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MaRussiya [10]
3 years ago
12

Find the quotient of 90 over -10

Mathematics
2 answers:
OLEGan [10]3 years ago
5 0
Hi! The answer is -9 because 90/-10 = -9
I hope this helps you, Goodluck! :)
vodka [1.7K]3 years ago
3 0

90/-10

= 9/-1

= -9

So, -9 is the quotient.

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If p:q = 5;2 and q:r = 3:4, what is the ratio of p to r?
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Given:

\sin x=-\dfrac{15}{17}

x lies in the III quadrant.

To find:

The values of \sin 2x, \cos 2x, \tan 2x.

Solution:

It is given that x lies in the III quadrant. It means only tan and cot are positive and others  are negative.

We know that,

\sin^2 x+\cos^2 x=1

(-\dfrac{15}{17})^2+\cos^2 x=1

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\cos x=\pm\sqrt{\dfrac{289-225}{289}}

x lies in the III quadrant. So,

\cos x=-\sqrt{\dfrac{64}{289}}

\cos x=-\dfrac{8}{17}

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\cos 2x=1-2\sin^2x

\cos 2x=1-2(-\dfrac{15}{17})^2

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\tan 2x=\dfrac{\sin 2x}{\cos 2x}

\tan 2x=\dfrac{-\dfrac{240}{289}}{-\dfrac{161}{289}}

\tan 2x=\dfrac{240}{161}

Therefore, the required values are \sin 2x=-\dfrac{240}{289},\cos 2x=-\dfrac{161}{289},\tan 2x=\dfrac{240}{161}.

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