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3 years ago

12

How to solve -7(x-1)^2-2 vertex form to standard form

1 answer:

ahrayia [7]3 years ago

4 0

Expand it

$-7(x-1)^2-2$
$-7(x^2-2x+1)-2$
$-7x^2+14x-7-2$
$-7x^2+14x-9$ is standard form
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3 years ago

Read 2 more answers

HELP WITH THESE QUESTIONS PART 2.

polet [3.4K]

Answer: (a) IV (b) positive (c) $\frac{\pi}{4}$ (d) C (e) $\sqrt{2}$

<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>

b) sec is $\frac{1}{cos}$ . Cos is the x-coordinate. The x-coordinate inQuadrant IV is positive.

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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