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Karolina [17]
3 years ago
11

When Θ = 5 pi over 6, what are the reference angle and the sign values for sine, cosine, and tangent?

Mathematics
2 answers:
stira [4]3 years ago
3 0
Note that (5π)/6 radians = 150°. Therefore the given angle is in quadrant 2.

Refer to the figure shown below.

Reference angles are measured relative to the horizontal axis.
Therefore the reference angle in each quadrant is π/6 radians or 30°.
Denote the reference angle as θ'.
Then, in quadrant 1,
cos θ' = √3/2,  sin θ' = 1/2,  tan θ' = √3.

Because we are in quadrant 2,
  sin θ' = π/6;
  sin(5π/6) is positive, but cos (5π/6) and tan (5π/6) are negative.

 Answer:
5π/6 is in quadrant 2.
The reference angle, θ' = π/6.
sin(5π/6) is positive, cosine and tangent are negative.

jarptica [38.1K]3 years ago
3 0

Answer:

5π/6 is in quadrant 2.

The reference angle, θ' = π/6.

sin(5π/6) is positive, cosine & tangent are both negative

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Multiply (1.4 x 10^2) x (2.0 x 10^-5)
Mars2501 [29]

Answer:

2.8 x 10^-3

Step-by-step explanation:

5 0
3 years ago
2. Suppose 27 blackberry plants started growing in a yard. Absent constraint, the blackberry plants will spread by 80% a month.
Marta_Voda [28]

Explanation

The question indicates we should use a logistic model to estimate the number of plants after 5 months.

This can be done using the formula below;

\begin{gathered} P(t)=\frac{K}{1+Ae^{-kt}};A=\frac{K-P_{0_{}}}{P_0}_{} \\ \text{From the question} \\ P_0=\text{ Initial Plants=27} \\ K=\text{Carrying capacity =140} \end{gathered}

Workings

Step 1: We would need to get the value of A using the carrying capacity and initial plants that started growing in the yard.

This gives;

\begin{gathered} A=\frac{140-27}{27} \\ A=\frac{113}{27} \end{gathered}

Step 2: Substitute the value of A into the formula.

P(t)=\frac{140}{1+\frac{113}{27}e^{-kt}}

Step 3: Find the value of the constant k

Kindly recall that we are told that the plants increase by 80% after each month. Therefore, after one month we would have;

\begin{gathered} P(1)=27+(\frac{80}{100}\times27) \\ P(1)=\frac{243}{5} \end{gathered}

We can then have that after t= 1month

\begin{gathered} \frac{140}{1+\frac{113}{27}e^{-k\times1}}=\frac{243}{5} \\ Flip\text{ the equation} \\ \frac{1+\frac{113}{27}e^{-k}}{140}=\frac{5}{243} \\ 243(1+\frac{113}{27}e^{-k})=700 \\ 243+1017e^{-k}=700 \\ 1017e^{-k}=700-243 \\ 1017e^{-k}=457 \\ e^{-k}=\frac{457}{1017} \\ -k=\ln (\frac{457}{1017}) \end{gathered}

Step 4: Substitute -k back into the initial formula.

\begin{gathered} P(t)=\frac{140}{1+\frac{113}{27}e^{\ln (\frac{457}{1017})t}} \\ =\frac{140}{1+\frac{113}{27}(e^{\ln (\frac{457}{1017})})^t} \\ P(t)=\frac{140}{1+\frac{113}{27}(\frac{457}{1017}^{})^t} \\  \end{gathered}

The above model is can be used to find the population at any time in the future.

Therefore after 5 months, we can estimate the model to be;

\begin{gathered} P(5)=\frac{140}{1+\frac{113}{27}(\frac{457}{1017}^{})^5} \\ P(5)=\frac{140}{1.07668} \\ P(5)=130.029\approx130 \end{gathered}

Answer: The estimated number of plants after 5 months is 130 plants.

8 0
1 year ago
5/3x−2y=−2 −x+y=2<br> What is the solution of the system of equations?
gtnhenbr [62]

Answer:

  (x, y) = (-6, -4)

Step-by-step explanation:

The solution can be found quickly using a graphing calculator. It is the point of intersection of the two lines.

_____

If you like, you can solve the system algebraically. The second equation can be rewritten to give ...

  y = x +2

Substituting this into the first equation gives ...

  5/3x -2(x +2) = -2

  -1/3x -4 = -2

Adding 4 and multiplying by -3 gives ...

  x = -3(-2+4) = -6

Then our equation for y tells us ...

  y = -6 +2 = -4

The solution is (x, y) = (-6, -4).

5 0
3 years ago
Help me of i die so pls help me
nekit [7.7K]

Answer:

10

100

1000

Step-by-step explanation:

I had a quiz on that the other day got 100%

7 0
3 years ago
I just need somebody to double check my answers. I’m pretty sure I’m right but I don’t really know.
Fantom [35]

Answer:

<h2>Yeah, all answer is right</h2>

<h2>I HOPE IT IS HELPFUL</h2>
4 0
3 years ago
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