If you're talking about the second part of the question where it asks you which facts you included in your answer, you can choose any of them and it will be counted right.

8 0

3 years ago

Point b has coordinates (-8,15) and lies on the circle whose equation is x^2+y^2=289. If an angles is drawn in standard position

Alenkinab [10]

Answer:

$\cos \theta=-\dfrac{8}{17}$

Step-by-step explanation:

Coordinates of Point b $=(-8,15)$

b lies on the circle whose equation is $x^2+y^2=289$

$x^2+y^2=17^2$

Comparing with the general form a circle with center at the origin: $x^2+y^2=r^2$

The radius of the circle =17 which is the length of the hypotenuse of the terminal ray through point b.

For an angle drawn in standard position through point b,

x=-8 which is negative

y=15 which is positive

Therefore, the angle is in Quadrant II.

$\cos \theta=\dfrac{Adjacent}{Hypotenuse} \\$Adjacent=-8\\Hypotenuse=17\\\cos \theta=\dfrac{-8}{17} \\\cos \theta=-\dfrac{8}{17}$

8 0

3 years ago

Let f(x) = 3x - 6 and g(x) = x - 2. Find f/g and its domain

Neko [114]

F(x) = 3x-6
g(x) = x-2

so f/g = 3(x-2)/(x-2) = 3

hope this will help you

7 0

3 years ago

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