Mathematics: Post your doubts here!

[2pac]

can anyone please help me with q6 part ii of p3 oct 2011.Have done the first part with ease but the second part has 2 theeta instead of normal recurring theeta,sp please help me out.
thanks

 

[Dug]

The answer is 38500 cm^3.

Ahh...I forgot to square my answer.
r = √530.5
r = 23

Inserting r=23 into the equation gives V = 38400 to 3s.f
This is because i rounded values to make it easy.
But the method is the same so...cheers

 

[Dug]

LOL my board exam is on tuesday ... atleast help a little -.-

Did you do it? Sorting the questions? Ofc not!!

 

[Jimmy Russels]

For Q3)
dy/dx = 8(1-x^3)^(-1-1) (-1) (-3x^2)
dy/dx = 24x^2/(1-x^3)^2
Inserting x = -1
dy/dx = 24(-1)^2 / (1-(-1)^3)^2
dy/dx = 24/4
dy/dx = 6

Eq of the normal :-
Gradient of the normal = -1/6
4=-1(-1/6) + c
c = 23/6
y = -1/6x + 23/6

Again, the accuracy of my answers is not guaranteed.

100% accurate. Thank you so much, btw, Dug. I knew how to do these questions but I just now realized I made the stupidest mistakes and messed them up and then got frustrated. Thank you.

Btw, do you have any tips to avoid making silly errors in CIE? I know the syllabus... I just keep making small mistakes. Not improving much with practice...

 

[hm12]

The answer is 38500 cm^3.

put dV/dr=0 you get r=23.0329 put this value of r into V=1/2 r(5000-pi r^2) you get V=38388.23

 

[hm12]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_qp_12.pdf
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_ms_12.pdf
please explain question 3 part (ii)

anyone please help?

 

[Dug]

100% accurate. Thank you so much, btw, Dug. I knew how to do these questions but I just now realized I made the stupidest mistakes and messed them up and then got frustrated. Thank you.

Btw, do you have any tips to avoid making silly errors in CIE? I know the syllabus... I just keep making small mistakes. Not improving much with practice...

Some people might say practice makes you perfect and they are not wrong but at times... its just the quirk of fate.

 

[Dug]

LOL my board exam is on tuesday ... atleast help a little -.-

Do you have skype?

 

[hm12]

@dug please help me with the question i posted above thanks

 

[Esme]

Are you implying that i am wasting his time?

Lol no !! infact you'r really helping.
but cumon... understanding one whole section of the syllabus is best done from the book esp in limited time!! you agree right ?

 

[Dug]

@dug please help me with the question i posted above thanks

If you don't mind, I need to get a break. Post the questions here and someone will definitely try to help you when they get time. If not, I will be back not before too long to help you with them.

 

[Esme]

what this mean ??
Use the condition for the convergence of a geometric progression, and the formula for the sum to infinity of a convergent geometric progress.
This is from the syllabus.
What does the term 'convergent geometric progression' mean ?

 

[Dug]

Lol no !! infact you'r really helping.
but cumon... understanding one whole section of the syllabus is best done from the book esp in limited time!! you agree right ?

Exactly! Actually I also used that book. Its amazing!!

 

[Dug]

what this mean ??
Use the condition for the convergence of a geometric progression, and the formula for the sum to infinity of a convergent geometric progress.
This is from the syllabus.
What does the term 'convergent geometric progression' mean ?

I think it means the terms are decreasing with the progression. Just like converging rays and diverging rays

 

[roadtrip9o9]

yes i got skype ... but i cant message u on xpf

 

[Dug]

yes i do .. i cant send u a message on xpf ... u send me ill reply my skype id

I will have to make one. Why can't you message me?

 

[redapple20]

Can sum1 tel me that gradient formula involving tan?

 

[Bassel.Lm]

Re: Maths help available here!!! Stuck somewhere?? Ask here!

Hope this helps!

Thank you so much, ur answer to this doubt helped me alot.
btw, do u have any short notes for mechanics p4. i am facing some troubles in that paper and my exam is on 14th of may. Please reply as soon as possuble!
thank you

 

[Dug]

Can sum1 tel me that gradient formula involving tan?

tan theta = m
tan theta = rise/run or perp/base

 

[hm12]

If you don't mind, I need to get a break. Post the questions here and someone will definitely try to help you when they get time. If not, I will be back not before too long to help you with them.

no one has been replying its the 3rd time i posted that question i saw you active on this page so asked you its okay anyways thanks for your help in the previous questions