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Sonja [21]
3 years ago
13

___,36 ,108 ,324 ,972

Mathematics
1 answer:
hjlf3 years ago
3 0
12

First figure out of it is a arithmetic or geometric sequence, in this case it is a geometric sequence because you have to multiply, not add to find the next number.

Divide one number, I'll pick 324 by the number before it, 108, you get three.

Now divide 36 by 3, you get 12.

To check, multiply 12 by 3, you get 36, multiply that by 3, you get 108, and so on and so forth.
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4 years ago
Please help me for the love of God if i fail I have to repeat the class
Elena-2011 [213]

\theta is in quadrant I, so \cos\theta>0.

x is in quadrant II, so \sin x>0.

Recall that for any angle \alpha,

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

Then with the conditions determined above, we get

\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35

and

\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}

Now recall the compound angle formulas:

\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta

\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta

\sin2\alpha=2\sin\alpha\cos\alpha

\cos2\alpha=\cos^2\alpha-\sin^2\alpha

as well as the definition of tangent:

\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}

Then

1. \sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}

2. \cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}

3. \tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}

4. \sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}

5. \cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}

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\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2

\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2

\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}

Because x is in quadrant II, we know that \dfrac x2 is in quadrant I. Specifically, we know \dfrac\pi2, so \dfrac\pi4. In this quadrant, we have \tan\dfrac x2>0, so

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6 0
4 years ago
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Answer:

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3        3       3

<u>3 y </u> =  <u>x  </u> +  <u>2  </u>                                                    

3        3       3

y =  <u>x </u>  +  <u>2   </u>        replace y with f ^ -1(x)

   <em> </em>  3        3                                            

f ^-1 (x) =    <u>x</u>   +  <u>2 </u>        

                 3       3      

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