NotThatSmart

Here's the entire expansion (I have no life)

\(729y^{12}+2916y^{11}-2430y^{10}-19980y^{9}+135y^{8}+59976y^{7}+6364y^{6}-99960y^{5}+375y^{4}+92500y^{3}-18750y^{2}-37500y+15625\)

The answer is 375

Thanks! :)

Let's focus on triangle AHB. 

It's made up of 3 angles. We have \( \angle AHB, \angle BAH, \angle HBA\)

We also have that \(\angle AHB = 144^\circ,\angle BAH = 66^\circ\)

However, note that this would mean triangle AHB has interior angkes summing to  

\(144 + 66 = 210°\)

Since a triangle can't sum to above 180, this triangle is therefore impossible and invalid. 

Thanks! :)

I can only think of two possible combinations to satisfy the conditions given.

We have Combinations of 2, 2, 4, 5, 5, 5 and combinations of 1, 4, 4, 5, 5, 5    

There might be others, but I can't think of any other ones. 

The number of permutations of 2, 2, 4, 5, 5, 5 is \(6! = 720   \)

The number of permutations of 1, 4, 4, 5, 5, 5 is \(6! = 720   \)

So the total number of sequences would be \(720 + 720  =  1440 \)sequences   

So 1440 is the answer. 

Thanks! :)

Since AK is the altitude, so we can use the pythaogrean thereom. 

We have'

\(AK = \sqrt { AC^2 - CK^2 } = \sqrt { 8^2 - 3^2 } = \sqrt { 55 } \\ AB = \sqrt { AK^2 + BK^2 } = \sqrt { 55 + 4 } = \sqrt { 59 }\)

Thanks! :)

https://web2.0calc.com/questions/right-triangles_35

First, let's find the total number of cases we work through. We can actually just list them all out. 

We have

\((HHT,HTH,HTT, THH, THT, TTH, TTT)\)

The only possibility where a head doesn't appear is TTT. That means that

\((HHT,HTH,HTT, THH, THT, TTH)\) all work. 

Since probability is essentially \(\text{Probability} = \frac{\text{# of Favorable Outcomes}}{\text{Total Outcomes}}\)

So we just have \(\frac{6}{7}\)

So 6/7 is our answer. 

Thanks! :)

We can use the great pythaogrean theorem to our advantage. 

First off, let's set some variables. 

Let's let BC = a and AC = b. 

Now, using pythagoras, let's right some equations. We have

\(\overline{AM} = \sqrt{b^2 + \left(\dfrac a2\right)^2} = \dfrac12 \sqrt{4b^2 + a^2}\) since AM is the median and we also have

\(\overline{BN} = \sqrt{a^2+ \left(\dfrac{b}{2}\right)^2} = \dfrac12 \sqrt{4a^2 + b^2}\) since BN is also the median of the triangle. 

Now, we're trying to isolate a^2+b^2 because that's how we find the hypotenuse of the triangle. 

Thus, we now have

\(4b^2 + a^2 = 2^2 \\4a^2 + b^2 = 2^2\)

Now, adding the two equations up with each other and dividing both sides by 5, we find that

\(a^2 + b^2 = \dfrac{4+4}{5} = 8/5\)

So the hypotenuse is that square rooted, which is \(\sqrt{\frac{8}{5}}\)

So \(\sqrt{\frac{8}{5}}\) is our final answer. 

Thanks! :)