Subtract and simplify<br> (y^2 – 3y – 5) - (-y^2

What is 8/9 to the second power

densk [106]

Answer:

64/81

The drawing will help

7 0

3 years ago

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Which type of transformation preserves the symmetry of an even function f(x) but does not preserve the symmetry of an odd functi

Alexus [3.1K]

The type of transformation that preserves the symmetry of an even function f(x) but does not preserve the symmetry of an odd function g(x) is vertical translation.

<h3>What is vertical translation?</h3>

Vertical translation of a graph is done by moving the base graph up or down in the y-axis direction. Each point on a graph is moved k units vertically to translate the graph by that many units.

The vertical translation is the movement of the curve along the y-axis by a certain number of units without altering the function's shape or domain.

The function's form is preserved in the case of vertical translation.

Learn more about Vertical Translation here:

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7 0

2 years ago

EXPERTS/ACE/TRUSTED HELPERS

LekaFEV [45]

The sign of each Y coordinate changed

3 0

3 years ago

Please help? I’m super lost...

babunello [35]

Answer:

Step-by-step explanation:

In all of these problems, the key is to remember that you can undo a trig function by taking the inverse of that function. Watch and see.

a. $sin2\theta =-\frac{\sqrt{3} }{2}$

Take the inverse sin of both sides. When you do that, you are left with just 2theta on the left. That's why you do this.

$sin^{-1}(sin2\theta)=sin^{-1}(-\frac{\sqrt{3} }{2} )$

This simplifies to

$2\theta=sin^{-1}(-\frac{\sqrt{3} }{2} )$

We look to the unit circle to see which values of theta give us a sin of -square root of 3 over 2. Those are:

$2\theta =\frac{5\pi }{6}$ and

$2\theta=\frac{7\pi }{6}$

Divide both sides by 2 in both of those equations to get that values of theta are:

$\theta=\frac{5\pi }{12},\frac{7\pi }{12}$

b. $tan(7a)=1$

Take the inverse tangent of both sides:

$tan^{-1}(tan(7a))=tan^{-1}(1)$

Taking the inverse tangent of the tangent on the left leaves us with just 7a. This simplifies to

$7a=tan^{-1}(1)$

We look to the unit circle to find which values of <em>a</em> give us a tangent of 1. They are:

$7\alpha =\frac{5\pi }{4},7\alpha =\frac{\pi }{4}$

Dibide each of those equations by 7 to find that the values of alpha are:

$\alpha =\frac{5\pi}{28},\frac{\pi}{28}$

c. $cos(3\beta)=\frac{1}{2}$

Take the inverse cosine of each side. The inverse cosine and cosine undo each other, leaving us with just 3beta on the left, just like in the previous problems. That simplifies to:

$3\beta=cos^{-1}(\frac{1}{2})$