Homework Help/Tutoring

[iEATnoobs]

O Gawd im so glad im done with trig. Them identities made me want to slit my wrists.

[Spook]

That's why i dropped trig. It just didn't click in my mind so i took something more real life applicable.

[ViciouZ]

Ok, circle theorems: anyone got any good mnemonics for remembering these?












i said mnemonics

[Cameron]

Hello old thread.

The function f(x) = ax^3 - 7x^2 - bx - 20 is divisible by x^2 - 3x - 10.  Find the values of a and b.

I've spent more than the past hour working on this question with a friend, and we just can't get anything out of it.  The only place I could find any other question like this was in my textbook, but it had no help and description of how to do it, just the answer to it only. 

[MOFO]

whats 2 + 2 .... last time someone said it was 22.

[Gamabunta]

Whoa, think I may be a little to late but still.
@Cam
x^2 - 3x - 10 = (x - 5)(x + 2), let's call it w(x). w(x) zeroes for x = 5 and x = -2, and so does f(x). Put 5 and -2 on place of x in f(x) and it equals zero, like:

Code: [Select]

{ a*5^3 - 7*5^2 - b*5 - 20 = 0
{ a*(-2)^3 - 7*(-2)^2 - b*(-2) - 20 = 0
This leads to an answer that a=3 and b=36, unless I made a mistake.
Hope that solves the problem, if it still exists.

[P!nk]

Post removed

[Cameron]

Whoa, think I may be a little to late but still.
@Cam
x^2 - 3x - 10 = (x - 5)(x + 2), let's call it w(x). w(x) zeroes for x = 5 and x = -2, and so does f(x). Put 5 and -2 on place of x in f(x) and it equals zero, like:

Code: [Select]

{ a*5^3 - 7*5^2 - b*5 - 20 = 0
{ a*(-2)^3 - 7*(-2)^2 - b*(-2) - 20 = 0
This leads to an answer that a=3 and b=36, unless I made a mistake.
Hope that solves the problem, if it still exists.

Ahh it leaves 2 equations which can be solved simultaneously.  I didn't think of doing it like that.  Thanks for that, gets that out of my head .

[battle]

Permutation and Combinations

1. How many 13 card hands are there having exactly nine cards from any suit? Answer: 235 237 860

2. How many ways are there of picking 2 cards, one after the other, from a deck of 52 cards if the
A) First card is replaced?      Answer: 2704
B) First card is not replaced?   Answer: 2652

3. In how many ways can a 5 question test be answered if each question is to be answered true or false? Answer:32

I need help on how to do it, please. I'm betting i could get these questions if i took my time, but i don't feel like doing it right now.

[Cameron]

Haha, probability.

1) Don't understand the question.  Is it how many 13 card hands can be made up of 9 cards of the same suit?

2a) nCr(52,1) x nCr(52,1) = 2704.  Basically thats 52 ways of getting the first card AND (in probability X is always and, + is always or) 52 ways of getting the next card.
b) nCr(52,1) x nCr(51,1) = 2652, quicker still is nPr(52,2), use that when its not being replaced .

3) 2^5 = 32.  Which is essentially nCr(5,0) + nCr(5,1) + nCr(5,2) + nCr(5,3) + nCr(5,4) + nCr(5,5).  Meaning there is 1 way of giving no answers, 5 ways of giving only one answer, 10 ways of giving 2 answers, etc.  Add em together because its not giving 1 answer AND 2 answers, its 1 answer OR 2 answers.

[battle]

Oh okay, thanks I had a test on it today and i think i failed it, woot

[Cameron]

Haha, I aced probability last year, hoping to do the same this year

[battle]

I got the test back today and i got 31 out of 44. Wow, i guess i do kinda know what i'm doing.