The first polynomial is $y -4yz^2-3$ and let the second polynomial be P(y,z).

If the sum of two polynomials is $-yz^2-3z^2-4y+4$ , then $y-4yz^2-3+P(y,z)=-yz^2-3z^2-4y+4$ .

Take the first polynomial from the left side to the right side:

$P(y,z)=-yz^2-3z^2-4y+4 -(y-4yz^2-3),\\ P(y,z)=-yz^2-3z^2-4y+4 -y+4yz^2+3,\\ P(y,z)=3yz^2-3z^2-5y+7$ .

Answer: the second polynomial is $3yz^2-3z^2-5y+7$ .

3 0

4 years ago

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Why is circle 1 similar to circle 2

Kobotan [32]

Your answer is letter A.

Hope this helps :)

8 0

3 years ago

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The length of a rectangle is 12 in. and the perimeter is 56 in. find the width of the rectangle

pentagon [3]

The perimeter p of the rectangle with length l and width w is:
p = 2l + 2w
that is, two times its length and two times its width, lets solve for the width w and substitute known data:
p = 2l + 2w
2w = p - 2l
w = (p - 2l)/2
w = (56 - (2*12))/2
w = (56 - 24)/2
w = 32/2
w = 16
therefore the width of the rectangle is 16 in

8 0

3 years ago