SoulKingBrook

Haha! Love how confident this explanation. Cos(90) in right triangle trig literally just destroys the concept of a triangle lol

Using sticks and stones strategy, you will have 7 sticks and 10 stones, so seventeen different objects. If we have 17 slots, we can first place the 7 sticks in 17c7= \(\boxed{19448}\) ways.

Yeah if you spent the time to write all this out just copy paste it into a calcuator, but unfortunately you can't becaus eof some formatting issues. Good luck however!

In degrees cos(90)=0.

+Let us convert both of the following numbers into base 10: 

We can express \(1001\) as:

\(10^3+10^0\), then putting in 2 instead of 10, we get 9. So 1001 base 2 is equal to 9 in base 10.

We can express 200 (I know you put 123, but 123 isn't a valid number in base 3, so I did the regrouping for you) as:

\(2*10^2\), then plugging in 3, we get 2*9 or 18. (

Multiplying 18 and 9 we get \(\boxed{162}\)

Your question doesn't make sense unfortunately, you say that if you would have driven at a higher rate of speed, then you would have arrived 45 minutes later?! Could you please edit your question, thanks so much!

Oh you're right @guest, completely forgot about that part lol. Good job!

 Let this be the rows of chairs:

[] [] []

[] [] []

Let their be siblings \(x\) and \(x\), \(y\) and \(y\), and \(z\) and \(z\). Beginning with placing the \(x\) siblings, there are 6 options for the first sibling obviously, and 4 options for the second sibling, giving us 24 ways of placing them (6*4):

[x1]                      [] []

[x2 not allowed]   [] []

Let's say we go with this arrangement:

[x1] []  []

[]  [x2] []

We must place one of the y siblings in the rightmost column, or else the z siblings will be front and behind each other (there are two ways to do this):

[x1] []   []

[]  [x2] [y]

Then there are two ways of placing the other y sibling (row 1 column 2 or row 2 column 1), giving us 4 ways of placing the y siblings (2*2), but then we could also swap it so that sibling y2 could be in the right most column and y1 is in the other spots, givings us 8 ways of placing the y siblings:

[x]   []   []

[y2]  [x] [y1]

And then the z siblings go in automatically, with 2 ways, as we can swap z1 and z2. this gives us \(24*8*2=\boxed{384}\) ways. Sorry if my explanations a bit long, this problem I found hard to explain without going super in depth.

It think there may be an error in your question. Could you please double check that it's correct? It seems that the 2nd and 3rd sentence don't really fit together based off of the contex that Alice and Bob have a set starting amount of money through out each scenario.

Explanation: The prime factorization of \(2500\) is \(2^2*5^4\), so Catherine will roll 2 twos and 4 fives. Now we just need to deal with the amount of ways to reorganize those rolls. If we have 6 slots:

[] [] [] [] [] [], and place the 2 twos in them, the other 4 will be automatically placed.

And there are \(\frac{6*5}{2}\) 15 ways of doing this (you can also use the combination formula for this). That gives us \(\boxed{15}\) ways of doing so.