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Sedaia [141]
3 years ago
14

Consider point M.

Mathematics
1 answer:
Ksenya-84 [330]3 years ago
6 0

Answer: someone put a answer

Step-by-step explanation:

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a successful music app tracked the number of downloads each day for 4 artists, represented by lines l, j, m and d over the cours
Olegator [25]

Answer:

l

Step-by-step explanation:

5 0
3 years ago
Cos(theta) = - 3/4 and is in the 3rd quadrant, find the following:
elena-14-01-66 [18.8K]

Answer:

\begin{gathered} sin(\theta)=\frac{-\sqrt{7}}{4} \\ cos(\theta)=-\frac{3}{4} \\ tan(\theta)=\frac{\sqrt{7}}{3} \\ csc(\theta)=-\frac{4}{\sqrt{7}} \\ sec(\theta)=-\frac{4}{3} \\ cot(\theta)=\frac{3}{\sqrt{7}} \end{gathered}

Step-by-step explanation:

If theta is in the third quadrant, draw the diagram to easily identify the other trigonometric relations:

Solve for the missing leg of the triangle, using the Pythagorean theorem:

\begin{gathered} \text{ adjacent}^2+\text{ opposite}^2=\text{ hypotenuse}^2 \\ -3^2+\text{ opposite}^2=4^2 \\ \text{ opposite=}\sqrt{16-9} \\ \text{ opposite=}\sqrt{7} \end{gathered}

Therefore, for the trigonometric relationships:

\begin{gathered} \text{ sin\lparen}\theta)=\frac{opposite}{hypotenuse} \\ \text{ cos\lparen}\theta)=\frac{adjacent}{hypotenuse} \\ tan(\theta)=\frac{opposite}{adjacent} \\ csc(\theta)=\frac{hypotenuse}{opposite} \\ sec(\theta)=\frac{hypotenuse}{adjacent} \\ cot(\theta)=\frac{adjacent}{opposite} \end{gathered}

Now, substitute and solve for the relations:

\begin{gathered} sin(\theta)=\frac{-\sqrt{7}}{4} \\ cos(\theta)=-\frac{3}{4} \\ tan(\theta)=\frac{\sqrt{7}}{3} \\ csc(\theta)=-\frac{4}{\sqrt{7}} \\ sec(\theta)=-\frac{4}{3} \\ cot(\theta)=\frac{3}{\sqrt{7}} \end{gathered}

7 0
1 year ago
Monkeeeeeeeeeeeeeeeeeeeeeee also help
alex41 [277]

\frac{9}{27}  =  \frac{1}{3}

This is the simplest form of the fraction.

7 0
3 years ago
Read 2 more answers
I really need to know how long it takes the large pipe on its own
Andreyy89

s = hours it takes the small pipe on its own.


L = hours it takes the large pipe on its own.


we know the small pipe can do it in 11 hours total, so s = 11, so, how much of the whole job has the small done in 1 hour only?


well, if it takes 11 hours to do the whole thing, in 1 hour it has done only 1/11 of the whole job.


we know both pipes working together can do it in 6 hours flat. So in 1 hour only, both of them have done only 1/6 of the whole thing.


in that 1 hour, how much has the large one done? well, it has only done 1/L of the job.


\bf \stackrel{\textit{how much has it been done by both in 1 hour}}{\stackrel{\stackrel{small}{rate}}{\cfrac{1}{s}}+\stackrel{\stackrel{large}{rate}}{\cfrac{1}{L}}~~=~~\stackrel{job}{\cfrac{1}{6}}} \\\\\\ \cfrac{1}{11}+\cfrac{1}{L}=\cfrac{1}{6}\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{66L}}{6L+66=11L}\implies 66=5L \\\\\\ \cfrac{66}{5}=L\implies 13\frac{1}{5}=L\impliedby \textit{13 hours and 12 minutes}


and you can round that up as needed.

7 0
4 years ago
Where is the vertex of y=3(x-26)^2+14?
Morgarella [4.7K]
The vertex is (26,14)

Because it’s written in vertex form which y=a(x-h)^2 + k. H is the x for the vertex And h is changed to it’s opposite. So since 26 is negative it would actually be positive 26. While k ls the y of the vertex which is 14 in our case
3 0
3 years ago
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