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finlep [7]
3 years ago
9

The point (−3, 1) is on the terminal side of angle θ, in standard position. What are the values of sine, cosine, and tangent of

θ? Make sure to show all work.
Mathematics
1 answer:
Ivenika [448]3 years ago
6 0

The length of the hypotenuse = √(-3^2 +1^2) = √10

The point -3,1 tell us the length of y is 1 and the length of x is 3.

This would make opposite = 1 and adjacent = -3

Sinθ = opposite/hypotenuse =  1/√10  = √10/10

Cosθ = adjacent/hypotenuse = -3/√10  = - 3√10/10

Tanθ = opposite/adjacent = 1/-3 = -1/3

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Use photomath babe and if that doesn’t work idk i ain’t smart tbh
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Determine u-x and o-x from the given parameters of the population and the sample.size. round the answer to the nearest thousandt
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The u-x and o-x from the given parameters of the population and the sample size will be u =28 and standard deviation is 5.

<h3>How to calculate the values?</h3>

From the information given about the population and sample mean, the values include:

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The standard deviation will be:

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= 5/3.74

= 1.34

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2 years ago
a virus infected 400 people in the town and it spread and 25% more people each day. write an exponential function to model this
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f(10)= 400(1+0.25)^10

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3 0
4 years ago
Which of the following points would be a solution to this system of linear inequalities? ​
pychu [463]

Let's plug in both x and y values in the equation and check if the inequality is true...

\red{ \rule{35pt}{2pt}} \orange{ \rule{35pt}{2pt}} \color{yellow}{ \rule{35pt} {2pt}} \green{ \rule{35pt} {2pt}} \blue{ \rule{35pt} {2pt}} \purple{ \rule{35pt} {2pt}}

(4,1) \\ y \leqslant x + 1 \\ y <  -  \frac{x}{2}  - 1

Substitute x as 4 and y as 1

1 \leqslant 4 + 1 \\ 1 <  -  \frac{4}{2}  - 1

0  \leqslant  4 + 1 \\ 1 <  - 2 - 1

0 \leqslant 5 \\ 1 <  - 3

  • <em>This is not a solution because both are not true, Only one equation is giving true as answer which is not correct!~</em>

\purple{ \rule{300pt}{3pt}}

(0, - 3) \\ y \leqslant x + 1 \\ y <  -  \frac{x}{2}  - 1

Substitute x as 0 and y as -3

- 3 \leqslant 0 + 1 \\  - 3 <  -  \frac{0}{2}  - 1

- 3 \leqslant 1 \\  - 3 <  - 0 - 1

- 3 \leqslant 1 \\  - 3 <  - 1

  • <em>This is a solution because both are true, </em><em>Both the </em><em>equations</em><em> </em><em>are</em><em> </em><em>true</em><em> </em><em>which</em><em> </em><em>means</em><em> </em><em>the</em><em> </em><em>ordered</em><em> </em><em>pair</em><em> </em><em>is</em><em> </em><em>a</em><em> </em><em>solution</em><em>!</em><em>~</em>

<u>Thus</u><u>,</u><u> </u><u>Option</u><u> </u><u>B</u><u> </u><u>(</u><u>0</u><u>,</u><u>-</u><u>3</u><u>)</u><u> </u><u>is</u><u> </u><u>the</u><u> </u><u>correct</u><u> </u><u>choice</u><u>.</u><u>.</u><u>.</u><u>.</u>

6 0
3 years ago
A) f(-1)= -4<br> B) f(-1)= 3<br> C) f(-1)= 1/4<br> D)f(-1) = 1/3
rodikova [14]

Answer:

c  f(-1) = 1/4

Step-by-step explanation:

f(x) = x-1

       -----------

      x^2 -9

Let x=-1

f(-1) = -1-1

        -----------

         (-1)^2 -9

f(-1) = -2

        -----------

         1 -9

     =-2/-8   = 1/4

5 0
3 years ago
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