Caribexams math exam guide for the CXC CSEC May/June 2010 exams is posted
Caribexams math exam guide for the CXC CSEC May/June 2010 exams is posted!
Even though I didn't have much to do with it, I am still announcing that the caribexams math exam guide for the CXC CSEC May/ June 2010 exam is now posted in the community.
This math exam guide is meant to guide students sitting the CXC CSEC math exam in May/June 2010 and forward.
The new CXC CSEC math exam marking scheme for May/June 2010 is posted, and so are the topics for paper 1 and paper 2 of the May/June 2010 math exam.
Since the new (May/June 2010 and forward) math guide is NOT for students sitting the CXC CSEC math exam in January 2010, the old caribexams CXC CSEC math guide will remain posted for their use.
So for now, caribexams will provide guides for the two math exams, January 2010 (the old guide) and May/June 2010 (the new guide).
The time management guide for the CXC CSEC math exam stays the same since the number of questions to be answered in the exam remains the same.
I hope our members and guests find this updated CXC math exam guide useful.
i am having a problem with maths like a problem like 5p2+9pq-2p2 i don't really understand
I NEED SOME SERIOUS HELP WITH THESE SUBJECTS... CAN ANYONE HELP ME????
I need someone to help me write a program using pascal for my SBA
I CAN'T BELIEVE ITS ALMOST HERE,I'VE BEEN DREAMIN' OF THIS A LONG TIME NOW 'N' I AM HOPIN 2 DO MY BEST. MATHS 'N' ACCOUNTS HAVE ALWAYS BEEN MY CHALLENGE BT I'LL HAVE 2 TRY MY BEST 'N' PUT MY BEST FOOT FOWARD.I WISH GUD LUCK 2 ALL 'D' OTHER CANDIDATES OF DIS EXAM. I KW DAT WE ALL CAN DO WELL IF WE PUT SOME TRUST 'N' CONVEIDENCE IN OURSELF.GUD LUCK ONCE AGAIN 2 ALL U WONDERFUL CANDIDATES.
i really need your help to pass my cxc maths exam in june in ja
Can some demonstrate to me how to solve:
Question: Solve the following pair of equations for x&y:
y+4x=27
xy+x=40
y+4x=27.. equation 1
xy+x=40.. equation 2
from eq 1 y=27-4x
so x[27-4x]+x=4o
27x-4x*+x=40 {*means squared}
28x-4x*=40
-4x*+28x-40=o....quadratic eq
-4x*+20x+8x-40=0
-4x[x-5] +8[x-5]=o
[-4x+8][x-5]=o
-4x+8=0 x-5=0
x=-8/-4 x=5
either x=2 or x=5
when x=2 in eq 1 when x=5 in eq 1
y+4[2]=27 y+4[5]=27
y=19 y=7
...theres probably a shorter way to do it
Recent comments
7 hours 46 min ago
8 hours 57 min ago
9 hours 55 sec ago
9 hours 29 min ago
12 hours 10 min ago
12 hours 16 min ago
15 hours 47 min ago
18 hours 53 min ago
19 hours 31 min ago
19 hours 38 min ago