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Thank you guys for helping me! I really apreciate the help!

good job! you're on the right track! the numerator is correct, but the denominator isn't.

you got the 12 choose 3 part right, but not the rest.

think about it, if you've already chosen 3 cards, then there would be 12-3=9 cards left to choose from, not 8. 

you're really close! you got this!

:)))

There are 10C3 = 120 possibilities for the 3 white balls that are drawn (since we don't care about the order), and 10 possibilities for the SuperBall.

Hope this helps, whymenotsmart^m^. Guest, you can take it from here.

Ways to choose  3 balls out of 10 =  10C3  =  120

Ways to choose 1 ball from 10  =   10

Does that help????

I tried to calculate the number of possibilities of Aces, 2s and 3 in the same hand as (4^3\times3^3\times2^3\times1^3)=13824, and the number of possibilities as \dbinom{12}{3}\dbinom{8}{3}\dbinom{4}{3}=49280. 13824/49280=\boxed{\frac{108}{385}}

Sorry, the latex doesn't work here...

hi guest!

so the first guy can draw 3 cards in \(\dbinom{12}{3}\) ways, and there are 4*4*4 possible draws that contain an ace, a 2, and a 3.

the second guy can draw 3 cards in \(\dbinom{9}{3}\) ways, and there are 3*3*3 possible draws that contain an ace, a 2, and a 3.

the third guy can draw 3 cards in \(\dbinom{6}{3}\) ways, and there are 2*2*2 possible draws that contain an ace, a 2, and a 3.

and, finally, the last guy just gets the remaining cards, which are an ace, a 2, and a 3.

so, the answer is \(\frac{4^3\cdot3^3\cdot2^3}{\binom{12}{3}\binom{9}{3}\binom{6}{3}}=\boxed{\frac{72}{1925}}\)

I hope this helped you!

:)

Alright give me a minute

week 7 problem 5 part a

What week and question number?

I will try to find it in one of my saved PDFs of the course.

"The square of an integer is 812 greater than the integer itself.  What is the sum of all integers for which this is true?"

Let n be the integer.

Then we have  n² - 812 = n  or  n² - n - 812 = 0

This factors as  (n + 28)(n - 29) = 0 

You can take it from here.

But can you help? Im stuck on this.

You said you've been working on it, in a few hours you should make a bit of progress, can you share what you have done so far?

This is from AoPS Intro To Counting, do not give full answers, partial ok.

I know because I did this one before.

umm whymenotsmart, i think you're referring to a different problem?

We  have  that

3a  + (a - b)  +  3b   =  86 

4a + 2b  =  86       divide through by 2

2a  + b =  43

b  =  43 - 2a       (1)

     and

a^2  + b^2  =  386      (2)        sub (1) into (2) for b

a^2  + (43 - 2a)^2  =  386     simplify

a^2  + 4a^2  - 172a  + 1849  =  386         subtract   386  from both sides

5a^2  - 172a  + 1463   =   0      this will factor as

(5a  - 77)  (a - 19)  = 0

The second factor set to 0 and solved for "a"  gives us an integer solution....

a =19

And  b =  43 - 2(19)  =  5

So.....a + b  =   24

This question has appeared several times previously.  Search the answers of heureka or myself.

for some reason, it's not letting me edit the question:(

Let me try again. 

Difference of first pair is

2ad+d^2+4dn

Second pair

2ad-3d^2+4dn

So we seem to subtract 4d^2 every iteration.

For the third pair it's 

2ad-7d^2+4dn

Great!

As I said there are 2n terms in this series, so we have n pairs.

Thus we have 2adn + 4d(n^2), but also...

The x(d^2). So thus we have (1 - 3- 7- 11 - 15... - (1-4n))(d^2.)

That's a start.

Loki I don't know how to solve the question but perhaps this may help you , I don't know.

I was using

https://www.symbolab.com/solver/expand-calculator/expand%20

This for expansions.

Possibly. I self learn algebra and I don't think this is in a textbook, but I don't take AoPS courses on algebra so idk.

Besides I gave a partial answer that idk is correct or not.

Answer:

k = 13The smallest zero or root is x = -10

===============================

Work Shown:

note: you can write "x^2" to mean "x squared"

f(x) = x^2+3x-10

f(x+5) = (x+5)^2+3(x+5)-10 ... replace every x with x+5

f(x+5) = (x^2+10x+25)+3(x+5)-10

f(x+5) = x^2+10x+25+3x+15-10

f(x+5) = x^2+13x+30

Compare this with x^2+kx+30 and we see that k = 13

Factor and solve the equation below

x^2+13x+30 = 0

(x+10)(x+3) = 0

x+10 = 0 or x+3 = 0

x = -10 or x = -3

The smallest zero is x = -10 as its the left-most value on a number line.

Hope this helps, whymenotsmart^m^.

hehehe:)

hi guest!

So, if the Business class takes up 25%, that means it takes up \(\frac{1}{4}\) of all seats. Economy takes up \(\frac{2}{3}\). So, \(\frac{1}{4}\text{ Bsuiness seats}+\frac{2}{3} \text{ Economy seats}=\frac{11}{12}\)

That means, First class takes up \(\frac{1}{12}\) of the seats.

\(24=\frac{1}{12} \text{total seats}\)

\(\text{total seats}= \boxed{288}\)

I hope this helped you!

:)

are you sure, 'cause usually hugo would have said that first

So we have this where we have

(a * (x)) ^2.

x ranges from 2n+ 1 to 1, so there are thus 2n terms in this sequence.

(a + 2nd+d) ^ 2= a^2 + 2and + ad + 4n^2d^2 + 2and + 2nd^2 + d^2 + 2nd^2 + ad.

Simplifying that is a^2 + (2nd) ^2 + d^2 + 2ad + 4and + 4(nd)^2 + 2n(d)^2.

The next term is 
 

(a + 2nd)^2 = a^2 + 4and + 4(nd)^2

So we subtract and get... 

a^2 + (2nd) ^2 + d^2 + 2ad + 4and + 4(nd)^2 + 2n(d)^2

                           d^2 + 2ad                              + 2n(d)^2??
 

Yeah...

Imma use a expander caculator.

-2ad+4adn+4d^2n^2-4d^2n+d^2+a^2

and the next term...

-4ad+4adn+4d^2n^2-8d^2n+4d^2+a^2

So we have now...

-2ad+4adn+4d^2n^2-4d^2n+d^2+a^2

-4ad+4adn+4d^2n^2-8d^2n+4d^2+a^2

That's -2ad -4d^2n + 3d^2 

Last time was

2ad + 2d^2n +2ad

-2ad -4d^2n + 3d^2 

So minus -4ad ...

I give up

sorry