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In-s [12.5K]
3 years ago
8

3 Ms. Jones teaches 6 classes a day. Each class has 30 students. After school she coaches

Mathematics
1 answer:
antiseptic1488 [7]3 years ago
7 0

Answer: 190

Step-by-step explanation:

30x6=180+10=190

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Double angle (or half angle, depending how you look at it) identities:

\cos^2\dfrac x2=\dfrac{1+\cos x}2

\sin^2\dfrac x2=\dfrac{1-\cos x}2

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So we have

\cot^215^\circ=\dfrac{1+\cos30^\circ}{1-\cos30^\circ}=\dfrac{1+\frac{\sqrt3}2}{1-\frac{\sqrt3}2}

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Note that when taking the square root, we should take into account that that could yield two possible solutions, but we know \cos15^\circ>0 and \sin15^\circ>0, so it's also the case that \cot15^\circ>0.

Also, the reason we have equality in the last step can be explained like so:

7+4\sqrt3=4+4\sqrt3+3=4+4\sqrt3+(\sqrt3)^2=(2+\sqrt3)^2

(not unlike the process used to complete the square)

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