What is anything to the zero power

For the following problem, assume that all given angles are in simplest form, so that if A is in QIV you may assume that 270° &l

d1i1m1o1n [39]

Answer:

Step-by-step explanation:

Given that angle A is in IV quadrant

So A/2 would be in II quadrant.

sin A = -1/3

cos A = $\sqrt{1-sin^2 A} =\frac{2\sqrt{2} }{3}$

(cos A is positive since in IV quadrant)

Using this we can find cos A/2

$cosA = 2cos^2 \frac{A}{2} -1\\Or cos \frac{A}{2} =-\sqrt{\frac{1+cosA}{2} } =-\sqrt{\frac{3+2\sqrt{2} }{6} }$

3 0

3 years ago

Tan(-212°)=____. 1. -cot 32° 2. tan 32° 3. cot 32° 4. -tan 32°

Marrrta [24]

Answer:

4. $-\tan 32$ °

Step-by-step explanation:

Given:

The tangent of the angle is given as:

$\tan (-212\°)$

Since the angle is negative, it means that it is measured in the clockwise direction as angles measured in clockwise direction are negative and that measured in counter clockwise directions are positive.

-212° when measured from counter clockwise direction will be equal to:

$-212\°(Clockwise)=360-212=148\°(CCW)$

Therefore, $\tan (-212\°)$ = $\tan (148\°)$

Now, we have the identity for tan as:

$\tan(\pi-\theta)=-\tan \theta$

Here, $\theta=148\°$

Therefore,

$-\tan (148\°)=\tan(180-148)\\-\tan(148\°)=\tan(32\°)\\\textrm{Multiplying both sides by -1, we get:}\\-1(-\tan(148\°)=-1\times \tan(32\°)\\\\\tan(148\°)=-\tan(32\°)$

Hence, $\tan (-212\°)$ = $\tan (148\°)$ = $-\tan 32$ °

5 0

3 years ago

Write an ordered pair for this point (The point is circled in hot pink)

aleksley [76]

Answer:

(-4,0)

Step-by-step explanation:

The circled point is 4 spaces away from the center and it is on the the x-axis so that's why the answer would be (-4,0)

Your welcome! Brainliest???

:D

8 0

3 years ago

Solve the equation 56 ÷ (-8)= ?

Hunter-Best [27]

Answer:

-7

Step-by-step explanation:

Because for dividing one positive and one negative number, the rule is that the answer is going to be negative. So, what you can do is just do 56÷8 which equals 7 then add the negative sign to make it -7.

3 0

3 years ago

Find center,foci, and vertices of ellipse (x+3)^2/21+(y-5)^2/25=1

sasho [114]

I don't know if we can find the foci of this ellipse, but we can find the centre and the vertices. First of all, let us state the standard equation of an ellipse.

(If there is a way to solve for the foci of this ellipse, please let me know! I am learning this stuff currently.)

$\frac{(x-x_{1})^2}{a^2}+ \frac{(y-y_{1})^2}{b^2}=1$

Where $(x_{1},y_{1})$ is the centre of the ellipse. Just by looking at your equation right away, we can tell that the centre of the ellipse is:

$(-3,5)$

Now to find the vertices, we must first remember that the vertices of an ellipse are on the major axis.

The major axis in this case is that of the y-axis. In other words,

$b^2>a^2$

So we know that b=5 from your equation given. The vertices are 5 away from the centre, so we find that the vertices of your ellipse are:

$(-3,10)$
& $(-3,0)$

I really hope this helped you! (Partially because I spent a lot of time on this lol)

Sincerely,

~Cam943, Junior Moderator

6 0

3 years ago