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blagie [28]
4 years ago
9

Find the area of the regular polygon. Give the answer to the nearest tenth.

Mathematics
2 answers:
k0ka [10]4 years ago
7 0
I'm taking radius to be half of the side length.Then one side is 26 cm and the area of a square is the length of a side squared.A=26^2=676So,C) 676
Alenkinab [10]4 years ago
7 0

Answer:

The answers updated for 2019 for TCAH Unit 5 Lesson 7 Quiz (with 11 questions)

Step-by-step explanation:

1. C

2. A

3. D

4. C

5. A

6. D

7. B

8. B

9. A

10. B

11. C

I just took the quiz 100% ♥

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Answer:

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Step-by-step explanation:

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4 0
3 years ago
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Find the arc length of the given curve between the specified points.
yuradex [85]

Answer:

4.25

Step-by-step explanation:

Given:

f(x) = \frac{x^3}{12} + \frac{1}{x} \\\\Arc Length = \int\limits^a_b {\sqrt{1 + (f'(x))^2}  } \, dx \\f '(x) = \frac{x^2}{6} - \frac{1}{x^2}\\\\f'(x)^2 = (\frac{x^2}{6} - \frac{1}{x^2})^2 = \frac{x^4}{36} +  \frac{1}{x^4} - \frac{1}{3}\\\\1 + f'(x)^2 = \frac{x^4}{36} +  \frac{1}{x^4} + \frac{2}{3}\\\\\= \frac{x^8 + 36 + 12x^4}{36x^4}\\\\= \frac{(x^4 + 6)^2}{36x^4}\\\\=\sqrt{1 + f'(x)^2}  = \sqrt{ \frac{(x^4 + 6)^2}{36x^4}}\\\\= \frac{x^2}{6} + \frac{1}{x^2} \\\\

ArcLength = \int\limits^4_1 {\frac{x^2}{6} + \frac{1}{x^2}  } \, dx \\= (\frac{x^3}{18} -  \frac{1}{x})\limits^4_1\\\\= (\frac{64}{18} - \frac{1}{4}) - (\frac{1}{18} - 1)\\\\= \frac{17}{4}= 4.25

8 0
3 years ago
The output is 2 less than<br> the cube of the input.
Margarita [4]

Answer:

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Step-by-step explanation:

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3 years ago
Help please with these questions
Sonja [21]
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Hope this helps!
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What's the expanded form of 32.043
jok3333 [9.3K]
<span>32.043=30+2+0.04+0.003</span>
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