Mathematics: Post your doubts here!

[David Hussey]

rearrange the equation such that it is in the form:
lny = lnA - kx^2
-->
lny = -kx^2 + lnA

from this u can see that ur gradient will be -k and ur y-intercept is lnA

find the gradient using the coordinates of the two given points and equate it to -k .... then put in the values of lny , k, and x^2 from one of the points in the equation and find c i.e. lnA

got it?

Brilliant!

 

[Thought blocker]

Suchal Riaz

 

[David Hussey]

rearrange the equation such that it is in the form:
lny = lnA - kx^2
-->
lny = -kx^2 + lnA

from this u can see that ur gradient will be -k and ur y-intercept is lnA

find the gradient using the coordinates of the two given points and equate it to -k .... then put in the values of lny , k, and x^2 from one of the points in the equation and find c i.e. lnA

got it?

but in the exam how am i supposed to know that i should rearrange and do such stuff?

 

[sitooon]

Second part of vectors


From mj2013 , va'31
A star

 

[daredevil]

NOTE i am considering @ as thita

as u found out in part i ..
4Cos@ + 3Sin@ = 5Cos(@-0.6435)

5Cos(@-0.6435) = 2
Cos(@-0.6435) = 2/5
@ = Cos^-1(2/5) + 0.6435

@= 1.8028 radians
and
2(pie)-1.8028 = 4.48 radians

can't seem to grasp onto the last part tho

 

[daredevil]

but in the exam how am i supposed to know that i should rearrange and do such stuff?

bcz u have y as lny in the graph
and ur x from the graph is x^2

just rearrange the equation in the form of
y=mx+c

 

[daredevil]

X^2 + 5X = 8 in the form of (ax^2 +bx) +C
Pls help.

I'm sorry i dont get ur question :`)
is it a new question or something from the one that i answered??

 

[daredevil]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_32.pdf
guys for q5 is the answer
(y^2+3x)/6X=a
(id understand ms language -__- )

ummm no .. the answer is
x = (sqrt)3a

 

[David Hussey]

ummm no .. the answer is
x = (sqrt)3a

ufff
my dy/dx = (-3x^2-y^2+6ax)/(2yx+2ay)
is this correct? please say yes!

 

[ashcull14]

Two particles P and Q move on a line of greatest slope of a smooth inclined plane. The particles start
at the same instant and from the same point, each with speed 1.3ms−1. Initially P moves down the
plane and Q moves up the plane. The distance between the particles t seconds after they start to move
is d m.
(i) Show that d = 2.6t.

 

[robinhoodmustafa]

I'm sorry i dont get ur question :`)
is it a new question or something from the one that i answered??

new question

 

[daredevil]

ufff
my dy/dx = (-3x^2-y^2+6ax)/(2yx+2ay)
is this correct? please say yes!

yes it's ryt (Y)

 

[daredevil]

X^2 + 5X = 8 in the form of (ax^2 +bx) +C
Pls help.

from wat i can gather it will be

x^2 + 5x - 8 = 0

where a=1 ; b=5, c=-8

 

[itallion stallion]

Plz help me with these questions.

May June 2012 paper 61 maths q6 part iii.maths

October November 2011 paper 33 q10 maths

May June 2011 paper 33 q7 part iii.
Thanks aloooot!

 

[daredevil]

Plz help me with these questions.

May June 2012 paper 61 maths q6 part iii.maths

October November 2011 paper 33 q10 maths

May June 2011 paper 33 q7 part iii.
Thanks aloooot!

can u post the paper or the link to the paper?

 

[yuv2404]

hi can someone help me with may/june 2008 paper 3 question 8 (i) plz
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s08_qp_3.pdf

 

[asd]

Plz help me with these questions.

May June 2012 paper 61 maths q6 part iii.maths

October November 2011 paper 33 q10 maths

May June 2011 paper 33 q7 part iii.
Thanks aloooot!

There's no Q6 part (iii) in may june 2012 P61..

 

[asd]

hi can someone help me with may/june 2008 paper 3 question 8 plz
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s08_qp_3.pdf

Gradient = dy/dx = PN/TN= tanx = 1/2(PN)(TN)

For part (i): Rearrange PN/TN = tanx to give TN= PN/tanx
Substitute this into Area of the triangle (1/2 PN*TN) which gives 1/2* (PN)*(PN/tanx)
Note that PN= y.
So it goes down to 1/2. y^2. cotx

Now for part (ii): Separate the variables and integrate.
int(2/y^2) dy = int(cotx)dx
= -2/y = int(cosx/sinx) dx
= -2/y = ln(sinx) + c

Plug in (Pi/6, 2) to evaluate c. c= -0.31 Idk why they have it as 0.3 in the ms.
then rearrange to express y interms of x.

 

[Thought blocker]

This thread is really helpful Thanks all and creator XPFMember

 

[asd]

This thread is really helpful Thanks all and creator XPFMember

Thanks! and No problem bro.