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Ksju [112]
4 years ago
12

If the length of the square is increased by 2 and the width is decreased by 2, by how many units is the area of

Mathematics
1 answer:
mrs_skeptik [129]4 years ago
3 0

Answer:

4 units

Step-by-step explanation:

Let x represent the length of the square

Area of the square = x^2

So, the dimension of the rectangle formed is:

length = x + 2

width = x - 2

Area of the rectangle = ( x + 2 ) * ( x - 2 )

solve the parenthesis

x^2 - 2x + 2x - 4

Area of the rectangle = x^2 - 4

subtract this area from that of the square

x^2 - ( x^2 - 4 )

=x^2 - x^2 + 4

= 4 units

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laila [671]

It's a quadratic equation in disguise. If you let y=x^{-1}, then y^2=x^{-2}, and we can rewrite the equation as

5y^2-17y+6=0

Solve for y however you like; we get y=\dfrac25 and y=3.

But we want to solve for x, so we have

y=x^{-1}=\dfrac25\implies x=\dfrac52

y=x^{-1}=3\implies x=\dfrac13

5 0
4 years ago
lisa is painting a fence that is 26 feet long and 7 feet tall. a gallon of paint will cover 350 aquare feet. write and slove a p
vaieri [72.5K]
The fence is 26*7=182 square feet. 1 gallon is enough for 350 square feet  and x for 182 square feet. => x=182/350=0.52 gallons. I hope I'm right.
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3 years ago
PLEASE HELP I NEED HELP!! 25 POINTS
Marysya12 [62]

60x+80=100x combine like terms

80=100x-60x or 40x divide to get x by itself

80/40=40x/40

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after 20 hrs

3 0
3 years ago
Which sign makes this statement true​
Vesna [10]

Answer:

The = sign

Step-by-step explanation:

1 11/20 = 31/20 = 1.55 and 1.55 =1.55

3 0
4 years ago
Read 2 more answers
Solve (x - 3)^2+7=97, where X is a real number. Round your answer to the nearest hundredth.​
Eddi Din [679]

Answer:

x ≈ - 6.49, x ≈ 12.49

Step-by-step explanation:

Given

(x - 3)² + 7 = 97 ( subtract 7 from both sides )

(x - 3)² = 90 ( take the square root of both sides )

x - 3 = ± \sqrt{90} ( add 3 to both sides )

x = 3 ± \sqrt{90}

Then

x = 3 - \sqrt{90} ≈ - 6.49 ( to the nearest hundredth )

x = 3 + \sqrt{90} ≈ 12.49 ( to the nearest hundredth )

6 0
3 years ago
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