Algebra2

[Gengar_mybb_import19743]

Not sure how to solve these.

 

[tu y tu mama]

Don't know what this is but, WolfRamAlpha ftw

But just for the first one, I don't know what the #2 or 3 are. Oh well, going back to school again tomorrow, will probably see this later in like a week.

 

[Gengar_mybb_import19743]

Don't know what this is but, WolfRamAlpha ftw

But just for the first one, I don't know what the #2 or 3 are. Oh well, going back to school again tomorrow, will probably see this later in like a week.

Ily

Still can't seem to find out how to solve the 3rd and 4th, I've tried googling it but I haven't had any luck.

 

[Deathcrow]

Just so you know how to actually solve the first one, it isn't that hard. They are factorials. For example, 5! is the same thing as saying 5*4*3*2*1. And 3! is the same thing as 3*2*1.

So to solve the first one you simply do:

(8*7*6*5*4*3*2*1)/((6*5*4*3*2*1)*(2*1)) = 40320/1440 = 28.

 

[tu y tu mama]

Just so you know how to actually solve the first one, it isn't that hard. They are factorials. For example, 5! is the same thing as saying 5*4*3*2*1. And 3! is the same thing as 3*2*1.

So to solve the first one you simply do:

(8*7*6*5*4*3*2*1)/((6*5*4*3*2*1)*(2*1)) = 40320/1440 = 28.

My canadian teacher would have taken probably 2 classes to teach us that. Thanks, learned something today.

 

[Deathcrow]

Just so you know how to actually solve the first one, it isn't that hard. They are factorials. For example, 5! is the same thing as saying 5*4*3*2*1. And 3! is the same thing as 3*2*1.

So to solve the first one you simply do:

(8*7*6*5*4*3*2*1)/((6*5*4*3*2*1)*(2*1)) = 40320/1440 = 28.

My canadian teacher would have taken probably 2 classes to teach us that. Thanks, learned something today.

Those damn Canadians!

Lol. No problem, Daniel. I hated it in school when teachers wouldn't just tell us how to do something. I swear they would try to make things hard on us. I like stuff straight to the point.

Yesterday, my first day of class, I learned how to convert and do math with binary numbers, octal numbers, and hexadecimal numbers. Hexadecimal is the most odd because it involves letters and numbers. Hexadecimal is base 16 meaning that there are 16 unique digits: 0-9 and A-F. That means you would count like this: 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F,10,11,12,13,14,15,16,17,18,19,1A,1B,1C,1D,1E,1F,20....etc.

EDIT: Forgot to say what class it was. It was my fundamentals of computer engineering class.

 

[Gengar_mybb_import19743]

Just so you know how to actually solve the first one, it isn't that hard. They are factorials. For example, 5! is the same thing as saying 5*4*3*2*1. And 3! is the same thing as 3*2*1.

So to solve the first one you simply do:

(8*7*6*5*4*3*2*1)/((6*5*4*3*2*1)*(2*1)) = 40320/1440 = 28.

Do you have any idea about the other 2?

 

[tu y tu mama]

Just so you know how to actually solve the first one, it isn't that hard. They are factorials. For example, 5! is the same thing as saying 5*4*3*2*1. And 3! is the same thing as 3*2*1.

So to solve the first one you simply do:

(8*7*6*5*4*3*2*1)/((6*5*4*3*2*1)*(2*1)) = 40320/1440 = 28.

Do you have any idea about the other 2?

Well, you can type in C_3 in WolfRamAlpha and it will interpret it as the C with the little 3 below it, but I don't know how you would put the other 6, _6C_3 doesn't work.

 

[Gengar_mybb_import19743]

Just so you know how to actually solve the first one, it isn't that hard. They are factorials. For example, 5! is the same thing as saying 5*4*3*2*1. And 3! is the same thing as 3*2*1.

So to solve the first one you simply do:

(8*7*6*5*4*3*2*1)/((6*5*4*3*2*1)*(2*1)) = 40320/1440 = 28.

Do you have any idea about the other 2?

Well, you can type in C_3 in WolfRamAlpha and it will interpret it as the C with the little 3 below it, but I don't know how you would put the other 6, _6C_3 doesn't work.

Hmm. I'll try to mess around with it, this is the final test I take before my final and I have a 71 average currently. So this could either make or break me passing or failing.

 

[Deathcrow]

Just so you know how to actually solve the first one, it isn't that hard. They are factorials. For example, 5! is the same thing as saying 5*4*3*2*1. And 3! is the same thing as 3*2*1.

So to solve the first one you simply do:

(8*7*6*5*4*3*2*1)/((6*5*4*3*2*1)*(2*1)) = 40320/1440 = 28.

Do you have any idea about the other 2?

6C3 = [6!/(6-3)!*3!] = 6!/[3!*3!] = 20.

7P2 = 7!/(7-2)! = 7!/5! = 42.

 

[Gengar_mybb_import19743]

Just so you know how to actually solve the first one, it isn't that hard. They are factorials. For example, 5! is the same thing as saying 5*4*3*2*1. And 3! is the same thing as 3*2*1.

So to solve the first one you simply do:

(8*7*6*5*4*3*2*1)/((6*5*4*3*2*1)*(2*1)) = 40320/1440 = 28.

Do you have any idea about the other 2?

6C3 = [6!/(6-3)!*3!] = 6!/[3!*3!] = 20.

7P2 = 7!/(7-2)! = 7!/5! = 42.

I love you. Please make children with me in the near future.

 

[Deathcrow]

Do you have any idea about the other 2?

6C3 = [6!/(6-3)!*3!] = 6!/[3!*3!] = 20.

7P2 = 7!/(7-2)! = 7!/5! = 42.

I love you. Please make children with me in the near future.