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2 years ago

PLEASE HELP

Mathematics

1 answer:

erastovalidia [21]2 years ago

5 0

The necessary solution that would be used to fill in the blanks would be

perpendicular
is the center of
Circumscribed

<h3>How to find the center of the circle</h3>

The way that Quintin would be able to do this would be the center of the circumferential triangle is equidistant from the vertices.

The center is going to be the intersection of the bisectors of the three sides of the given triangle.

Read more on the center of a circle here:

brainly.com/question/25938130

#SPJ1

perpendicular bisectors, is the center of, circumscribed

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<u>Step-by-step explanation:</u>

a) $\frac{15\pi}{4}$ - $\frac{8\pi}{4}$ = $\frac{7\pi}{4}$ , which is located in Quadrant IV <em>per the Unit Circle. </em>

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c) the reference angle is the angle from $\frac{7\pi}{4}$ to 2π = $\frac{\pi}{4}$

d) since the angle is below the x-axis and the reference angle is $\frac{\pi}{4}$ , then the angle is equal to $-sec(\frac{\pi}{4}$ )

e) sec = $\frac{1}{cos}$ ⇒ sec $\frac{7\pi}{4}$ = $\frac{2}{\sqrt{2} }$ = $\frac{2}{\sqrt{2} }$ $(\frac{\sqrt{2}} {\sqrt{2}})$ = $\frac{2\sqrt{2} }{2}$ = $\sqrt{2}$

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Answer: (a) $\frac{3\pi}{2}$ (b) (0, -1) (c) A (d) -1

<u>Step-by-step explanation:</u>

a) $\frac{11\pi}{2}$ - $\frac{4\pi}{2}$ = $\frac{7\pi}{2}$ - $\frac{11\pi}{2}$ = $\frac{3\pi}{2}$

b) $\frac{3\pi}{2}$ is on the Unit Circle at (0, -1)

c) sin = $\frac{opposite}{hypotenuse}$ which equals $\frac{y}{r}$ on the Unit Circle.

d) sin is the y-coordinate. sin $(\frac{3\pi}{2})$ = -1

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