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# Mathematics HL/SL/Studies Help

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### #1 Desy Glau Posted Feb 04, 2011 - 09:33

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Hello people!

I have seen Biology, Chemistry and Physics help threads; I figured I would create one for Mathematics too.

I'm willing to help any Mathematics student with their problems or doubts. I am sure there would also be some other helpers, so if you have any Math question just post here unless they are questions regarding IA tasks. Questions regarding IA tasks must be posted in the correct IA threads and not here. Questions from textbooks, worksheets or even past year papers may be asked here no matter whether they are for Math HL, SL or even Studies.

Please also include which Math level you are taking so the solution and explanation would be suitable with the extent that you need to know because Math HL, SL and Studies are very different.

Don't be afraid to ask! Enjoy!

### #2 timtamboy63 Posted Feb 04, 2011 - 09:49

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How do I divide by 0?

### #3 Desy Glau Posted Feb 04, 2011 - 10:03

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That is not even in the syllabus

Division by zero is an operation for which you cannot find an answer, so it is disallowed.

### #4 Markee Posted Feb 04, 2011 - 10:22

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Is there a quicker way to solve Binomial Expressions?

### #5 Desy Glau Posted Feb 04, 2011 - 10:45

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Use the Binomial Theorem?

That is the quickest way that I know and if you are too lazy to calculate nCr , just refer to the Pascal Triangle

### #6 Markee Posted Feb 04, 2011 - 12:10

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Oh wow, that looks horrendously complex! It is a good thing we have done this already, because to the untrained eye, their mind would just blow .gif' class='bbc_emoticon' alt=':)' />

THANKS DESY! You are always the best! ^^

### #7 genepeer Posted Feb 04, 2011 - 12:26

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How do I divide by 0?

By multplying with (-1)!

On a serious note: MathsChallenge.Net: Division by Zero Great math site, by the way. Owned by the same person who runs ProjectEuler.

Edited by Gene-Peer, Feb 04, 2011 - 12:47.

### #8 Ezeh Posted Feb 04, 2011 - 12:46

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For some reason, i always struggle with these types of questions:

(b) Find the term in x3 in (3x+4)(x-2)4

And also, is it just me, or does it sound like it's worded wrong?

Anywho, i always seem to forget how to do these, and i'm sure they're fairly easy

Edited by Ezeh, Feb 04, 2011 - 12:48.

### #9 Desy Glau Posted Feb 04, 2011 - 13:05

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This thread might be a hot topic within a few days lol
You are welcome Mark!

For some reason, i always struggle with these types of questions:

(b) Find the term in x3 in (3x+4)(x-2)4

And also, is it just me, or does it sound like it's worded wrong?

Anywho, i always seem to forget how to do these, and i'm sure they're fairly easy

It is asking for the coefficient of x3 when you expand (3x+4)(x-2)4

And you don't need to expand (3x+4)(x-2)4 to get the coefficient of x3! Just use the Binomial Theorem! I always suck at explaining though and English is not my first language so I hope I can explain it well. The last time I learned this is in IGCSE though which is about 8 months ago so I need to recall the Binomial Theorem

Since (x-2)4 is multiplied by (3x+4) and you just want the coefficient of x3, you will need the coefficient of x3 and of x2 in the expansion of (x-2)4.

(x-2)4 = ax4+bx3+cx2+dx+e

(3x+4)(x-2)4 = 3x(ax4+bx3+cx2+dx+e) + 4(ax4+bx3+cx2+dx+e)

Coefficient of x3 = 3c+4b

(x-2)4 = ax4 + 4(-2)x3 + 6(-2)2x2 + dx+e
= ax4 - 8x3 + 24x2 + dx+e

So b=-8 and c=24!

Coefficient of x3 = 3*24+4(-8)
= 40

### #10 genepeer Posted Feb 04, 2011 - 13:06

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x3 can be formed when: 1) the 3x in the first brackets combines with the "x2" in the second or 2) the 4 in the first combines with the "x3" in the second.

For the second bracket.
x2: 4C2x2(-2)2 = 24x2
x3: 4C3x3(-2)1 = -8x3

So we have (3x + 4)(... - 8x3 + 24x2 + ...) = ... + 72x3 - 32x3 + ... = ... + 40x3 + ...

I think "the term in x3" means coefficient of x3, I'm not sure. So the answer is either 40x3 or just 40.

### #11 Markee Posted Feb 05, 2011 - 13:23

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I need help with this guys, if you could help me with how to answer these ones I might be able to finish my homework within the day, i will just use the ones you answered as examples to assist me

This is on Binomial Expressions.

expand

a) (3x+2y)^3
b) (2+1/2x)^4

Find the coefficient of the term indicated in square brackets in the expansion of each of these expressions.

a) (7-4x)^5 [x4]
b) (2/3-2/5x)^3 [x]
c) (1/3+3/2)^6 [x^3]

Expand

a) (2+x-4x^2)^2
b) (1+3x^2)^3
c) (2+3x-x^2)^2

### #12 nametaken Posted Feb 05, 2011 - 13:42

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I need help with this guys, if you could help me with how to answer these ones I might be able to finish my homework within the day, i will just use the ones you answered as examples to assist me

This is on Binomial Expressions.

expand

a) (3x+2y)^3
b) (2+1/2x)^4

Find the coefficient of the term indicated in square brackets in the expansion of each of these expressions.

a) (7-4x)^5 [x4]
b) (2/3-2/5x)^3 [x]
c) (1/3+3/2)^6 [x^3]

Expand

a) (2+x-4x^2)^2
b) (1+3x^2)^3
c) (2+3x-x^2)^2

I'll give you an example for part a. If you had, a) (4x+3y)^3, you would first start off by recognising that the first expansion is 4x^3. The second expansion is 4x^2 * 3y (the powers of 4x are decreasing, while the powers of 3y are increasing from 0 to 3). The next expansion is 4x*3y^2, and the last one is simply, 3y^3. So the overall expansion is 4x^3 + 4x^2*3y+4x*3y^2+3y^3. However, this is without the coefficients, which you'd find by drawing out pascals triangle.

### #13 Markee Posted Feb 05, 2011 - 15:49

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I'm guessing the coefficient is the ones with them C's,right?

And thanks btw

### #14 nametaken Posted Feb 05, 2011 - 16:26

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I'm guessing the coefficient is the ones with them C's,right?

And thanks btw

The coefficient is the number in front of the 'x' or the 'r' or whichever arbitary letter is used. I'm not sure what you mean by the C's.

### #15 Markee Posted Feb 06, 2011 - 09:10

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I'm guessing the coefficient is the ones with them C's,right?

And thanks btw

The coefficient is the number in front of the 'x' or the 'r' or whichever arbitary letter is used. I'm not sure what you mean by the C's.

Ohh! So this is only when expanding SOrry, I think i was trying to mention the other exercise that has that and not the 'just expand'

### #16 Desy Glau Posted Feb 06, 2011 - 09:20

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Are you guys classmates?

Did you mean the nCr? That must be included as well. Dont forget to include the coefficient of the x and also the constant!

### #17 Markee Posted Feb 06, 2011 - 09:46

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Are you guys classmates?

Did you mean the nCr? That must be included as well. Dont forget to include the coefficient of the x and also the constant!

YES! That one! But, is that needed for part 1???
Or is it just for part 2?

Are you guys classmates?

Did you mean the nCr? That must be included as well. Dont forget to include the coefficient of the x and also the constant!

AND Yes, we go to the same school. We are only classmates in Chemistry HL. haha

### #18 openur3eye Posted Feb 06, 2011 - 09:52

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i need help with this question on vectors
it says: Gina drives along a straight highway for 4.2km in a north-westerly direction and then along another road for 5.3km in a north-easterly direction. Find her displacement from her starting position

Thanks

### #19 nametaken Posted Feb 06, 2011 - 10:10

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i need help with this question on vectors
it says: Gina drives along a straight highway for 4.2km in a north-westerly direction and then along another road for 5.3km in a north-easterly direction. Find her displacement from her starting position

Thanks

The vector shown is, 'v' shaped, with north east going left, north west going right. So I think that the displacement would be 5.3^2+4.2^2.

Edited by Derek, Feb 06, 2011 - 10:20.

### #20 Markee Posted Feb 06, 2011 - 10:13

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Ahh, in all honesty, just by expanding them it IS that easy, but utterly confusing .gif' class='bbc_emoticon' alt=':drake:' /> Thanks Derek and Desy!