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How many solutions does the system have?
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4 0
3 years ago
Write a rule that represents the function. <br> (1,1/2),(2,1/4),(3.1/8),(4,1/16),(5,1/32)
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5 0
3 years ago
find the vertex of the quadratic passing through the point f(3/2)=9, has an x-intercept of 3 and the axis of symmetry is x=2
3241004551 [841]

The vertex of the quadratic is (2 , 12)

Step-by-step explanation:

The vertex form of the quadratic function is f(x) = a(x - h)² + k, where

  • (h , k) are the coordinates of its vertex
  • Its axis of symmetry is a vertical line at x = h
  • its minimum value is f(x) = k at x = h

∵ The function of quadratic

∴ f(x) = a(x - h)² + k

∵ The axis of symmetry of the function is x = 2

∵ The axis of symmetry of a quadratic function is a vertical line at x = h

∴ h = 2

- Substitute h by 2 in f(x)

∴ f(x) = a(x - 2)² + k

∵ The quadratic passing through the point f(3/2) = 9

- That means the coordinates of the point are (\frac{3}{2},9)

∴ x = \frac{3}{2}  and y = 9

- Substitute x and y in the equation by the values above

∴ 9 = a( \frac{3}{2} - 2)² + k

∴ 9 = a( \frac{-1}{2} )² + k

∴ 9 = \frac{1}{4} a + k

- Multiply each term by 4

∴ 36 = a + 4 k

∴ a + 4 k = 36 ⇒ (1)

∵ x intercept of f(x) = 3

- That means the graph of f(x) intersects x-axis at point (3 , 0)

∴ x = 3 and y = 0

- Substitute x and y in the equation by the values above

∴ 0 = a(3 - 2)² + k

∴ 0 = a(1)² + k

∴ 0 = a + k

∴ a + k = 0 ⇒ (2)

Now we have system of equation let us solve it to find k

∵ a + 4 k = 36 ⇒ (1)

∵ a + k = 0 ⇒ (2)

- Subtract equation (2) from equation (1) to eliminate a

∴ 3 k = 36

- Divide both sides by 3

∴ k = 12

∵ (h , k) are the coordinates of the vertex of the quadratic function

∴ (2 , 12) are the coordinates of the vertex of the quadratic function

The vertex of the quadratic is (2 , 12)

Learn more:

You can learn more about quadratic in brainly.com/question/9390381

#LearnwithBrainly

4 0
4 years ago
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