EvanWei123

Wow thanks a lot Eletric Pavlov you are very smart

Thanks for the help Rom!

This is a Artofproblemsolving alcumas problem....

Is there anyone?

Thank you ElectricPavlav that was alot of help Keep it up!

Hi Rudram, I can't really understand your question. May you use the latex thanks

-Evan

No problem :D

Hi maccabresubwoofer, 

First, $\frac{1^1}{3} = \frac{1}{3}$

next, $\frac{2^3}{4} = \frac{8}{4} = 2$

Finally, $2+\frac{1}{3} = 7/3$

but thats not a choice. And second, A and B are the same and D and E are the same...

I think you might mean $1^{\frac{1}{3}} + 2^{\frac{3}{4}}$

 But whatever.

-Evan

Oops CPhill was more detailed Sorry!

Hi Guest, 

First, 

$\frac{\sqrt{338}}{\sqrt{288}}+\frac{\sqrt{150}}{\sqrt{96}} = \frac{13}{2^2\cdot \:3}+\frac{\sqrt{150}}{\sqrt{96}}$

Next, $\frac{13}{2^2\cdot \:3}+\frac{\sqrt{150}}{\sqrt{96}} = \frac{13}{2^2\cdot \:3}+\frac{5}{2^2}$

Simplifying again, $\frac{13}{2^2\cdot \:3}+\frac{5}{2^2} = \frac{13}{12}+\frac{5}{2^2}$

Lastly, $\frac{13}{12}+\frac{5}{2^2} = \frac{13}{12} + \frac{5}{4} = \frac{13}{12}+\frac{15}{12} = \frac{28}{12} = \frac{7}{3}$

Your welcome :P, Evan

Hi Guest , 

You can assume that the bracelet only has 5 spot because the problem didn't say anything about the length.

Then, you think hm. How many ways can I put the first bead into the bracelet? 

Well because we have 5 beads there are 5 choices.

Next, you think after you put the first bead how many choices are there to put the second bead into the bracelet?

Because you already put one bead into the bracelet, there are only 4 choices.

You find that the patatern is 5, then 4, then 3, then 2, then finally 1 final choice.

Multiply them all together you get 120. Notice that $120 = 5!$

But since rotations and flips are not the same, you have to divide by 5, $\frac{120}{5}=24$

then you divide by 2 because flips aren't the same so the answer is $\boxed{12}$

-From Evan