Maths A level Core 1 Last resit 2019 May 15th Watch

vx038
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#61
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yep, i got d as 12 and u17 as 208
(Original post by patm47)
Then did everyone get d in the series question as 12 the U17 as 208. I remember for the sum of the sequence being 140 but don’t know if all of these are right
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Virolite
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#62
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this what i got, idk where the x coords for c and b = 3 is coming from :/
(Original post by Ksg5465)
The value of c was 8/9.
T was 11/2
Coordinayes of B was 4,8
And c was -4,2
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Ksg5465
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#63
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Also k was -4
D =12
U17 = 208
Equation of normal was x-3y-1=0
Dx2 question was 44
(Original post by vx038)
yep, i got d as 12 and u17 as 208
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Virolite
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#64
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boys i reckon it's gonna be around 67/68 for an A
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Ksg5465
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#65
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#65
Was dx2 question when x=9
44
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Evil Homer
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#66
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Just create a new thread here with the list of answers you guys have already:

link to creating a new thread
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Ksg5465
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#67
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Dx2 question was 44
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vx038
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#68
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The gradient for the straight line was 3/4 something like that, cuz its perpendicular to the curve, you set the reciprocal -4/3 equal to the differential -12/x^2

(Original post by Virolite)
this what i got, idk where the x coords for c and b = 3 is coming from :/
Last edited by vx038; 1 week ago
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Ksg5465
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#69
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The gradient was parallel to the line 4y=3x+5 so set dy/dx = to 3/4. The 3x2 = 48.
48/3 =16. The x is 4 or -4
(Original post by vx038)
The gradient for the straight line was 3/4 something like that, cuz its perpendicular to the curve, you set the reciprocal -4/3 equal to the differential -12/x^2
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H.Ash999
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(Original post by Ksg5465)
The gradient was parallel to the line 4y=3x+5 so set dy/dx = to 3/4. The 3x2 = 48.
48/3 =16. The x is 4 or -4
Dy/dx was -12x^-2 so you can’t set it to 3/4 - as you can’t square root a negative number. So set dy/dx to -4/3 as this is the normal, and you get the x- values at 3 & -3
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dumbasslmao
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#71
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How many marks was that
(Original post by Ksg5465)
Dx2 question was 44
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Virolite
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#72
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parallel not perpendicular
(Original post by vx038)
The gradient for the straight line was 3/4 something like that, cuz its perpendicular to the curve, you set the reciprocal -4/3 equal to the differential -12/x^2
this is correct.
(Original post by Ksg5465)
The gradient was parallel to the line 4y=3x+5 so set dy/dx = to 3/4. The 3x2 = 48.
48/3 =16. The x is 4 or -4
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Virolite
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#73
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unofficial mark scheme lads

https://www.thestudentroom.co.uk/sho...4#post83213634
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ian.blanch
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#74
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#74
Area was 6
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Virolite
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#75
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has anyone got changes to unofficial mark scheme?
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sdx47x
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#76
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is any of it wrong lol
(Original post by Virolite)
has anyone got changes to unofficial mark scheme?
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Virolite
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(Original post by sdx47x)
is any of it wrong lol
nah but maybe the questions i forgot people can remember?
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mariya_mx
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#78
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Well I realized what I did wrong for question 10 and the values for a and k. I feel I could have done more if I checked properly with my working out. But other than that I think it was a good paper
(Original post by gapyhgal)
hi everyone im pretty sure ive not done that well on that paper anymore. i didnt get t or the area ( 5) i dont remember doing the y intercept bc im stupid and then i got something stupid so pretty sure ive got 65/75 which will most deffo be a C. how do you all think it went
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vx038
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#79
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sorry, it wasn't perpendicular. I am pretty sure the question was something like the line that passes through point B and the line that passes through point C are both normals to curve H. The lines are parellel to the straight line with equation 4y=3x+5.
(Original post by Virolite)
parallel not perpendicular

this is correct.
Last edited by vx038; 1 week ago
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sdx47x
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#80
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my unofficial markscheme from earlier has just disapperared :/
(Original post by Evil Homer)
Just create a new thread here with the list of answers you guys have already:

link to creating a new thread
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