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Is from now till the 2nd of November adequate time to prepare for the MAT? Had a change of heart recently and decided to take MAT with hope of offer from Imperial. I have scored above 90 ums in all my maths and further modules. Will 3 weeks be enough time to prepare?
Original post by KloppOClock
Also, is anyone here going to that MAT support event thing tomorrow at manchester uni?


how was the MAT thingy in manchester, did they talk about anything helpful?
Original post by DylanJ42
how was the MAT thingy in manchester, did they talk about anything helpful?


not really, kinda **** tbh. the only new things we did were modulus graphs and floor function graphs which only really appear with step. although, they said they were gonna email some additional resources to our maths teachers to give us. if any of the stuff they give us is useful, ill pass it on.
Original post by KloppOClock
not really, kinda **** tbh. the only new things we did were modulus graphs and floor function graphs which only really appear with step. although, they said they were gonna email some additional resources to our maths teachers to give us. if any of the stuff they give us is useful, ill pass it on.


PRSOM, thanks :smile:

Spoiler

Original post by DylanJ42
PRSOM, thanks :smile:

Spoiler


you didnt miss out on anything :smile:
Original post by Aggirus
Is from now till the 2nd of November adequate time to prepare for the MAT? Had a change of heart recently and decided to take MAT with hope of offer from Imperial. I have scored above 90 ums in all my maths and further modules. Will 3 weeks be enough time to prepare?


Work hard and find out tbh.
Not tryna be a dick btw just saying it how it is no one can say whther it is or not enough time for u but Mat aint possible just do all the papers and see what happens. Tips are all over thread so read thru


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Hey guys, I'm trying to leave aside a few MAT papers for later on so i'm thinking of doing some STEP 1/2 questions. I dont think i'll have time to do full papers especially since some topics aren't included in the MAT. with the spreadsheet on the STEP thread showing the different topics of the questions, are there any topics that i should focus on and do for STEP which would help for MAT? Thanks!
Original post by Aggirus
Is from now till the 2nd of November adequate time to prepare for the MAT? Had a change of heart recently and decided to take MAT with hope of offer from Imperial. I have scored above 90 ums in all my maths and further modules. Will 3 weeks be enough time to prepare?


3 weeks is not much but if you're good at maths you don't need to study too much.
BTW, search for DrFrost, he is your lord and saviour :smile:.
IMG_4871.jpgCan someone tell me where I went wrong? This is MAT 2013 Q5 last part
Reply 289
Can anyone help with question C on the 2008 MAT please? I know you have to rearrange them so they equal to x and then set both equations equal to each other, but i'm not getting the same arrangement as the mark scheme.
Reply 290
Original post by Mitchb777dotcom
What's everyone scoring on the 2008 paper? I just got 67 but it's 2 years before the score data, as it's still a relatively recent paper I hope 65-75 is still a very competitive score?


How did you do question C on the 2008 paper? I know you have to rearrange both equations so they equal to x or y and then set them to each other, but im struggling to get the same answer as the mark scheme. How did you do tis question?
Screen Shot 2016-10-14 at 19.08.32BST.png

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How do they get (A+B) +B=1
Original post by J1998P
Can anyone help with question C on the 2008 MAT please? I know you have to rearrange them so they equal to x and then set both equations equal to each other, but i'm not getting the same arrangement as the mark scheme.


Think of the two equations as equations of lines (sin and cos are just numbers/cooefficients and re-arrange to make y\displaystyle y the subject;
[br](sinθ)y=(cosθ)x2EquationOne[br](cosθ)y=1(sinθ)xEquationTwo[br][br]Eqn1:y=((cosθ)x2)/sinθ[br]Eqn2:y=(1(sinθ)x)/cosθ[br]\displaystyle[br](sin \theta )y=(cos \theta )x-2 \gets Equation One[br](cos \theta )y=1-(sin \theta )x \gets Equation Two[br][br]Eqn1: y= ((cos \theta )x -2) /sin\theta[br]Eqn2: y= (1-(sin \theta )x)/cos\theta[br]

As these are both straight lines, they will only have solutions if they intersect.

Gradient of Eqn1 is Cos/Sin =1/Tan
gradient of Eqn2 is -Sin/Cos = -Tan

As those gradient are different, they both are not parallel, so they intersect at some point, so they always have a solution
(edited 7 years ago)
For someone that hasnt prepared at all yet, what would be the best way to prepare?
Original post by jamestg
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How do they get (A+B) +B=1


Let the area of 0-1 be A.
Let the area of 1-2 be B.

The first equation says 3 times the area A plus 2 the area B = 7
The second equation says the area of 0-2 (which is the area of 0-1+1-2) which is A+B, plus the area B = 1.

So you get 3A+2B=7
2A+B=1
and solve for A and B then add them together.
Original post by BurdMan
Can someone explain MAT15 Q3 Part VI?

I have absolutely no idea where they got anything in their solution from.


bump
Reply 296
Original post by KloppOClock
Think of the two equations as equations of lines (sin and cos are just numbers/cooefficients and re-arrange to make y\displaystyle y the subject;
[br](sinθ)y=(cosθ)x2EquationOne[br](cosθ)y=1(sinθ)xEquationTwo[br][br]Eqn1:y=((cosθ)x2)/sinθ[br]Eqn2:y=(1(sinθ)x)/cosθ[br]\displaystyle[br](sin \theta )y=(cos \theta )x-2 \gets Equation One[br](cos \theta )y=1-(sin \theta )x \gets Equation Two[br][br]Eqn1: y= ((cos \theta )x -2) /sin\theta[br]Eqn2: y= (1-(sin \theta )x)/cos\theta[br]

As these are both straight lines, they will only have solutions if they intersect.

Gradient of Eqn1 is Cos/Sin =1/Tan
gradient of Eqn2 is -Sin/Cos = -Tan

As those gradient are different, they both are not parallel, so they intersect at some point, so they always have a solution


Thank You!
On the syllabus, under differentiation one of the points is "derivative of a sum of functions". What does this mean? When i search it up it comes up with a different form of differentiation so are we supposed to learn this
Original post by blurred
On the syllabus, under differentiation one of the points is "derivative of a sum of functions". What does this mean? When i search it up it comes up with a different form of differentiation so are we supposed to learn this


AFAIK, it just means that the derivative of f(x) + the derivative of g(x) = the derivative of (f(x) + g(x)). So it means that if you want to differentiate 2x + 5x^2, you'd differentiate 2x (2), differentiate 5x^2 (10x), and add them together (2 + 10x)
Hi, can anyone shed any light on 2013 1J multi choice, i can't make sense of some of their justifications in the mark scheme thank you