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Radda [10]
3 years ago
12

Find the slope of the line that contains (−3,−3) and (−4,−8).

Mathematics
1 answer:
irga5000 [103]3 years ago
5 0

Answer:

y=5x+12 it is rise over run so 5/1 but you don't put the 1 so its just 5x

Step-by-step explanation:

I did it using y=mx+b

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Question in picture<br><br>A) 16/63<br><br>B)-16/63<br><br>C) 63/16<br><br>D) -63/16
Orlov [11]
ANSWER
\tan(x + y) =  -  \frac{63}{16}


EXPLANATION


We were given that,

\csc(x)  =  \frac{5}{3}

This implies that,

\sin(x)  =  \frac{3}{5}

We use the Pythagorean identity

\sin^{2} (x)  +  \cos^{2} (x)= 1
to get,


\cos(x)  =  \sqrt{1 - ( { \frac{3}{5} })^{2}}  =  \frac{4}{5}


We were also given that,


\cos(y)  =  \frac{5}{13}

This means that,


\sin(y)  =  \sqrt{1 -  {( \frac{5}{13}) }^{2} }  =  \frac{12}{13}

This is because,


0 <  \: x \:  <  \frac{\pi}{2}


0 <  \: y \:  <  \frac{\pi}{2}

This angles are in the first quadrant so we pick the positive values.

\tan(x + y)  =  \frac{ \sin(x + y) }{ \cos(x + y) }


\tan(x + y)  =  \frac{ \sin(x ) \cos(y)   +  \sin(y)  \cos(x) }{ \cos(x) \cos(y)  -  \sin(x)  \sin(y) }



\tan(x + y)  =  \frac{  \frac{3}{5}   \times  \frac{5}{13}  +   \frac{12}{13}   \times  \frac{4}{5}  }{  \frac{4}{5}  \times  \frac{5}{13}   -   \frac{3}{5}  \times  \frac{12}{13}  }



\tan(x + y) =  -  \frac{63}{16}

The correct answer is D
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3 years ago
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