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Thank you!

That is pretty Good going!

Congratulations!!

We can say x+11=y and then write log(y)=(2^10)/(2^y). From this we can observe that lince log is log_10, that if y is 10, then both sides have a value of 1. Since x+11=y, x=y-11, and x=-1. This is the way I would've done it without graphing but I am sure there might be an easier version to understand for solving this.

What I would do is take the cases with the red shirts and the cases with the green shirts seperately. For the red shirts, you have 5*8*7 possibilities of outfits and for the green shirts, you have 5*8*8 possibilities of outfits. The total is 280+320=600 total outfits.

So your answer is 600 total outfits.

Hope this helps you understand!

I know what formula you are after but I can never remember  them of the top of my head so...

Let

\(\theta = acos(\frac{-1}{9})\\ \quad note: cos \theta = \frac{-1}{9}\\ find \;\;sin\frac{\theta }{2}\)

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\(cos(\frac{\theta}{2}+\frac{\theta}{2})=cos^2\frac{\theta}{2}-sin^2\frac{\theta}{2}\\ cos(\frac{\theta}{2}+\frac{\theta}{2})=1-sin^2\frac{\theta}{2}-sin^2\frac{\theta}{2}\\ cos\theta=1-2sin^2\frac{\theta}{2}\\ \frac{1-cos\theta}{2}=sin^2\frac{\theta}{2}\\ sin\frac{\theta}{2}=\pm \sqrt{\frac{1-cos\theta}{2}}\\ sin\frac{\theta}{2}=\pm \sqrt{\frac{1-\frac{-1}{9}}{2}}\\ sin\frac{\theta}{2}=\pm \sqrt{\frac{\frac{10}{9}}{2}}\\ sin\frac{\theta}{2}=\pm \sqrt{\frac{10}{18}}\\ sin\frac{\theta}{2}=\pm\sqrt{\frac{5}{9}}\\ sin\frac{\theta}{2}=\pm \frac{\sqrt5}{3}\\ sin\left[\frac{acos(\frac{-1}{9})}{2}\right]=\pm\frac{\sqrt5}{3}\\\)

Theta has to be in the second quadrant but theta/2 can be in the first or second quadrent .

Either way.   sin(theta) is positive!

\(\boxed{sin\left[\frac{acos(\frac{-1}{9})}{2}\right]=\frac{\sqrt5}{3}}\)

LaTex:

cos(\frac{\theta}{2}+\frac{\theta}{2})=cos^2\frac{\theta}{2}-sin^2\frac{\theta}{2}\\
cos(\frac{\theta}{2}+\frac{\theta}{2})=1-sin^2\frac{\theta}{2}-sin^2\frac{\theta}{2}\\
cos\theta=1-2sin^2\frac{\theta}{2}\\
\frac{1-cos\theta}{2}=sin^2\frac{\theta}{2}\\
sin\frac{\theta}{2}=\pm \sqrt{\frac{1-cos\theta}{2}}\\
sin\frac{\theta}{2}=\pm \sqrt{\frac{1-\frac{-1}{9}}{2}}\\
sin\frac{\theta}{2}=\pm \sqrt{\frac{\frac{10}{9}}{2}}\\
sin\frac{\theta}{2}=\pm \sqrt{\frac{10}{18}}\\
sin\frac{\theta}{2}=\pm\sqrt{\frac{5}{9}}\\
sin\frac{\theta}{2}=\pm \frac{\sqrt5}{3}\\
sin\left[\frac{acos(\frac{-1}{9})}{2}\right]=\frac{\sqrt5}{3}\\

Nice job. :DDD

=^._.^=

I figured it out! Thanks for your help guys. I figured out that I could use the trig identity that sin(theta/2)=+-sqrt((1-cos(theta))/2). Then, I substitued cos(theta) with -1/9 and got sin(theta/2)=+-sqrt(5/9). 

P.S. I am so relieved I got this right because I spent so long trying to figure it out!

I could use a unit circle and then use the Pythagorean Theorem to say sin(theta)^2+(1/9)^2=1^2 or sin(theta)^2=80/81. This leads to sin(theta)=(sqrt80)/9.

Am I correct because I feel like I did something wrong? 

And if I got sin(theta) right, how can I evaluate sin(theta/2)?

Find all values of c such that c/(c - 5) = 4/(c + 4).  

Cross multiply                     c • (c + 4)  =  4 • (c – 5)  

Multiply it out                           c² + 4c  =  4c – 20  

Subtract 4c from both sides            c²  =  –20

Square root of both sides                  c  =  +sqrt(–20)  

Find all values of c such that c/(c - 5) = 4/(c + 4).

Hello Guest!

\(\frac{c}{c-5}=\frac{4}{c+4}\\ c^2+4c=4c-20\\ c^2=-1\cdot 2^2\cdot 5\)

\(c=2i \sqrt{5}\)

  !

thank you, may I just ask how you got to know that the side of the second triangle is 4/cos 15

First triangle:

A = a²(√3 / 4) 

Second triangle: 

Side = 4 / cos15º

Area = a²(√3 / 4)

The rest of this problem is entirely up to you. Good luck! 

I gave you a plus for your concept. Kudos! 

There is a circle in the coordinate plane. If the radius doubles, the area increases by 10 units. Find the new radius of the circle.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Radius => r             New radius => 2r

A = r²pi                    A_n = (2r)²pi             (A_n = A + 10)

A_n - A = 10

(2r)²pi - r²pi = 10

Responded in first post. 

=^._.^=

Adding on to what Melody said. :))

Let theta = arccos(-1/9)

cos(theta) = -1/9

Note that the range of arccos is [-pi/2, pi2].

Can you figure out what sin(theta) is?

Hint: look at Melody's hint with the triangle and see if you can construct another one. 

=^._.^=

Thank you, it's always nice to see the problem in a graph. :))

=^._.^=

I don't have time to answer now. (even for myself) but I would probably start by letting 

theta = acos (-1/9)

I will get back to it later/

Oh sorry.

I didn't notice the brackets properly.  My fault.