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ipn [44]
4 years ago
8

A regular polygon inscribed in a circle can be used to derive the formula for the area of a circle. The polygon area can be expr

essed in terms of the area of a triangle.
Let s be the side length of the polygon,

let r be the hypotenuse of the right triangle,

let h be the height of the triangle, and

let n be the number of sides of the regular polygon.



polygon area = n(1/2sh)



Which statement is true?

A. - As h increases, ns gets closer to 2πr​.

B. - As s increases, ns gets closer to 2πr​.

C. - As r increases, ns gets closer to 2πr

D. - As n increases, ns gets closer to 2πr​.

Mathematics
2 answers:
Alex Ar [27]4 years ago
6 0
D. as n increases, ns --> 2πr
 h/r cannot increase it is the circle radius
s, side length we want to get smaller and smaller to make a circle

gregori [183]4 years ago
4 0

Answer:

D let N be the number of sides of the regular polygon

Step-by-step explanation: i just took the test

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Answer:

46

Step-by-step explanation:

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8 0
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ASAP! GIVING BRAINLIEST! Please read the question THEN answer correctly! No guessing.
alekssr [168]

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x > 5

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<em><u>Subtract 7 from both sides, like so:</u></em>

x + 7 > 12

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3 years ago
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trapecia [35]
I believe it’s c but yeah
5 0
3 years ago
I need to solve this using l'hopital's rule and logarithmic diferentiation.
arlik [135]

Yo sup??

For our convenience let h=x+1

therefore

when x tends to -1, h tends to 0

hence we can rewrite it as

\lim_{h \to \ 0 } (cos(h))^{(cot(h^2 )}

This inequality is of the form 1∞

We will now apply the formula

e^(^g^(^x^)^(^f^(^x^)^-^1^)^)

plugging in the values of g(x) and f(x)

e^{lim_{h \to \ 0}{(cot(h)^2(cos(h)-1))}

express coth² as cosh²/sinh² and also write cosh-1 as 2sin²(h/2)

(by applying the property that cos2x=1-sin²x)

After this multiply the numerator and denominator with h² so that we can apply the property that

\lim_{x \to \ 0 } sinx/x =1

Now your equation will look like this.

e^{lim_{h \to \ 0}{((cos(h)^2(2sin^2(h/2)*h^2)/(sin(h)^2*h^2)}

We will now apply the result

\lim_{x \to \ 0 } sinx/x =1

where x=h²

we get

e^{lim_{h \to \ 0}{((cos(h)^2(2sin^2(h/2))/(h^2)}

we now multiply the numerator and denominator with 4 so that we can say

\lim_{h^2 \to \ 0 } sin^2(h/2)/(h^2/4) = 1

e^{lim_{h \to \ 0}{((cos(h)^2(2sin^2(h/2))/(h^2*4/4)}

=e^{lim_{h \to \ 0}{((cos(h)^2*2)/(4)}

Apply the limits and you will get

e^{cos(0)^2*2/4

=e^{1/2}

Hope this helps.

7 0
3 years ago
Find the area of the shaded region. All angles are right angles.
Tanzania [10]

Answer: 147

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Again, 9 x 4 = 36

Finally 15 x 5 = 75

So, 36 + 36 + 75 = 147

8 0
4 years ago
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