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Edexcel C2 20th May 2015 *Official Thread* watch

1. (Original post by frozo123)
Have you got it now? Sub a back into any of the equations
then to understand what you've done
Solve

sin(ax-b)=0 0<x<360 ( the inequality sign meant to be or equal to )

so let (ax-b)= theta

so values are 0, pi, 2pi
you should get what the coordinates are in the question
Ahh, that makes sense! So it'll be 0+b/2 which must equal pi/10..which is why b is pi/5? I don't think I'll be able to do this in an exam though :/

Thanks
2. (Original post by userxx990)
So guys, ive summed up my very dumb maths doubts, very simple and quick to answer like weird equations etc, can any one show me a paper work solution of these, if not, one of these? Its hell important, as my basics realllyyy sucks! Please help!
I'm not sure whether because the sun is blinding my view or your pdf file is upside down
Ahh, that makes sense! So it'll be 0+b/2 which must equal pi/10..which is why b is pi/5? I don't think I'll be able to do this in an exam though :/

Thanks
I can't remember whether a is 2 but if it was then yeah that's it
just work backwards
Idk if you need any help just ask
4. (Original post by frozo123)
I'm not sure whether because the sun is blinding my view or your pdf file is upside down
Sorrry! is it finee nooww?
Attached Images
5. 4444444.pdf (373.6 KB, 112 views)
6. (Original post by userxx990)
So guys, ive summed up my very dumb maths doubts, very simple and quick to answer like weird equations etc, can any one show me a paper work solution of these, if not, one of these? Its hell important, as my basics realllyyy sucks! Please help!
okay first one, I don't know where the 25 comes from
but the square root of y
because log5y= b
1/2log5y=1/2b
and 1/2log5y is the same as log5y^1/2

second one you could put x1/3 as the denominator and then multiply everything by x1/3 to get rid of the negative power
so you'll get -x^1/3=-2
then solve for x

third one well cubing it is (2root2)(2root2) which is 8 so then another 2root2 is 16root2 then you can subtract as the roots are the same
fourth one it's an arithmetic sequence so use the formula Sn= blah blah
there will be 21 terms if you think about it because the 10 and 30 are inclusive
so Sn=10.5(54+(21-1)2)

3tan^2theta -1=0
you put the 1 on the other side then divde by 3 and solve it how you would
if you don't know how to solve it let me know

next one you put the -1/2 as the denominator
then multiply every term by 1/2
so you'll get x^2-8=0
then solve for x
next one turn it into a top heavy fraction so (49/9)^-1/2
flip it because the power is negative so (9/49)^1/2 then square root each term

final one, you multiply every term by x to get rid of the denominator then multiply by 2 to get rid of the 2
although I don't get how that equals what you've written down

hope that helps!

are you an international student or something? cus half of that is c1
Not looking forward to this
Hope everyone's well though
8. (Original post by frozo123)
okay first one, I don't know where the 25 comes from
but the square root of y
because log5y= b
1/2log5y=1/2b
and 1/2log5y is the same as log5y^1/2

second one you could put x1/3 as the denominator and then multiply everything by x1/3 to get rid of the negative power
so you'll get -x^1/3=-2
then solve for x

third one well cubing it is (2root2)(2root2) which is 8 so then another 2root2 is 16root2 then you can subtract as the roots are the same
fourth one it's an arithmetic sequence so use the formula Sn= blah blah
there will be 21 terms if you think about it because the 10 and 30 are inclusive
so Sn=10.5(54+(21-1)2)

3tan^2theta -1=0
you put the 1 on the other side then divde by 3 and solve it how you would
if you don't know how to solve it let me know

next one you put the -1/2 as the denominator
then multiply every term by 1/2
so you'll get x^2-8=0
then solve for x
next one turn it into a top heavy fraction so (49/9)^-1/2
flip it because the power is negative so (9/49)^1/2 then square root each term

final one, you multiply every term by x to get rid of the denominator then multiply by 2 to get rid of the 2
although I don't get how that equals what you've written down

hope that helps!

are you an international student or something? cus half of that is c1
bro, firstly thank you so much for such a great help!

Just the 3tan^2 one, i solved it properly but they are asking to give the answer in terms of Pie, now how do you express this number in terms of pie?

Next, the one you told for - x^1/3 = -2 how to solve this equation? i really dont know :P

and the 2 under root 2 one if you cube it out how will you end up with 8?

And yes you guessed right, im international have C12 on wednesday..
9. (Original post by userxx990)
bro, firstly thank you so much for such a great help!

Just the 3tan^2 one, i solved it properly but they are asking to give the answer in terms of Pie, now how do you express this number in terms of pie?

Next, the one you told for - x^1/3 = -2 how to solve this equation? i really dont know :P

and the 2 under root 2 one if you cube it out how will you end up with 8?

And yes you guessed right, im international have C12 on wednesday..
Cube both sides

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10. (Original post by userxx990)
bro, firstly thank you so much for such a great help!

Just the 3tan^2 one, i solved it properly but they are asking to give the answer in terms of Pie, now how do you express this number in terms of pie?

Next, the one you told for - x^1/3 = -2 how to solve this equation? i really dont know :P

and the 2 under root 2 one if you cube it out how will you end up with 8?

And yes you guessed right, im international have C12 on wednesday..
was it tan^2theta=1/3 ?
square root both sides
so you get tantheta= +-root3/3
it should work use a graph of tan to work it out
or CAST but graph is better
and its aite
11. (Original post by frozo123)
was it tan^2theta=1/3 ?
square root both sides
so you get tantheta= +-root3/3
it should work use a graph of tan to work it out
or CAST but graph is better
and its aite
ohh so radians mode was needed, after that ill simply put the answer and thats in terms of pie right?
12. (Original post by physicsmaths)
Cube both sides

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yeep thankss!
13. 1 more question i had forgot, how to factorise, 2x^3 - 16 ?
14. Why is arn used instead of arn-1 in question 6 January 2010?

Question Paper
Mark Scheme
Examiner's Report
15. (Original post by userxx990)
1 more question i had forgot, how to factorise, 2x^3 - 16 ?
You can't
16. (Original post by Cosmi)
Why is arn used instead of arn-1 in question 6 January 2010?

Question Paper
Mark Scheme
Examiner's Report
hard to explain but it says 3 years after it was purchased, one year after it was purchased would be x0.8, 2 years after would be x0.8^2 etc.
17. [QUOTE=userxx990;55935917]1 more question i had forgot, how to factorise, 2x^3 - 16 ?[/QUOTE

Hi you can try factor theorem and see if anything can factor in to it . Then factorise x-2 is a solution and then you can divide by it hope that helps
18. (Original post by Cosmi)
You can't
Yes you can.

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19. (Original post by userxx990)
1 more question i had forgot, how to factorise, 2x^3 - 16 ?
X=2 is asolution so divide by x-2

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20. (Original post by physicsmaths)
X=2 is asolution so divide by x-2

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how did you get x-2 from?
21. where can you get the info for 100UMS on each year

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