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3 years ago

Subtract 8 y^2 − 5 y + 7 from 2 y^2 + 7 y + 1 1 The answer is: −6y ^2 +12y+4

Mathematics

1 answer:

lys-0071 [83]3 years ago

4 0

Answer:

$\large\boxed{-6y^2+12y+4}$

Step-by-step explanation:

$(2y^2+7y+11)-(8y^2-5y+7)\\\\=2y^2+7y+11-8y^2-(-5y)-7\\\\=2y^2+7y+11-8y^2+5y-7\qquad\text{combine like terms}\\\\=(2y^2-8y^2)+(7y+5y)+(11-7)\\\\=-6y^2+12y+4$

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2 years ago

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2 years ago

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3 years ago

For what value of the constant c is the function fcontinuous on (−[infinity], [infinity])?

Arlecino [84]

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For $c=\frac{1}{7}$ the function f(x) is continuous on $(-\infty,\infty)$ .

Step-by-step explanation:

We have the following function

$f(x) = \left\{ \begin{array}{ll} cx^2+5x & \quad x$

For the function f(x) to be continuous on $(-\infty,\infty)$ it is sufficient to have continuity at x = 6, we need to ensure that as x approaches 6, the left and right limits match, this means that

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2 years ago

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3 years ago