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kondor19780726 [428]
3 years ago
6

Algebra problem below.

Mathematics
1 answer:
umka21 [38]3 years ago
3 0
A= \frac{1}{2}bh= \frac{1}{2}*18*11=9*11=99 \ ft^2
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I need help proving this ASAP
Ket [755]

Answer:

See explanation

Step-by-step explanation:

We want to show that:

\tan(x +  \frac{3\pi}{2} )  =  -   \cot \: x

One way is to use the basic double angle formula:

\frac{ \sin(x +  \frac{3\pi}{2} ) }{\cos(x +  \frac{3\pi}{2} )}  =  \frac{ \sin(x)  \cos( \frac{3\pi}{2} )  +   \cos(x)  \sin( \frac{3\pi}{2}) }{\cos(x)  \cos( \frac{3\pi}{2} )   -    \sin(x)  \sin( \frac{3\pi}{2}) }

\frac{ \sin(x +  \frac{3\pi}{2} ) }{\cos(x +  \frac{3\pi}{2} )}  =  \frac{ \sin(x) ( 0)  +   \cos(x) (  - 1) }{\cos(x) (0)   -    \sin(x) (  - 1) }

We simplify further to get:

\frac{ \sin(x +  \frac{3\pi}{2} ) }{\cos(x +  \frac{3\pi}{2} )}  =  \frac{ 0  -   \cos(x) }{0 +    \sin(x) }

We simplify again to get;

\frac{ \sin(x +  \frac{3\pi}{2} ) }{\cos(x +  \frac{3\pi}{2} )}  =  \frac{- \cos(x) }{ \sin(x) }

This finally gives:

\frac{ \sin(x +  \frac{3\pi}{2} ) }{\cos(x +  \frac{3\pi}{2} )}  =  -  \cot(x)

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#6 is incorrect
it's 6 7/8
#7 is correct
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the other side would be 15

Step-by-step explanation:

because 15 is bigger than 12 and the 12 is the bottom line and then the other line going up is 15 so it's equal.

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Answer:

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Step-by-step explanation:

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