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3 years ago

9

Jake scored 269 points on three math tests. To get an A for the class, his average number of points on four tests must be no les

s than 90. If Jake hopes to get an A, how many points must he score on the fourth test?

1 answer:

nadya68 [22]3 years ago

3 0

Jake must score a minimum of a 94%

$269 + 94 = 363 \\ 363 \div 4 = 90.75$

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Given that ABC is a right triangle.

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<u>Option A</u>: $\sin \left(45^{\circ}\right)=\frac{BC}{9}$

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$\sin \left(45^{\circ}\right)=\frac{BC}{9}$

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<u>Option B</u>: $\sin \left(45^{\circ}\right)=\frac{9}{BC}$

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$sin\ \theta=\frac{opp}{hyp}$

where $\theta=45^{\circ}$ , $opp= BC$ and $hyp = 9$

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$\sin \left(45^{\circ}\right)=\frac{BC}{9}$

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where $\theta=45^{\circ}$ , $opp= BC$ and $adj=AC$

Substituting the values, we get;

$tan \ 45^{\circ}=\frac{BC}{AC}$

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<u>Option D</u>: $(A C) \sin \left(45^{\circ}\right)=B C$

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$sin\ \theta=\frac{opp}{hyp}$

where $\theta=45^{\circ}$ , $opp= BC$ and $hyp = 9$

Hence, substituting the values, we get;

$\sin \left(45^{\circ}\right)=\frac{BC}{9}$

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Hence, Option D is not the correct answer.

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Substituting the values, we get;

$\cos \left(45^{\circ}\right)=\frac{AC}{9}$

Hence, the given equation $\cos \left(45^{\circ}\right)=\frac{BC}{9}$ is not possible.

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