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lesya692 [45]
3 years ago
9

For what values of x is the area of the rectangle greater than the perimeter?

Mathematics
1 answer:
soldier1979 [14.2K]3 years ago
4 0
Area = 7(x + 2) = 7x + 14
Perimeter = 2(7 + x + 2) = 2(9 + x) = 18 + 2x
7x + 14 > 18 + 2x
7x - 2x > 18 - 14
5x > 4
x > 4/5
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How do you solve this?
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Answer:

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Length: 14

Step-by-step explanation:

The area of a rectangle can be found my multiplying the length by the width. We do not know neither the length nor the width. However, we do know that the length is 7 feet longer than the width. Because we do not know the value of either side, let's let x represent the width.

Since the length is 7 feet longer, we can represent the length by x+7.

Now that we have our values for the length and the width we can multiply them together to find our area.

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Now we have our equation, so we can address the changes made by the original question. The original question states that when 7 is added to both the length and width the area becomes 3 times larger. To do so simply increase each side by 7 by adding 7 to the original values.

(x+7)(x+14)

And multiply our area by 3.

3(x^{2} +7x)

set these equal to each other to find your new equation.

(x+7)(x+14)=3(x^{2} +7x)

Now you need to solve for x. To do this first muliply (x+7) and (x+14).

(x+7)(x+14)=\\x^{2} +14x+7x+98=\\x^{2} +21x+98

Then multiply 3(x^{2} +7x)

3(x^{2} +7x)=\\3x^{2} +21x

Now you can begin solving for x.

x^{2} +21x+98=3x^{2} +21x

Subtract x^{2} from both sides.

21x+98=2x^{2} +21x

Subtract 21x from both sides.

98=2x^{2}

Divide by 2.

49=x^{2}

And finally take the sqaure root of both sides.

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Remember, because we are dealing with length, there cannot be negative. So while normally we would get both +7 <em>and</em> -7, in this case we <em>only </em>get +7.

Now that we have the value of x, we can plug it into our original values.

For width we simply get 7.

For length we get 7+7=14

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