Y² - 12y + 32

The middle term must be written in such a way that the sum of two numbers is -12y and the product is 32y²

The two such numbers are -4y and -8y

So, we can write:

y² - 4y - 8y + 32

= y(y - 4) - 8(y - 4)

= (y - 4)(y - 8)

So, option B gives the correct answer

4 0

3 years ago

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Simplify <img src="https://tex.z-dn.net/?f=sin%20x%5E%7B2%7D%20%28x%29cos%20x%5E%7B2%7D%20%28x%29-cos%20x%5E%7B2%7D%20%28x%29" i

zhenek [66]

$sin^2(x)cos^2(x)-cos^2(x)\\\\=sin^2(x)\cdot cos^2(x)-1\cdot cos^2(x)\\\\=cos^2(x)\cdot(sin^2(x)-1)\\\\=cos^2(x)\cdot[-(1-sin^2(x)]^{(*)}\\\\=cos^2(x)\cdot[-cos^2(x)]^{(*)}\\\\\boxed{=-cos^4(x)}\\------------------\\(*)\ used:\\\\sin^2(x)+cos^2(x)=1\to cos^2(x)=1-sin^2(x)$

4 0

3 years ago

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61 = m + - 8 what would be m

Zigmanuir [339]

Answer:

m would be 69!

Step-by-step explanation:

61 = m - 8

61 + 8 = m - 8 + 8

69 = m

7 0

3 years ago

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