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tensa zangetsu [6.8K]
3 years ago
11

To compare and order fractions,the fraction can be written with a​

Mathematics
1 answer:
Greeley [361]3 years ago
5 0
Common denominator i believe. i’m not sure. hope this helps
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Express answer in exact form. Show all work for full credit.
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The chord length is the same as the radius, so R^2 = 9.
Area = (1/2)(9)(pi/3 - sqrt(3)/2) = (3/4) (2 pi - 3 sqrt(3))
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Jayden and Sheridan both tried to find the missing side of the right triangle. A right triangle is shown. One leg is labeled as
stellarik [79]

Answer:

Sheridan's Work is correct

Step-by-step explanation:

we know that

The lengths side of a right triangle must satisfy the Pythagoras Theorem

c^{2}=a^{2}+b^{2}

where

a and b are the legs

c is the hypotenuse (the greater side)

In this problem

Let

a=7\ cm\\c=13\ cm

substitute

13^{2}=7^{2}+b^{2}

Solve for b

169=49+b^{2}

b^{2}=169-49

b^{2}=120

b=\sqrt{120}\ cm

b=10.95\ cm

we have that

<em>Jayden's Work</em>

a^{2}+b^{2}=c^{2}

a=7\ cm\\b=13\ cm

substitute and solve for c

7^{2}+13^{2}=c^{2}

49+169=c^{2}

218=c^{2}

c=\sqrt{218}\ cm

c=14.76\ cm

Jayden's Work is incorrect, because the missing side is not the hypotenuse of the right triangle

<em>Sheridan's Work</em>

a^{2}+b^{2}=c^{2}

a=7\ cm\\c=13\ cm

substitute

7^{2}+b^{2}=13^{2}

Solve for b

49+b^{2}=169

b^{2}=169-49

b^{2}=120

b=\sqrt{120}\ cm

b=10.95\ cm

therefore

Sheridan's Work is correct

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Step-by-step explanation:

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The correct answer is 24
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