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777dan777 [17]
3 years ago
14

Factorize this pls!Thank you!

Mathematics
2 answers:
Kitty [74]3 years ago
4 0

Let me write the polynomial in the usual order (decreasing powers), just for a matter of habit:

6s^6+18s^4-9s^3

As for the numeric part, we can factor a 3, because it's the largest common divisor of 6, 9 and 18:

6s^6+18s^4-9s^3 = 3(2s^6+6s^4-3s^3)

For the variable part, we can factor a s^3, because it's the power with the smaller exponent appearing in all terms:

3(2s^6+6s^4-3s^3) = 3s^3(2s^3+6s-3)

Alisiya [41]3 years ago
3 0
Can i have brainlieas if my answer true

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Hunter-Best [27]

Since \alpha lies in quadrant II and \beta lies in quadrant IV, we expect \sin\alpha>0, \cos\alpha, and \sin\beta.

Recall the Pythagorean identities,

\sin^2x+\cos^2x=1\iff1+\cot^2x=\csc^2x\iff\tan^2x+1=\sec^2x

It follows that

\sec\alpha=\dfrac1{\cos\alpha}=-\sqrt{\tan^2\alpha+1}=-\dfrac{13}5\implies\cos\alpha=-\dfrac5{13}

\sin\alpha=\sqrt{1-\cos^2\alpha}=\dfrac{12}{13}

\sin\beta=-\sqrt{1-\cos^2\beta}=-\dfrac45

Recall the angle sum identity for sine:

\sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha

So we have

\sin(\alpha+\beta)=\dfrac{12}{13}\dfrac35+\left(-\dfrac45\right)\left(-\dfrac5{13}\right)=\boxed{\dfrac{56}{65}}

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3 years ago
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Must click thanks and mark brainliest

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3 years ago
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