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It says incorrect 

We are looking for the smallest n such that n(n - 1)(n - 2)(n - 3) > 100.

Since 4*3*2*1 = 24 and 5*4*3*2 = 120, the smallest n that works is n = 5.

Ohhhh so triangle  CEB is similar to the congruent right triangles in the middle

That makes sense thanks!

The minimum value of a + b + c + abc is 4/3*sqrt(3).

There are 2385 possible positions for the centroid.

sin A = 4/5

$2 ==$1  x  e^(0.06)t,  where t == number of years

$2 ==e^(0.06)t, solve for t

Take the natural log (ln) of both sides

t ==Ln(2) / 0.06==11.552453 years to double your money at 6% compounded continuously.

The possible values of k are -6 <= k <= 6.

The area of triangle PYR is 8*sqrt(3).

The possible values of j are -5 and 12.

It says the deck is shown below.

I do not see it.

More words needed.

Because we are dealing with 4 coins, the easiest way (imo) is to list out all 16 cases and count the average

The 16 cases are: 

HHHH   HTHT   TTTT   THTT

HHHT   HTTH   TTTH   THHT

HHTH   HTTT   TTHT   THTH

HHTT   HTHH   TTHH   THTH

We see that 2 have a run of 4, 4 have a run of 3, 8 have a run of 2, and 2 have a run of 1.

So the expected value is \({{(2 \times 4) + (4 \times 3) + (8 \times 2) + (2 \times 1)} \over 16} = \color{brown}\boxed{2.375}\)

Now that I think of it, here's another way of reaching the same answer. 

We already know that \(CB = 12\sqrt2\) (look at the other solution to see how)

Now, let \(CE = a\) and \(AE = b\). We have \(a^2 + b ^2 = 81\), and \(a^2 + (b+9)^2 = ({12 \sqrt2})^2 = 288\). 

Subtracting the two equations gives us \(18b + 81 = 207\), meaning \(b = 7\). 

Plugging this into the first equation also gives us \(a = 4 \sqrt 2\)

First, we know that \(DB =DC = \sqrt{9^2 - 3^2} = 6\sqrt{2}\), which means \(CB = 12 \sqrt 2\)

Also, note that \(\triangle CEB \sim \triangle ADB\) by AAA. This is because both triangles have a right angle and share \(\angle B\), which means that all 3 traingles have the same angles, and thus, are similar. 

Now, by similiar triangles, we have \({AB \over CB} = {AD \over CE}\). Substituting what we know, we have \({9 \over 12 \sqrt 2} = {3 \over CE}\), meaning \(CE = \color{brown}\boxed{4 \sqrt2}\).

To solve for AE, we use the Pythagorean Theorem on \(\triangle AEC\), and find that \(AE = \sqrt{9^2 - {4\sqrt2}^2} = \color{brown}\boxed{7}\)

You can't

2 variables one equation, no way to solve the 2 variables unless ur looking for something else

Let's see what i can do..

So you need lengths AE and CE

Apparently Angle ADC is not a right angle?

thanks, I feel like I expressed this question incorrectly lol

going to restart the question thank you