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Mathematics: Post your doubts here!

robinhoodmustafa

robinhoodmustafa

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daredevil said:
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?
Click to expand...
X^2 + 5X = 8 in the form of (ax^2 +bx) +C
Pls help. :)
 
David Hussey

David Hussey

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daredevil said:
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?
Click to expand...
Brilliant! (y)
 
David Hussey

David Hussey

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daredevil said:
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?
Click to expand...
but in the exam how am i supposed to know that i should rearrange and do such stuff? :(
 
daredevil

daredevil

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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 :p
 
A

ashcull14

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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.
 
itallion stallion

itallion stallion

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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!
 
asd

asd

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yuv2404 said:
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
Click to expand...
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.
 
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