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algol13
3 years ago
12

Need help with these problems

Mathematics
1 answer:
Amanda [17]3 years ago
7 0
Use Photomath; it’s free for ios and android and shows you the steps to solve, too. i did 22
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Find all solutions to the equation: cos x = sin x
quester [9]
Divide both sides by cos[x] 

<span>You get 1 = Tan[x] </span>

<span>Tan[x] = 1 where x = 45 degrees {+ integer multiples of 180 degrees due to periodicity of tan} </span>

<span>x = 45 + 180*k OR Pi/4 + kPi, for k any integer</span>
7 0
3 years ago
Read 2 more answers
Please help as soon as possible. thanks in Advance
ivann1987 [24]

Answer:

Step-by-step explanation:

1.) (36-4) times 12-4= 380

2.) (7+4) times 6 divided by 3= 22

3.) 25 -3 times 7 divided by 3= 18

                  Not rocket science.

7 0
3 years ago
What's the answer to this question
zalisa [80]
This isn't a question without a question mark.
3 0
3 years ago
What value does the 2 represent in the number 52.3? ​
Vsevolod [243]

Answer:

2 in 52.3

Step-by-step explanation:

the two in 52.3 represents 2

4 0
3 years ago
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the point (3,-6) lies on the terminal side of angle 0. Find the exact value of the six trigonometric functions of 0. Sin cos tan
Shkiper50 [21]

ANSWER

See explanation

EXPLANATION

The point (3,-6) is the fourth quadrant.

In this quadrant only the cosine ratio and the secant ratio are positive.

The remaining four trigonometric ratios are negative.

The diagram is shown in the attachment.

We use the Pythagoras Theorem to find the hypotenuse, h.

{h}^{2}  =  {6}^{2}  +  {3}^{2}

{h}^{2}  = 36 + 9

{h}^{2}  = 45

h =  \sqrt{45}

h = 3 \sqrt{5}

\sin( \theta)  =   -  \frac{opp}{hyp}

\sin( \theta)  =   -  \frac{6}{3 \sqrt{5} }

\sin( \theta)  =   -  \frac{2 \sqrt{5} }{ 5 }

\sin( \theta)  =   -  \frac{2}{ \sqrt{5} }

\ cos(\theta)  =   \frac{adj}{hyp}

\ cos(\theta)  =   \frac{3}{3 \sqrt{5} }

\ cos(\theta)  =   \frac{1}{ \sqrt{5} }

\ cos(\theta)  =   \frac{ \sqrt{5} }{5}

\tan(\theta)= -  \frac{opp}{hyp}

\tan(\theta)= -  \frac{6}{3}  =  - 2

\csc(\theta)= -  \frac{hyp}{opp}

\csc(\theta)= -  \frac{3 \sqrt{5} }{6}  =  -   \frac{\sqrt{5} }{2}

\sec( \theta)  =   \frac{hyp}{adj}

\sec( \theta)  = \frac{3 \sqrt{5} }{3}  =  \sqrt{5}

\cot( \theta)  =  -  \frac{adj}{opp}

\cot( \theta)  =   - \frac{3}{6}  =  - \frac{1}{2}

6 0
3 years ago
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