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# Maths exam IGCSE - HIGHER TIER - PAPER 3H AND PAPER 4H POST DISCUSSION.

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1. (Original post by rapdog)
how do you find the gradient of the line x + 2y =6
x + 2y=6
2y= 6 - x
y = (6-x)/2 = 6/2 - x/2
y = -0.5 x + 3
y = mx + c
2. (Original post by gabriel 41)
Rearrange it in the form y=mx+c

so you get:

2y=-x+6

Divide the both sides by two to make y the subject:

y=-x/2+3

So your gradient is -x/2 (by the way which exam series question is this , for I want to confirm my answer, and please be quick in replying)

it is nov 2007 4H
3. (Original post by gabriel 41)
That just mean you give your above calculated answer to the upper bound of the particular situation in question.
(62/45) < ROOT OF 2 < (64/45)

Use these bounds to rite the value of root 2 to an app. degree of accuracy. You must show....and explain your answer (2)
4. (Original post by rapdog)
it is nov 2007 4H
Sorry I wrote -x/2 instead of -1/2 the correct answer is -1/2

5. (Original post by StUdEnTIGCSE)
(62/45) < ROOT OF 2 < (64/45)

Use these bounds to rite the value of root 2 to an app. degree of accuracy. You must show....and explain your answer (2)
May 2006 paper 3H question 17 right? Convert the both the fractions into decimals, and then take the higher valued decimal and write it to 2 significant answers as your final answer.
6. how do you do this question because im pretty sure its kinematics "a body is moving in a straight line which passes through a fixed point O. the displacement s metres of the body from O at time t is given by s=t(cubed)+4t(squared)-5t. (a) find the expression for the velocity, vm/s at time t seconds. (b) find the acceleration after 2 seconds. i think it pretty hard

this is question is in paper 3H may 2004
7. Anyone have the mark scheme for the 2 jan 2012 papers?
8. how is a recurring decimal converted to fraction?
9. (Original post by kish667)
how do you do this question because im pretty sure its kinematics "a body is moving in a straight line which passes through a fixed point O. the displacement s metres of the body from O at time t is given by s=t(cubed)+4t(squared)-5t. (a) find the expression for the velocity, vm/s at time t seconds. (b) find the acceleration after 2 seconds. i think it pretty hard

this is question is in paper 3H may 2004
Yeah, that's kinematics. You need to differentiate it so it becomes; v =3t^2+8t-5. (a) Then for (b) Substitue in the 2.
10. (Original post by kish667)
how do you do this question because im pretty sure its kinematics "a body is moving in a straight line which passes through a fixed point O. the displacement s metres of the body from O at time t is given by s=t(cubed)+4t(squared)-5t. (a) find the expression for the velocity, vm/s at time t seconds. (b) find the acceleration after 2 seconds. i think it pretty hard

this is question is in paper 3H may 2004
Okay, that's usually called "calculus" in simple terms.
11. (Original post by kish667)
how do you do this question because im pretty sure its kinematics "a body is moving in a straight line which passes through a fixed point O. the displacement s metres of the body from O at time t is given by s=t(cubed)+4t(squared)-5t. (a) find the expression for the velocity, vm/s at time t seconds. (b) find the acceleration after 2 seconds. i think it pretty hard

this is question is in paper 3H may 2004
It's calculus: displacement differentiated is velocity, so the answer to part a is simply that.
For part b, velocity differentiated is acceleration. To find the acceleration at 2 seconds, I think you substitute in the 2 for t. The equation should be 6t+8, so 6(2)+8= 20
Feel free to correct me if i'm wrong :/
12. 4MA0_3H_msc_20120307.pdf4MA0_4H_msc_20120307.pdf
(Original post by fajita.and.friends)
Anyone have the mark scheme for the 2 jan 2012 papers?
find them attached
13. Thank you
14. (Original post by kish667)
how do you do this question because im pretty sure its kinematics "a body is moving in a straight line which passes through a fixed point O. the displacement s metres of the body from O at time t is given by s=t(cubed)+4t(squared)-5t. (a) find the expression for the velocity, vm/s at time t seconds. (b) find the acceleration after 2 seconds. i think it pretty hard

this is question is in paper 3H may 2004
Basically, you multiply the number by the indices and then subtract one from the indices. For example:

t^3 becomes 3t^2.

5t is the equivalent of 5t^1. So that becomes just 5. (the t^1 becomes t^0 so its gone).

If it is a number with no letter after it such as just 10, then we just ignore it altogether as it becomes 10^0.

Note: 1/x is the equivalent of x^-1 . So to differentiate it would become -x^-2. To find the acceleration, simply differentiate again your answer to the velocity and then substitute in the given acceleration. Finally, remember the positive and negative signs!
15. If function graphs come up I'm fu!ked
16. yeh but i dont understand what it is or how you do it
(Original post by gabriel 41)
Okay, that's usually called "calculus" in simple terms.
17. simultaneous equations question: 3x+4y=10 and 5x-6y=53
i rearranged to 15x+20y=50 and 15x-18y=69 and somehow got two completely incorrect answers...how would you progress from there? :/
18. (Original post by XxRahulXx)
If function graphs come up I'm fu!ked
theyre actually easy once you understand them

look through papers like nov 2005 3H or more recent ones, theres some on that

its just like find f(3) on a graph
19. has anybody got the mark scheme for may 2006 3H ?
20. is this the mathematics b specification?

anyway we've got the first paper tomorrow for me its the issue of timing rather than knowledge and application

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