For the 3D pythagoras:(Original post by GCSESTUDENT5000)
Maybe... I'm not sure. I don't think there has been a decisive majority of what the answer for this question is; 2 people have got 42.8 and 2 have got 36.9 so I'm not sure yet
sqrt(16^{2 }+ 22^{2})= 2sqrt(185)
Divide this by 2 to get the value of the center of the rectangle to the point A (length of AX)
Therefore, AX is sqrt(185)
Therefore the height of the pyramid (VX) is sqrt(17^{2}  sqrt(185)^{2}) = 2 sqrt(26)
Then find the midpoint of AB = 8
Find the length of V to the midpoint of AB = sqrt(17^{2 } 8^{2}) = 15
Therefore angle of plane is sin^{1}(2 sqrt(26) / 15) = 42.83 (2dp)
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AQA Level 2 Certificate in Further Mathematics UNOFFICIAL MARKSCHEME Paper 2 Watch
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 24062016 13:50

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 24062016 13:52
(Original post by GCSESTUDENT5000)
How did you find the paper? 
GCSESTUDENT5000
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 24062016 13:53
(Original post by Redcoats)
For the 3D pythagoras:
sqrt(16^{2 }+ 22^{2})= 2sqrt(185)
Divide this by 2 to get the value of the center of the rectangle to the point A (length of AX)
Therefore, AX is sqrt(185)
Therefore the height of the pyramid (VX) is sqrt(17^{2}  sqrt(185)^{2}) = 2 sqrt(26)
Then find the midpoint of AB = 8
Find the length of V to the midpoint of AB = sqrt(17^{2 } 8^{2}) = 15
Therefore angle of plane is sin^{1}(2 sqrt(26) / 15) = 42.83 (2dp) 
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 24062016 13:55
(Original post by Redcoats)
Really easy. For the non calculator paper, I apparently got 70 / 70 according to the unofficial mark scheme and for this paper I got all the same answers apart from the 3D pythagoras so anything between 100  105 / 105
Well done 
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 24062016 13:57
(Original post by GCSESTUDENT5000)
I will try and remember everything. Please correct me if I'm wrong
1) 15 square units [3]
2) a) x=2 [1]
b) x=0.8, x=4.8 [2]
3) a) (c/a, 0) [1]
b) b/a [1]
I don't know the question number for these:
sin^2(x)  3 cos^2(x) = sin^2(x)  3(1sin^2(x)) = sin^2(x)  3 + 3sin^2(x)
= 4sin^2(x)  3 [2]
solutions of the above is equal to 0 are x = 60, 120, 240, 300 (all in degrees) [4]
y > 45 degrees for tan y [1]
sin y = (p+1)/(root (2p^2+2)) [4]
p = 9 for the coefficient is 23 [3]
For the pyramid, I got 36.9 degrees but I'm not sure [4]
The graph started in the left and went below the x axis on the right [3]
To prove y = 3x on the parallel lines, you had to do angles around a point make 360 degrees and then use supplementary angles [4]
x/y = 12 [3]
f(x+1)  f(x) = 1/(2x+1)(2x+3) [5]
The 2/5* root x equation was 25/4 [2]
x^3 = 5x^2 was x=0, x=5 [2]
The nth term was 2n+5 [2]
The quadratic nth term was 6n^2+13n5 [3]
The circle theorems was 37.5 degrees [4]
The matrix mapping the points was that they can be the same as long as a = 3 [4]
The matrix transformation was (3 0) so scale factor 3 [5]
0 3
The equation with 3/x2 + .... was something like x = 1.23, x= 2.77 [6]
The length of CB was 32 units [6]
The range of f(x) was 3 <= f(x) <= 6 [2]
The tick box question was left box, middle box, right box, middle box [4]
The quadratic graph and coordinates of P were (0,7) [4]
The coordinates of P with the line ratios were (9/2, 47/8) [3]
The simplify fraction was x+2/x+3 I think [2]
The simplify was something like w^3x^2y^2 (w^2+y) [2]
The make the subject of the formula was something like x = 8w/y+8 [4]
The coordinates of intersection were (1.4, 3.4) [3]
When you were given the gradient and had to find k, k was 1 [4]
I can't remember 2 marks, sorry. Please tell me if you can remember but I will try and remember myself!
Hope this helps
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 24062016 14:04
(Original post by ErinMei)
I think i got all of these apart from pyramid angle i got 48 or something
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Well, everyone seems to be getting different answers on this pyramid question so you might be right! 
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 24062016 14:07
(Original post by poohplop)
The length of CB was 35 
Ishan_2000
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 24062016 14:08
Was the gradient a/b or b/a?
It was:
ax + by = c
by = c  ax
y = c/b  ax/b
So, isn't that a/b times by x, which gives the gradient? 
Bulbasaur10
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 24062016 14:13
(Original post by GCSESTUDENT5000)
I will try and remember everything. Please correct me if I'm wrong
1) 15 square units [3]
2) a) x=2 [1]
b) x=0.8, x=4.8 [2]
3) a) (c/a, 0) [1]
b) b/a [1]
I don't know the question number for these:
sin^2(x)  3 cos^2(x) = sin^2(x)  3(1sin^2(x)) = sin^2(x)  3 + 3sin^2(x)
= 4sin^2(x)  3 [2]
solutions of the above is equal to 0 are x = 60, 120, 240, 300 (all in degrees) [4]
y > 45 degrees for tan y [1]
sin y = (p+1)/(root (2p^2+2)) [4]
p = 9 for the coefficient is 23 [3]
For the pyramid, I got 36.9 degrees but I'm not sure [4]
The graph started in the left and went below the x axis on the right [3]
To prove y = 3x on the parallel lines, you had to do angles around a point make 360 degrees and then use supplementary angles [4]
x/y = 12 [3]
f(x+1)  f(x) = 1/(2x+1)(2x+3) [5]
The 2/5* root x equation was 25/4 [2]
x^3 = 5x^2 was x=0, x=5 [2]
The nth term was 2n+5 [2]
The quadratic nth term was 6n^2+13n5 [3]
The circle theorems was 37.5 degrees [4]
The matrix mapping the points was that they can be the same as long as a = 3 [4]
The matrix transformation was (3 0) so scale factor 3 [5]
0 3
The equation with 3/x2 + .... was something like x = 1.23, x= 2.77 [6]
The length of CB was 32 units [6]
The range of f(x) was 3 <= f(x) <= 6 [2]
The tick box question was left box, middle box, right box, middle box [4]
The quadratic graph and coordinates of P were (0,7) [4]
The coordinates of P with the line ratios were (9/2, 47/8) [3]
The simplify fraction was x+2/x+3 I think [2]
The simplify was something like w^3x^2y^2 (w^2+y) [2]
The make the subject of the formula was something like x = 8w/y+8 [4]
The coordinates of intersection were (1.4, 3.4) [3]
When you were given the gradient and had to find k, k was 1 [4]
I can't remember 2 marks, sorry. Please tell me if you can remember but I will try and remember myself!
Hope this helps
I got pretty much the same, but a few of my answers were different. I may have done these completely wrong, but I'm pretty sure I got the leftmost box for the last 'tick' question. I also got completely different x values for the last trig identity question, and different coordinates of intersection. Also, I'm not sure whether it was b/a. I thought it was b/a, as the equation was ax+by = c, so it was the same as by = ax+c, and y = a/bx+c. I must have misread it because I thought you had to work out the gradient of a line perpendicular to it, and did the negative reciprocal of a/b, which is b/a. (It must have said parallel.... ) Sorry for the long post.... I'm probably wrong. 
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 24062016 14:17
(Original post by Ishan_2000)
Was the gradient a/b or b/a?
It was:
ax + by = c
by = c  ax
y = c/b  ax/b
So, isn't that a/b times by x, which gives the gradient?
Sorry, that's my bad memory  I'll edit the markscheme 
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 24062016 14:19
(Original post by Bulbasaur10)
I got pretty much the same, but a few of my answers were different. I may have done these completely wrong, but I'm pretty sure I got the leftmost box for the last 'tick' question. I also got completely different x values for the last trig identity question, and different coordinates of intersection. Also, I'm not sure whether it was b/a. I thought it was b/a, as the equation was ax+by = c, so it was the same as by = ax+c, and y = a/bx+c. I must have misread it because I thought you had to work out the gradient of a line perpendicular to it, and did the negative reciprocal of a/b, which is b/a. (It must have said parallel.... ) Sorry for the long post.... I'm probably wrong.
It said b>1 and 1<c<1
The statement was bc>1
So, if b = 2 and c = 0.5, bc=2.5
If b = 1.2 and c = 0.6, bc = 0.6
Therefore, it is only sometimes true I think
I might have remembered some of my answers wrong so you could be right, but, even if you're wrong, you still might get method marks as long as you have shown a valid method
Hope this helps 
Chittesh14
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 24062016 14:20
(Original post by GCSESTUDENT5000)
Yeah, I wasn't sure about the plane:
I got AC = root (740) using Pythagoras' theorem
I then got CX = root (185) and did cos1 (root(185)/17)= 36.9 degrees
I'm probably wrong ...
Please tell me your method
But it was the plane, there was 1 side which equals to 16 cm.
I found the midpoint connected it to the midpoint of the plane (ABCD or whatever it was) and then connected that line to the vertex or whatever it was V.
So, I think the midpoint of the base was X.
I labelled the midpoint half way through the 16 cm side as D.
So XD was 11 cm because it was 1/2 AB which was 22 cm.
I forgot what it was but I think VB = VC and they both equal to 17 cm.
So, VB or whatever it was.. VB^2  VD^2 (8 cm) = one of the side = 15 cm.
So, angle theta = Cos 1 (A/H) = Cos 1 (11/15) = 42.8...
Surprised how I remember it lmfao.
I also worked out an extra side just incase to crosscheck and it was O/A then so tan1 (10.2 or something/11) = 42.8 also
and then I also done sin lool sin1 (10.2/15) = 42.8 
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 24062016 14:20
(Original post by Bulbasaur10)
I got pretty much the same, but a few of my answers were different. I may have done these completely wrong, but I'm pretty sure I got the leftmost box for the last 'tick' question. I also got completely different x values for the last trig identity question, and different coordinates of intersection. Also, I'm not sure whether it was b/a. I thought it was b/a, as the equation was ax+by = c, so it was the same as by = ax+c, and y = a/bx+c. I must have misread it because I thought you had to work out the gradient of a line perpendicular to it, and did the negative reciprocal of a/b, which is b/a. (It must have said parallel.... ) Sorry for the long post.... I'm probably wrong. 
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 24062016 14:21
(Original post by MaxHSloan)
I also thought you had to find the perpendicular gradient, I got the same as you
I just went with b/a at the end.. 
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 24062016 14:23
(Original post by GCSESTUDENT5000)
The left tick box I think was always true
It said b>1 and 1<c<1
The statement was bc>1
So, if b = 2 and c = 0.5, bc=2.5
If b = 1.2 and c = 0.6, bc = 0.6
Therefore, it is only sometimes true I think
I might have remembered some of my answers wrong so you could be right, but, even if you're wrong, you still might get method marks as long as you have shown a valid method
Hope this helpsLast edited by Bulbasaur10; 24062016 at 14:24. 
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 24062016 14:23
So what is the answer to the 3D Pythagoras question

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 24062016 14:24
(Original post by MaxHSloan)
I also thought you had to find the perpendicular gradient, I got the same as you 
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 24062016 14:24
(Original post by GCSESTUDENT5000)
x
How did you get that, did you also get x = 12y and do 12y/y = 12? 
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 24062016 14:24
(Original post by Bulbasaur10)
Ah..... I thought it said more than 0!!! Thanks. 
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 24062016 14:26
(Original post by GCSESTUDENT5000)
x
So, you've got 1 question or something where you've wrote an extra mark lol.
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