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Vlad [161]
3 years ago
11

How many solutions per each equation?

Mathematics
1 answer:
Oksi-84 [34.3K]3 years ago
4 0
Assuming y=ax²+bx+c, calculate D = b²-4ac, D<1: no solutions; D=0: 1 solution; D>0: 2 solutions.

1-4*-3*12 = 145, so 2 solutions
36-4*2*5 = -4, no solutions
49-4*-11 = 5, so 2 solutions
64-4*-1*-16 = 0, so 1 solution
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Step-by-step explanation:

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3 years ago
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Use long division to determine the quotient of the polynomials. (x3 – 7x – 6) ÷ (x – 4) Which of the following options best desc
Lelechka [254]

Answer:

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Step-by-step explanation:

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4 0
3 years ago
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4 years ago
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