I'm sorry.  Yes, I will not do this again in the future. Oh ok nvm I got it. :)

thank you

your answer is real!

if your calculator can calculate one quadrlion factorial, please send download link and name of your calculator.

Please stop cheating on your homework and making alt accounts to upvote your posts. 

The angle bisector of angle BAC shouldn't meet AC at any otherpoint than A...

i dont think d will be positive...

(i got 1,12,3,-8)

(a) The length of the semi-major axis is 12.

(b) b = 12.

a = -1 - sqrt(3), b = 5*sqrt(3) - 5

He's right. You gotta ask specific questions and not post the whole problem .

\(1_3 = 1\\12_3=1*3+2=5\\212_3=2*9+1*3+2=23\\2121_3 =2*27+1*9+2*3+1=70\)

The angles in a triangle sum to 180° so you can find angle Q from this.  Then use the sin rule:  q/sin(Q) = p/sin(P) to find q.

Like so:

".... If  ?  students take only Spanish ..."  The number of students taking only Spanish is missing,

Hello, 
 

Let's assume that there are "x" students who take both Chinese and Spanish, and "y" students who take only Spanish.

We know that the total number of students in South High School is 97, and that every student takes at least one language. Therefore, the number of students taking Chinese is "97 - y".

Also, we know that "there are more students in Chinese than in Spanish", so:

97 - y > y

Simplifying this inequality, we get:

97 > 2y

y < 48.5

Since "y" is a whole number, the largest value it can take is 48.

Now, we can use this value of "y" to find the number of students who take both Chinese and Spanish:

x + y = 97 - (97 - y)

x + y = y

x = 0

Therefore, there are 0 students who take both languages, if 48 students take only Spanish.

yeah,

so 5x26x26=3380

so like theres 5 vowels, and then the other 2 letters can be from the 26 letters of the alphabet... right?

I'm not sure if I'm correct because im not very good at math. But:

12 (1+2 = 3), 21 (2+1 = 3), 111 (1+1+1)... 3

not sure lmao

5x26x26... how every many letters are in the words you are asking for

sure it is not that difficult.

I'll do it with a semi-specific example

\((a+b)^4= \binom{4}{0} a^4 * b^{0}\;+\;\binom{4}{1} a^3 * b^{1}\;+\;\binom{4}{2} a^2 * b^{2}\;+\; \binom{4}{3} a^1 * b^{3}\;+\;\binom{4}{4} a^0 * b^{4} \\~\\ (a+b)^4=\displaystyle \sum_{n=0}^4\;\;\binom{4}{n}a^{4-n}*b^n\)

Do you see the pattern?

If you want help with knowing where you went wrong then you should post what you have tried to do yourself and ask specific questions.

Also, if it is a direct AOPS question you should not post it here. Especially not in its entirity.

perracita, you could have just said it is divisible by 10 goddamit