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Artist 52 [7]
3 years ago
14

Trey needs to make four batches of this brownie recipe.

Mathematics
1 answer:
Korolek [52]3 years ago
5 0
We need the background info
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which function has real zeros at x = −10 and x = −6? f(x) = x2 16x 60 f(x) = x2 − 16x 60 f(x) = x2 4x 60 f(x) = x2 − 4x 60
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Answer:

Option A is correct

The function x^2+16x+60 has real zeroes at x =-10 and x =-6

Explanation:

Given: The real zeroes or roots are x = -10, and x = -6

To find the quadratic function of degree 2.

x^2- (\alpha+\beta)x + \alpha\beta =0 where α,β are real roots.   ....[1]

Here, α= -10  and β= -6

Sum of the roots:

α+β =  -10+(-6) = -10-6 = -16

Product of the roots:

αβ = (-10)(-6)= 60

Substitute these value in equation [1] we have;

x^2-(-16)x+60 = x^2+16x+60

Therefore, the quadratic function for the real roots at x =-10 and x =-6 ;

x^2+16x+60

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