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3 years ago

PLEASE HELP ME! :(

1 answer:

5 0

Here, I'll do the first one for you.
When they are talking about a number, use "x".
Since it says twice a number, you say 2x. Or 3x for three times the number
2x+12=3x-31
Then use algebra to find x. Get the numbers on one side and all the x's on the other.
2x+12+31=3x-31+31
2x+43-2x=3x-2x
x=43

Now do the rest on your own!

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Use the given information to find the exact value of the trigonometric function

eimsori [14]

$\begin{gathered} \csc \theta=-\frac{6}{5} \\ \tan \theta>0 \\ \cos \frac{\theta}{2}=\text{?} \end{gathered}$

Half Angle Formula

$\cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}}$ $\tan \theta>0\text{ and csc}\theta\text{ is negative in the third quadrant}$ $\begin{gathered} \csc \theta=-\frac{6}{5}=\frac{r}{y} \\ x^2+y^2=r^2 \\ x=\pm\sqrt[\square]{r^2-y^2} \\ x=\pm\sqrt[\square]{6^2-(-5)^2} \\ x=\pm\sqrt[\square]{36-25} \\ x=\pm\sqrt[\square]{11} \\ \text{x is negative since the angle is on the 3rd quadrant} \end{gathered}$ $\begin{gathered} \cos \theta=\frac{x}{r}=\frac{-\sqrt[\square]{11}}{6} \\ \cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}} \\ \cos \frac{\theta}{2}is\text{ also negative in the 3rd quadrant} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{1+\frac{-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{\frac{6-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}} \\ \\ \end{gathered}$

Answer:

$\cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}$

Checking:

$\begin{gathered} \frac{\theta}{2}=\cos ^{-1}(-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}) \\ \frac{\theta}{2}=118.22^{\circ} \\ \theta=236.44^{\circ}\text{ (3rd quadrant)} \end{gathered}$

Also,

$\csc \theta=\frac{1}{\sin\theta}=\frac{1}{\sin (236.44)}=-\frac{6}{5}\text{ QED}$

The answer is none of the choices

7 0

1 year ago

Find the domain: y= 6+9x 6− x−1

inna [77]

Answer:

Domain = {All real values of x EXCEPT x = -5 and x = 7}

Step-by-step explanation:

This is a rational function given as y=\frac{6+9x}{6-|x-1|}y=

6−∣x−1∣

6+9x

The domain is the set of all real value of x for which the function is defined.

For rational functions, we need to find which value of x makes the denominator equal to 0. We need to exclude those values from the domain.

Now

6 - |x-1| = 0

|x-1| = 6

x- 1 = 6

-(x-1) = 6

x = 6+1 = 7

and

-x+1=6

x = 1-6 = -5

So, the x values of -5 and 7 makes this function undefined. So the domain is the set of all real numbers except x = -5 and x = 7

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sukhopar [10]

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The <em>correct answer</em> is:

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