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There are $\binom{12}{5}$ ways to choose all the balls.

There are $\binom{3}{2}$ ways to catch the red ball, and $\binom{3}{1}$ ways to catch each of the other ones.

The answer is $\frac{\binom{3}{2}\left(\binom{3}{1}\right)^3}{\binom{12}{5}}=\frac{81}{792}=\boxed{\frac{9}{88}}$.

You cannot have any  ZERO denominators...

so    x^2   cannot = 7              x cannot   =  +- sqrt 7

         x^3 cannot equal 8        x  cannot  = 2

        x^4 cannot equal 9         x  cannot  = +- sqrt 3                sum = 2 

It's obviously tangent at $(k/2, k/2)$. So $\frac{k^2}{2}=2k+1\implies k^2=4k+2\implies k^2-4k-2=0\implies k=2\pm\sqrt{6}$. We take positive root, so $\boxed{k=2+\sqrt{6}}$.

At $3$, it must be continuous. So $2*3+a=3+2\implies \boxed{a=-1}$.

yes just like that

Let's look at the right hand side: $\frac{2x(x-1)-6(x+1)}{x^2-1}=\frac{2x^2-2x-6x-6}{x^2-1}=\frac{2x^2-8x-6}{x^2-1}$. So $2x^2-8x-6=-9x\implies 2x^2+x-6=0$, so the sum of the solutions is $\boxed{-\frac12}$ by Vieta's formulae.

The answer is actually 6.

We claim that it's $\boxed{x=1}$.

Notice that $f(1)=f(1)$, and $f(1+k)=f(1-k)$, implying the answer.

The minimum value of $x^2$ is $0$, the maximum is $81$. The answer is just $\boxed{81}$.

I'm not sure what the best method is either, but it says: each test must be made beforehand. Thus, your strategy wouldn't work because your tests are dependent on previous tests.

25?

I dont really understand your sentence

p.s plz use commas annd/or periods. 

What it is five times five quite a big question I want you to answer it using a calculated if you half to because sometimes it gets tricky to have a fun time with this equation don't forget to like it.

I think you wrote $x=2x+5$? 

If so, subtract $2x$ from both sides to get $-x=5$. From here, $\boxed{x=-5}$.

Yep, sorry about that.

$19:8^2=64$

$p:3^2=9$

We are multiplying by $\frac{9}{64}$, and the inverse is $\frac{64}{9}$.

So $p=\boxed{\frac{1216}{9}}$.

Let's have him get his first $4$ pairs from white, red, blue, rainbow. Then we have to pick as many socks as possible without getting another pair. I will take a green sock, a purple sock, and another white, red, blue, and rainbow sock. Now, the next sock I pick, I am guaranteed to get another pair. So far, I have picked $4*2+6=14$ socks. The answer is $14+1=\boxed{15}$ by the pigeonhole principle. 

I like $y=mx+b$ form better, so I'll use that. We have:

$11=m+b, 5=-4m+b$. Subtracting, $5m=6\implies m=\frac{6}{5}$

And $b=11-m=\frac{49}{5}$.

So $y=\frac{6}{5}x+\frac{49}{5}\implies -\frac{6}{5}x+y=\frac{49}{5}\implies -6x+5y=49\implies -\frac{6}{49}x+\frac{5}{49}y=1$. So $a=\boxed{-\frac{49}{6}}$. 

There is no diagram, so I am assuming $E\in ABCD$. 

We have $\angle PAE=60-\angle BAC=60-45=15$

And $\angle APE=180-60-15=180-75=105$.

By Vertical angles, $\angle APE=\angle BPC=\boxed{105^{\circ}}$.

Multiply all the ratios together:

\((\frac{a}{b})(\frac{b}{c})(\frac{c}{d}) = \frac{a}{d}\)

Hope this gets you started.

sorry i made a mistake not u

it wrong

Putting $x=-5$, $50-5p-15=0\implies 5p=35\implies p=7$. 

So $7(x^2+x)+k=0$ has equal roots. Let's expand: $7x^2+7x+k=0$ has equal roots. The discriminant is, therefore, $0$, so we have $7^2-4*7*k=0 \implies 49=28k\implies k=\boxed{\frac{7}{4}}$.

I see you are doing AoPS Prealgebra! Great course, I hope you have fun. On the alcumus forum, you can ask for help with alcumus problems. 

Notice that the yes portion is $\frac{24}{36+24}$ of the whole. Now, the whole is $360^{\circ}$. Can you use this to find $\angle ACB=\text{ the yes portion }$?

If you don't understand angles very well, ask your teacher or your peers on the tab that says "Message board". They will be happy to help 

For the first one - $f(n)=2*f(n-1)$, I'll leave you to find the starting condition.

A recursive formula is when you write $f(x)$ in terms of $f(x-1)$. Explicit is when you have some exact equation like $f(x)=3x+4$. Arithmetic is when we add a number to the last number - geometric is when we multiply. 

You should be able to do these now.