If sin θ = square root of three over two, which could not be the value of θ?
1 answer:
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Answer:
f(x) = 12x³ - 19x² + 17x - 17
Step-by-step explanation:
We can the numerator by multiplying the denominator by the quotient, and then adding the remainder.
We're given a denominator of 4x - 5
a quotient of 3x² - x + 3
and a remainder of -2
so the original numerator is:
(4x - 5)(3x² - x + 3) - 2
= 12x³ - 4x² + 12x - 15x² + 5x - 15 - 2
= 12x³ - 19x² + 17x - 17
Let's test that by dividing it by 4x -5 with long division:
That matches, so we know our answer's right, and:
f(x) = 12x³ - 19x² + 17x - 17
Answer:
a) (-9, 4)
b) (0, 0)
c) No
Step-by-step explanation:
