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Ymorist [56]
4 years ago
10

I'd sure hope others know how to do this..

Mathematics
1 answer:
ivann1987 [24]4 years ago
5 0

Answer:

omgg they literally deleted 10 x 5

Step-by-step explanation:

it's 50, I'll be the calculator fam

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Evaluate sin 300° without using a calculator.
Anni [7]

Answer:

-\sqrt{3} /2

Step-by-step explanation:

300 degrees is in the fourth quadrant (it's between 270 and 360); sine is negative in the fourth quadrant.

Given we're in the fourth quadrant, the reference angle is 360 - 300 = 60 degrees

sin(60°) = \sqrt{3} /2

And since sine is negative, this value turns negative:

sin(300°) = -\sqrt{3} /2

3 0
3 years ago
Please help me!!!!!!!!!!!!!!!!!!!!!!!
MAXImum [283]

Answer:

120(I'm not definitely sure)

Step-by-step explanation:

IQR= Q3- Q1= 170 - 50 = 120

8 0
2 years ago
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jekas [21]

Answer:

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4 0
2 years ago
Read 2 more answers
The angle θ lies in Quadrant II .
Andreyy89

let's keep in mind that, in the II Quadrant, cosine is negative and sine, is positive.

cosine is adjacent/hypotenuse, however the hypotenuse is simply a radius unit, and thus is never negative, so in the -(2/3) the negative must be the numerator, -2.


\bf cos(\theta )=\cfrac{\stackrel{adjacent}{-2}}{\stackrel{hypotenuse}{3}}\impliedby \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{3^2-(-2)^2}=b\implies \pm\sqrt{5}=b\implies \stackrel{\textit{II Quadrant}}{+\sqrt{5}=b}~\hfill tan(\theta )=\cfrac{\stackrel{opposite}{\sqrt{5}}}{\stackrel{adjacent}{-2}} \\\\\\ ~\hspace{34em}

6 0
4 years ago
Solve these two by using factor the polynomial by grouping
Kruka [31]

Answer:

15. (8x^3 + 27)(x + 1)

17. (x^2 + 3)(x + 2)

Step-by-step explanation:

8x^3(x + 1) + 27(x + 1)

x^2(x + 2) + 3(x + 2)

3 0
3 years ago
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