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Diano4ka-milaya [45]
3 years ago
14

Factor. x^3 + 64y^3 !!!!!!!! will mark brainliest!!!

Mathematics
1 answer:
lorasvet [3.4K]3 years ago
5 0

The answer is the last one !!:)

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Work out m and c for the line y=2x+1​
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Answer:

Again... <em>m</em> = 2, and <em>c</em> = 1

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Which of the following descriptions could represent the Venn diagram?
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The angle 0 lies in Quadrant II . <br>Cos 0 = <br><img src="https://tex.z-dn.net/?f=%20-%20%20%5Cfrac%7B2%7D%7B3%7D%20" id="TexF
mel-nik [20]

Answer:

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

Step-by-step explanation:

Since we are in quadrant 2, sine is positive.  Since sine is positive and cosine is negative, then tangent is negative.

Now I'm going to find the sine value of this angle given using one of the Pythagorean Identities, namely \sin^2(\theta)+\cos^2(\theta)=1.

If given \cos(\theta)=\frac{-2}{3}, then we have \sin^2(\theta)+(\frac{-2}{3})^2=1 by substitution of \cos(\theta)=\frac{-2}{3}.

Let's solve:

\sin^2(\theta)+(\frac{-2}{3})^2=1 for \sin(\theta).

\sin^2(\theta)+\frac{4}{9}=1

Subtract 4/9 on both sides:

\sin^2(\theta)=1-\frac{4}{9}

Simplify:

\sin^2(\theta)=\frac{5}{9}

Square root both sides:

\sin(\theta)=\sqrt{\frac{5}{9}}

\sin(\theta)=\frac{\sqrt{5}}{\sqrt{9}}

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

===========

\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}=\frac{\frac{\sqrt{5}}{3}}{\frac{-2}{3}}

Multiplying top and bottom by 3 gives:

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

I'm going to move the factor of -1 to the top:

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

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3 years ago
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Good luck  ..........

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A football is kicked at 40 yards away from a goal post that is 10 feet high. Its path is modeled by y = -0.03x 2 + 1.6x, where x
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16>10

Yes, the ball clears the goal by 6 feet.
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4 years ago
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