KnockOut

Consider similar triangles $ABE$ and $BCD$, then we have $\angle BDC=\angle EAB$.

For triangle $ABE$, 

$\tan(\angle EAB)=\frac{EB}{AB}$.

For triangle $BCD$,

$\tan(\angle BDC)=\frac{BC}{BD}$

Thus we have

$\large \frac{EB}{AB}=\frac{BC}{BD}\Rightarrow EB\cdot BD=AB\cdot BC$.

I'm not sure. I've done the same exact problem somewhere else before, but I do not recall the answer.

Here is a similar question: https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_8_Problems/Problem_20

Maybe like the question above, sort them by base lengths, and see how many triangles can be formed by connecting points from each distinct base length.

Since only the number of stickers on the sheets matters, we can list the possibilities systematically:

$\begin{align*} & 8-0-0-0 \\ & 7-1-0-0 \\ & 6-2-0-0 \\ & 6-1-1-0 \\ & 5-3-0-0 \\ & 5-2-1-0 \\ & 5-1-1-1 \\ & 4-4-0-0 \\ & 4-3-1-0 \\ & 4-2-2-0 \\ & 4-2-1-1 \\ & 3-3-2-0 \\ & 3-3-1-1 \\ & 3-2-2-1 \\ & 2-2-2-2 \end{align*}$

There are  $\boxed{15}$  possible arrangements of stickers on sheets of paper.

You want to distribute your 5 distinguishable balls into 3 indistinguishable boxes. Let $B(5,3)$denote the number of ways in which this can be done into exactly 3 indistinguishable non-empty boxes, and use the recurrence relation $B(n,k)=B(n−1,k−1)+kB(n−1,k)$with $B(n,1)=1$ and $B(n,n)=1$. You seek $B(5,1)+B(5,2)+B(5,3)$.

This has something to do with "Stirling numbers" although I am not too sure on this topic. You might consider searching it up!

$\text{let }P[\text{heads}] = p\\ E[V] = 2p -(1-p) = 3p-1$

That's what Rom came up with...

https://web2.0calc.com/questions/expected-value_5

You asked this question earlier today. I think Rom was helping you on this one.

For each pair of cards, the probability that they are both black is $\frac{26}{52} * \frac{25}{51} = \frac{25}{102}$, so since there are 52 pairs in the circle, the expected number of pairs which are both black is $\frac{25}{102} * 52 = \frac{650}{51}$.

Our answer should be $\boxed{\frac{650}{51}}$.

$\frac{1}{x} \geq 3 - 2\sqrt{x} \iff \frac{1+2x\sqrt{x} -3x}{x} \geq 0 \iff \frac{(\sqrt{x}-1)^2(\sqrt{x} + \frac{1}{2})}{x} \geq 0$

We can cancel off the $(\sqrt{x}-1)^2$, and the $x$ on the bottom, since $x$ must be positive, and squares are always positive.

This leaves us with $\sqrt{x} + \frac{1}{2} \geq 0$ which is true for all positive $x$. 

Hence this inequality also applies for all positive $x$.

Equality will apply when $x = 1$, since $\sqrt{x}+\frac{1}{2}$ is strictly positive, and $\frac{1}{x}$ is strictly positive for $x \neq 0$.

Source: 

1) https://math.stackexchange.com/questions/1929829/show-that-frac-1x-ge-3-2-sqrtx-for-all-positive-real-numbers-x-descri

2) https://math.stackexchange.com/questions/1929829/show-that-frac-1x-ge-3-2-sqrtx-for-all-positive-real-numbers-x-descri/1929838

If it is urgent, this is all I have. If you need something more detailed, I'm sure someone will be able to better assist you soon. 

Hello this is a math forum...