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Marizza181 [45]
3 years ago
5

Solve for b. −1/3b=9 What is the answer?

Mathematics
2 answers:
Ksivusya [100]3 years ago
5 0
-1/3b = 9
b = 9 * -3
b = -27
zalisa [80]3 years ago
5 0
<span>−1/3b=9
Divide -1/3 on both sides
b= 9 divided by -1/3
Convert the 9 into an improper fraction
b= 9/1 divided by -1/3
Convert the division sign to multiplication while flipping the fraction -1/3 to -3/1
b= 9/1*-3/1
Multiply
Final Answer: b = -27</span>
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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) }

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

6 0
3 years ago
What is the value of the expression
Inga [223]

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Step-by-step explanation:

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8 0
3 years ago
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