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The Edexcel C4 (21/06/12 - PM) Revision Thread

Hi can anyone help me on question 7 from the January '07 paper? It involved a vector by another vector and I just can't get my head around it :frown:

. Its only one mark so I'm sure it must be something really simple and stupid that I've missed. Thanks.

Reply 981

Brainiac

Are the solomon papers meant to be harder or easier than the actual past papers for c4?

Reply 982

JordanS94

16

Original post by Arva

Oh god I'm crap at differential equations... Trying to go through all the ones in the textbook but when I get one wrong there's no worked solution so all I can do is try it again. I think I'm going to start doing papers instead now.

I'm terrible at the water going into a cylinder questions :s-smilie:

I can't understand them, other than that and a tricky vector question I'm fine :smile:

Reply 983

avipsita

Original post by Brainiac

Are the solomon papers meant to be harder or easier than the actual past papers for c4?

i thought they where a bit easier than the edxcel papers i dunno why tho ?/
:confused:

Reply 984

Jack_Smith

8

what would you do if both ln are -
if one is negative you divide, if positive you multiply. So what do you do if both are negative?

Reply 985

avipsita

Original post by Belton42

Hi can anyone help me on question 7 from the January '07 paper? It involved a vector by another vector and I just can't get my head around it :frown:

. Its only one mark so I'm sure it must be something really simple and stupid that I've missed. Thanks.

all u have to do is add the vector too get vector c ...jst add it

Reply 986

Brainiac

Original post by avipsita

i thought they where a bit easier than the edxcel papers i dunno why tho ?/
:confused:

I thought that too, but i have read other people saying they find them harder...

Reply 987

Arva

14

Original post by Jack_Smith

what would you do if both ln are -
if one is negative you divide, if positive you multiply. So what do you do if both are negative?

-\ln a = \ln \dfrac{1}{a}[br][br]\therefore -\ln (a) - \ln (b) = \ln \dfrac{1}{a \times b}

Reply 988

Wilko94

6

Does anyone think they could help me interpreting a mark scheme?
I have the correct answer, but I'm a bit confused with something they've done.

I used the formula for a parallelogram:

area=base \times height[br]\sqrt 43 \times sin(71) \times \sqrt 14

Why does the mark scheme divide and multiply by 2?

(edited 11 years ago)

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ImageUploadedByStudent Room1339941775.646642.jpg127.5KB

Reply 989

ezioaudi77

Original post by Wilko94

Does anyone think they could help me interpreting a mark scheme?
I have the correct answer, but I'm a bit confused with something they've done.

I used the formula for a parallelogram:

area=base \times height[br]\sqrt 43 \times sin(71) \times \sqrt 14

Why does the mark scheme divide and multiply by 2?

In the markscheme, the areas of two triangles(ABD and ACD) are found. These two are then added to give the total area of the parallelogram. But your way is OK too.

Reply 990

raheem94

13

Original post by Arva

Oh god I'm crap at differential equations... Trying to go through all the ones in the textbook but when I get one wrong there's no worked solution so all I can do is try it again. I think I'm going to start doing papers instead now.

Don't you have the solution bank?

Reply 991

soempty

3

Why do we always use position vector of the line to find scalar product? Why cant we use two points on it?

Reply 992

raheem94

13

Original post by soempty

Why do we always use position vector of the line to find scalar product? Why cant we use two points on it?

Do you mean direction vector or position vector?

Reply 993

exe

Original post by soempty

Why do we always use position vector of the line to find scalar product? Why cant we use two points on it?

we don't? we use the direction vector, which is basically the same thing as two points on it

Reply 994

keerthi3

Hey guys, which past paper do you think is the hardest?

Reply 995

emmaaa88

12

I'm doing January 2007 Question 2(b), I'm fine with working out that 3/(0.75) is 4, but I don't understand why you do 4^3 when timesing it by the earlier volume? I would only have thought to multiply it by 4. Here's the question/mark scheme, hope my problem isn't too muddled:

Reply 996

raheem94

13

Original post by soempty

Why do we always use position vector of the line to find scalar product? Why cant we use two points on it?

Original post by exe

we don't? we use the direction vector, which is basically the same thing as two points on it

When we write the equation of a straight line, we write it as

y = mx + c

where

m

is the gradient.

In the vector case, the direction vector is like the gradient.

Reply 997

Jack_Smith

8

5 ln 4/5+ 2 ln 5/4
= 3 ln 4/5
= ln 64/125

how do you get 3ln 4/5?

Reply 998

fudgesundae

16

Original post by emmaaa88

I'm doing January 2007 Question 2(b), I'm fine with working out that 3/(0.75) is 4, but I don't understand why you do 4^3 when timesing it by the earlier volume? I would only have thought to multiply it by 4. Here's the question/mark scheme, hope my problem isn't too muddled:

Because volume is the length cubed, the volume scale factor must also be the length scale factor cubed.