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stellarik [79]
3 years ago
14

The area of a square quilt is 289 square inches. There is a smaller square patch on the quilt with an area of 121 square inches.

What is the difference between the overall perimeter of the quilt and the perimeter of the smaller square patch?
Mathematics
2 answers:
nadezda [96]3 years ago
8 0
The answer is 42, 289/4=72.25, then 121/4=30.25
so we subtract the answers, 72.25-20.25= 42
answer=42
alisha [4.7K]3 years ago
3 0
Area of a square quilt = 289 square inches
Area of a smaller square quilt = 121 square inches

Area of a square = a²

Side length of big square quilt = √289 in² = 17 inches
side length of small square quilt = √121 in² = 11 inches

Perimeter of big square = 17 inches * 4 = 68 inches
perimeter of small square = 11 inches * 4 = 44 inches

difference = 68 inches - 44 inches = 24 inches.
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Determine two pairs of polar coordinates for the point (5, 5) with 0° ≤ θ < 360°.
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Answer:

First part

The answer is (5 square root 2, 45°), (-5 square root 2, 225°) ⇒ answer (d)

Second part

The equation in standard form for the hyperbola is y²/81 - x²/19 = 1 ⇒ answer(b)

Step-by-step explanation:

First part:

* Lets study the Polar form and the Cartesian form

- The important difference between Cartesian coordinates and

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# In Cartesian coordinates there is exactly one set of coordinates

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# In polar coordinates there is an infinite number of coordinates

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# In the polar the coordinates the origin is called the pole, and

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# The angle measurement θ can be expressed in radians

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# r = ± √(x² + y²)

# θ = tan^-1 (y / x)

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- The point in the Cartesian coordinates is (5 , 5)

∵ x = 5 and y = 5

∴ r = ± √(5² + 5²) = ± √50 = ± 5√2

∴ tanФ = (5/5) = 1

∵ tanФ is positive

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∴ Ф in the third quadrant is 180 + 45 = 225°

* The answer is (5√2 , 45°) , (-5√2 , 225°)

Second part:

* Lets study the standard form of the hyperbola equation

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• the length of the transverse axis is 2a

• the coordinates of the vertices are (0 , ±a)

• the length of the conjugate axis is 2b

• the coordinates of the co-vertices are (±b , 0)

•      the coordinates of the foci are (0 , ± c),  

• the distance between the foci is 2c, where c² = a² + b²

* Lets solve our problem

∵ The vertices are (0 , 9) and (0 , -9)

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∵ The foci at (0 , 10) , (0 , -10)

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∵ c² = a² + b²

∴ (10)² = (9)² + b² ⇒ 100 = 81 + b² ⇒ subtract 81 from both sides

∴ b² = 19

∵ The equation is  y²/a² - x²/b² = 1

∴  y²/81 - x²/19 = 1

* The equation in standard form for the hyperbola is y²/81 - x²/19 = 1

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