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Year 13 Maths Help Thread

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Maths question.jpg

can someone help me with question as well please (last one sorry!)
the answer is supposed to be 129.6 degrees and 230.4 degrees to 1dp
Original post by cuppycake333
can someone help me with question as well please (last one sorry!)


You have values for cosx, not x. One is not possible, as |cosx|<=1.
Original post by Palette

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Don't know if I've missed the boat here, but I'm doing physics at uni (second year atm) and defo enjoy helping people out with maths
Original post by cuppycake333
Maths question.jpg

can someone help me with question as well please (last one sorry!)


As @RogerOxon said, you've solved a quadratic equation in cos(x), hence your values are for cos(x). When you reject the impossible solution, the remaining one gives the correct answers :smile:
Original post by Darth_Narwhale
My mistake, you are correct, k is pi/6. This gives the correct answer :smile:


oh ok that does work now thanks! :h:
M1 - Scale Pan.png

For part a, resolving the mass vertically, why don't you have to consider a reaction force of R on Q, the mark scheme says 15g-T=15a, but I wrote 15g-T-R=15a and I couldn't solve as there were three unknowns.
Original post by NotNotBatman


For part a, resolving the mass vertically, why don't you have to consider a reaction force of R on Q, the mark scheme says 15g-T=15a, but I wrote 15g-T-R=15a and I couldn't solve as there were three unknowns.

When you resolve forces, all the forces should be acting on the same particle. The reaction force on Q only acts on Q so you could use it in your working but only if you're considering the forces on Q only.

So you could do this if you're really careful : resolve for Q, R and the triangle separately but most students are bound to make mistakes if they try this.

There is a much easier way...

In M1 always look for connected particles whose forces all act in the same straight line. You can treat this kind of system as one particle.

So here you can treat Q, R and the (light) triangle as one particle of mass 15kg. So the only forces acting on this combined particle are tension and 15g.
(edited 7 years ago)
Screenshot_20170218-113543.pngs2 solomon d q4

for q4 part but why is the x~Po (8/6) as it does say given its on the day it is open so why don't u do 8/7 as there's 7 days in a week and it doesn't say given the shop is open so how are you supposed to know it is referring to the days it is open?is it a badly worded q?
Also can someone explain part c as I let s be the number in stock and worked out
P (X>s)<0.05 and found some but it was wrong
This is s2 edexcel solomon d
Original post by notnek
When you resolve forces, all the forces should be acting on the same particle. The reaction force on Q only acts on Q so you could use it in your working but only if you're considering the forces on Q only.

So you could do this if you're really careful : resolve for Q, R and the triangle separately but most students are bound to make mistakes if they try this.

There is a much easier way...

In M1 always look for connected particles whose forces all act in the same straight line. You can treat this kind of system as one particle.

So here you can treat Q, R and the (light) triangle as one particle of mass 15kg. So the only forces acting on this combined particle are tension and 15g.


Thanks for the help.
Original post by Youngkeynes
Screenshot_20170218-113543.pngs2 solomon d q4

for q4 part but why is the x~Po (8/6) as it does say given its on the day it is open so why don't u do 8/7 as there's 7 days in a week and it doesn't say given the shop is open so how are you supposed to know it is referring to the days it is open?is it a badly worded q?
Also can someone explain part c as I let s be the number in stock and worked out
P (X>s)<0.05 and found some but it was wrong
This is s2 edexcel solomon d


Yeah this is potentially a badly worded question, this sort of thing will get noticed and changed before the actual exams are released. Solomon papers don't go through the same rigorous processes as actual exam papers.

For part (c) would you mind showing your working? I'd prefer to get you to work out the correct answer rather than just giving you a correct method...
Reply 1591
16722842_1441489155869281_3216599602954038530_o.jpg
Can anyone please help explain question 15 ii?
Original post by arwaak
16722842_1441489155869281_3216599602954038530_o.jpg
Can anyone please help explain question 15 ii?


Use you sketch. 15(i) says to find the maximum value for |z|, give the condition that |z-w| = 2. i.e. what point on that locus of z has the greatest magnitude (is furthest from the origin)?
Reply 1593
Original post by Darth_Narwhale
Use you sketch. 15(i) says to find the maximum value for |z|, give the condition that |z-w| = 2. i.e. what point on that locus of z has the greatest magnitude (is furthest from the origin)?


Okay so for part a, lzl I got √10 +2 for the maximum value and
10 - 2 for the minimum value.
I did that assuming that the distance from the origin through the center would give me the furthest distance.
But what I don't get is part b?
Original post by arwaak
Okay so for part a, lzl I got √10 +2 for the maximum value and
10 - 2 for the minimum value.
I did that assuming that the distance from the origin through the center would give me the furthest distance.
But what I don't get is part b?


Part b is actually almost exactly the same. Part a asks you to find the maximum and minimum distance from the locus to the origin, or |z-0| which we write as just |z|. Part b asks you to do the same but not for the origin, rather the point x where x= 3. Do you understand why?
Reply 1595
Original post by Darth_Narwhale
Part b is actually almost exactly the same. Part a asks you to find the maximum and minimum distance from the locus to the origin, or |z-0| which we write as just |z|. Part b asks you to do the same but not for the origin, rather the point x where x= 3. Do you understand why?


Ahhhhhh that makes sense! Thank youuuu!
See I'm just starting this module now but I know it's got to do with vectors and the standard equation lz1-z2l which shows the difference between the locus z1 and the coordinate z2, in this case.
1487542378724812849240.jpg
Original post by Darth_Narwhale
Yeah this is potentially a badly worded question, this sort of thing will get noticed and changed before the actual exams are released. Solomon papers don't go through the same rigorous processes as actual exam papers.

For part (c) would you mind showing your working? I'd prefer to get you to work out the correct answer rather than just giving you a correct method...
14875425085391856693066.jpg
Original post by Darth_Narwhale
Yeah this is potentially a badly worded question, this sort of thing will get noticed and changed before the actual exams are released. Solomon papers don't go through the same rigorous processes as actual exam papers.

For part (c) would you mind showing your working? I'd prefer to get you to work out the correct answer rather than just giving you a correct method...
You may have noticed that this thread is no longer stickied at the top of the maths forum. It has been temporarily removed but can still be found in the Maths Forum Index which is a sticky.
Hi guys :smile:


Another core 3 question about a modulus exponential graph


I cant do 7)iii) at all:redface:

it says attempt method with x=0 to find a but how would I do that. I cant seem to interpret the graph well enough and the main technique we were taught in class was to square both sides if mod appears?

Thanks for any help!





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