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3 years ago

11

HELP HELPP! GIVING LOTS OF POINTS!!!! nonsense-report

1 answer:

jonny [76]3 years ago

8 0

We can organize the data by making a board (i can't draw one so i'll do it in another way)

First, we have to see how many different values are there :
19 ; 21 ; 22; 23 ; 24 ; 25 ; 26 ; 27 ; 28 ; 29 ; 30 ; 31 ; 32 ; 34 ; 35 ; 36 ; 37 ; 39 ; 40 ; 43 ; 44 ;45 ; 46

Frequency :
19 : 1 time
21 : 3 times
22 : 1 time
23 : 1 time
24 : 5 times
25 : 3 times
26 : 2 times
27 : 4 times
28 : 3 times
29 : 2 times
30 : 2 times
31 : 3 times
32 : 2 times
34 : 2 times
35 : 4 times
36 : 2 times
37 : 1 time
39 : 1 time
40 : 1 time
43 : 2 times
44 : 1 time
45 : 1 time
46 : 1 time
47 : 1 time

(that was so long, now you can construct the board yourself right xd)

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Option D.

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3 years ago

Find out the answers to this

meriva

Answer:

$\tan \theta = - \frac{1}{5} = - 0.2$

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Step-by-step explanation:

It is given that $\cot \theta = - 5$ and $\theta$ is in the fourth quadrant.

So, only $\cos \theta$ will have positive value and $\sin \theta$ , $\tan \theta$ will have negative value.

Now, $\cot \theta = - 5$

⇒ $\tan \theta = \frac{1}{\cot \theta} = -\frac{1}{5}$ (Answer)

We know, that $\sec^{2} \theta - \tan^{2} \theta = 1$

⇒ $\sec \theta = \sqrt{1 + \tan^{2} \theta } = \sqrt{1 + (- \frac{1}{5} )^{2} } = 1.019$

{Since, $\cos \theta$ is positive then $\sec \theta$ will also be positive}

⇒ $\cos \theta = \frac{1}{\sec \theta} = \frac{1}{1.0198} = 0.98$ (Answer)

We know, that $\csc^{2} \theta - \cot^{2} \theta = 1$

⇒ $\csc \theta = - \sqrt{1 + \cot^{2} \theta } = - \sqrt{1 + (- 5 )^{2} } = - 5.099$

{Since, $\sin \theta$ is negative then $\csc \theta$ will also be negative}

⇒ $\sin \theta = \frac{1}{\csc \theta} = \frac{1}{- 5.099} = - 0.196$ (Answer)

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3 years ago