Αand β are angles in standard position whose terminal sides lie in quadrant ii. if cosα = -24/25 and sinβ = 3/5, find cos(α - β).
-4/25
3/5
4/5
117/125

The answer is 117/125

Step-by-step explanation:

The identity for the difference of 2 angles as far as cos goes is

cos(α-β) = cosα cosβ + sinα sinβ

We have both alpha and beta in QII.  In order to find the sinα, we need the missing leg, the one opposite the reference angle.  Applying Pythagorean's Theorem to that right triangle we get that the missing leg is +7 (y values are positive here so the 7 is positive since it is above the y = 0 line).  We also have to find the missing leg in beta so we can find the cosβ.  Applying Pythagorean's Theorem to that right triangle we get that the missing leg is -4 (negative because x is negative to the left of the origin).  Now that we have everything we need to fill in the identity, it looks like this:

Multiplying that out and then adding gives you

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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