A polynomial has zeros of 7, 6. What is the minimum degree of the polynomial?

1 answer:

4 0

If the polynomial has zeros at 7 and 6 the equation will be
(X-7)(x-6)
When we distribute that we get
x^2-13x+42

A degree is the highest exponent of a polynomial. In this case the x^2. So the answer is 2

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Which inequality would result in the shaded solution on the unit circle to the right?

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<h3>Answer: Choice B</h3>

Explanation:

Cosine is positive in quadrants I and IV, but quadrant IV isn't shaded in so we can rule out choice A.

Sine is positive in quadrants I and II. So far it looks like choice B could work. In fact, it's the answer because sin(pi/6) = 1/2 and sin(5pi/6) = 1/2. So if 0 ≤ sin(x) < 1/2, then we'd shade the region between theta = 0 and theta = pi/6; as well as the region from theta = 5pi/6 to theta = pi.

Choice C is ruled out because tangent is positive in quadrants I and III, but quadrant III isn't shaded.

Choice D is ruled out for similar reasoning as choice A. Recall that $\sec(x) = \frac{1}{\cos(x)}$