You are Here: Home >< Maths

# Edexcel IAL Core Mathematics - C12 - WMA01 - JUNE 2016

Announcements Posted on
Why bother with a post grad course - waste of time? 17-10-2016
2. (Original post by Blazyy)
11/64
Yea I think it was 11/64

Posted from TSR Mobile
3. [QUOTE=hasi.99;65147757]Yea I think it was 11/64
no it will be 71/64
4. wht finished ???
(Original post by TeeEm)
5. i got 0.33 as the area
6. how did u guys solve 100-3r sum question ?
7. (Original post by azizmansoor)
wht finished ???
The search for Elvis
8. (Original post by ToxicPhantom007)
how did u guys solve 100-3r sum question ?
You replace r with 1,2,3 to get a sequence. Then you'll realize that it's arithmetic with d=-3. Use the formula and that's all.
9. (Original post by hasi.99)
Yea I think it was 11/64

Posted from TSR Mobile
I'm pretty sure it's 11/64. My friend got the same answer.
You replace r with 1,2,3 to get a sequence. Then you'll realize that it's arithmetic with d=-3. Use the formula and that's all.
and what did you do for area in sector question ?
11. was it shaded area or whole area of shape ?
12. [QUOTE=Imran hasan;65147825]
(Original post by hasi.99)
Yea I think it was 11/64
no it will be 71/64
Area of triangle is 0.5*3/8*1.5 which is 0.28. Your answer is greater than the area of the triangle lol.
13. (Original post by ToxicPhantom007)
and what did you do for area in sector question ?
I got it wrong because j didn't subtract 0.4 from radius. But you do r-0.4 to get the side of the triangle. Find the area of the triangle. Find the area of sector. Subtract triangle from sector.
14. in that sector question what was the area supposed to be ? for the shaded part or for the whole shape
15. (Original post by ToxicPhantom007)
in that sector question what was the area supposed to be ? for the shaded part or for the whole shape
huh thank god
one of friends calc for the whole shape and it confused me a bit
17. im 100% sure it was 27/64 in the last one.
it was area under line- area under curve.
calculation was like this:integ(limit 2 to .5)[-.25x+.5]-integ(limit 2 to .5)[x^3-3x^2+2x]
18. (Original post by EUSHA)
im 100% sure it was 27/64 in the last one.
it was area under line- area under curve.
calculation was like this:integ(limit 2 to .5)[-.25x+.5]-integ(limit 2 to .5)[x^3-3x^2+2x]
Are you sure it is 100%? Because I got that answer too and I'm getting freaked out if I got that wrong too in addition to all the other silly mistakes I've done.
19. (Original post by I <3 WORK)
Are you sure it is 100%? Because I got that answer too and I'm getting freaked out if I got that wrong too in addition to all the other silly mistakes I've done.
um pretty sure
btw what was your values of angles where sin_theta was (1/3) and (-1/2)?
20. (Original post by EUSHA)
um pretty sure
btw what was your values of angles where sin_theta was (1/3) and (-1/2)?
I got 30 and -19.5 I think

Posted from TSR Mobile

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

Updated: May 28, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

### A level maths

Which modules should I take?

Poll
Useful resources