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# AQA A Level Maths Core 1 - 18th May 2016 [Exam Discussion] Watch

1. (Original post by money-for-all)
ooooooo , check you fancy pants , desmos and ****, you still clapped tho
Naw bruv, not into the alcohols and that kinda stuff - yolo remember. I'm glad the desmos image helped you. Feel free to PM me if you have more questions.
2. (Original post by craigie435)
you had to use pythagaros which got you route 64
It was √65:

3. What did we all get on the last integration question?
4. (Original post by Frazzles10)
That was easy it was 9
I got 9 aswell, annoying thing was if u got the first part of the question wrong you would've mucked the rest up
5. (Original post by Heidirennie)
What did we all get on the last integration question?
The integral came out to be 81/4 or 20.25.

The shaded area was 45/4 or 11.25

This has been answered before in the forum.
6. Do you think i will lose mark just because i did not simplify the answer 243/12 into 81/4？
7. oh gosh that paper went down hill all because of that circle question and the integration one, i literally couldn't find a trapezium or a triangle but as soon as people told me about the triangle i knew i failed
8. (Original post by goofyygoober)
Do you think i will lose mark just because i did not simplify the answer 243/12 into 81/4？
i doubt it unless the question said "simplify" or "in its simplest form" otherwise u would be able to get the OE (or equivalent) mark
9. How many marks would I get on the intergration question if i did it all right but somehow i managed to get the area of the triangle as 6 because i did 24 x 2/4 /2 instead of 24 x 3/4 /2 (i think this is what you were supposed to do?)

Also for the length of CT I added it up wrong and got root80 instead of root81 so would i get method marks or no marks at all?
10. (Original post by beanigger)
i doubt it unless the question said "simplify" or "in its simplest form" otherwise u would be able to get the OE (or equivalent) mark
thx!!
11. (Original post by Fambox)
I got the same I hope we are right !!! Mine was in fraction form like 135/12
Same for some reason I never bothered to convert them in decimals and thought I was wrong...thanks the mental peace lol

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12. did anyone get -5/4 as the x co-ordinate?
13. (Original post by aliyaahh)
did anyone get -5/4 as the x coordinate?
Yes, -5/4 was the x-coordinate of Q
14. (Original post by ella-1997)
Exam was easy overall, but i just couldn't work out question 5)b where you had to find the coordinates of b?? how did everyone do this?
I worked out that C was the midpoint, because AB was the diameter, so the distance AC was half of AB. Then just double the change in coordinates.
15. (Original post by aliyaahh)
did anyone get -5/4 as the x co-ordinate?

Yes. The Co-ordinates of Q were (-0.8,0) but you only needed to state the x.
16. (Original post by jakeheart291)
how many marks for Q8?????/
It was 7/8 marks
17. (Original post by SamCoulthard)
I worked out that C was the midpoint, because AB was the diameter, so the distance AC was half of AB. Then just double the change in coordinates.
You could do this, but I think it is easier and quicker to use the formula for a midpoint and sub in numbers you knew and worked out unknowns from this.
18. what was the equation that we had to integrate?
19. Does anyone remember what the integration equation after integrating was? 3x4/4 was one of them, right?
20. (Original post by SamCoulthard)
I worked out that C was the midpoint, because AB was the diameter, so the distance AC was half of AB. Then just double the change in coordinates.
Yeah I did that as well, seemed too simple at the time after thinking about it for a while. I said straight away the diameter would be 2*radius so using midpoint formula seemed than doing the midpoint thing below. Plus helped on the next question as I had already considered the gradient as -7/4 which made it easier to get the equation of the tangent to the circle. It was something like y-1=4/7(X+2) which simplified to 7x-4y+18=0

(Original post by BenAtkinson)
You could do this, but I think it is easier and quicker to use the formula for a midpoint and sub in numbers you knew and worked out unknowns from this.

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