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MrRa [10]
3 years ago
11

Find the dimensions of a rectangle whose perimeter is 22 meters and whose area is 28 square meters

Mathematics
1 answer:
kondaur [170]3 years ago
8 0
You can solve this by finding all of the factors of 28 and seeing which ones add up to 22 when doubled.

The factors of 28 are as follows:

28 x 1      28(2) + 1(2) = 56 + 2 = 58 ≠ 22
14 x 2      14(2) + 2(2) = 28 + 4 = 32 ≠ 22
 7 x 4         7(2) + 4(2) = 14 + 8 = 22

The dimensions of the rectangle are 4 x 7.


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Answer:

QR = 65.4 m

Step-by-step explanation:

a. Apply Law of Cosines to find QR:

p² = q² + r² - 2qr × Cos P

p = QR = ?

q = PR = 150 m

r = PQ = 120 m

P = 25°

Plug in the values

p² = 150² + 120² - (2)(150)(120) × Cos(25°)

p² = 22,500 + 14,400 - 36,000 × 0.9063

p² = 36,900 - 32,626.8

p² = 4,273.2

p = √4,273.2

p ≈ 65.4 m (nearest tenth)

QR = 65.4 m

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3 years ago
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3 years ago
What is the root of the quadratic equation <br> y= (x-4) (x+7)
pishuonlain [190]
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X=4. X=-7
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kirill115 [55]

Answer:

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the Answear is B,D,A,C

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