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- 37
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knowitall10 here you go..
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can u show me the question pls ?oops.. wait, i just checked the marking scheme.. i think it's wrong. then i don't know how to do it, i'm sorry
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_31.pdfcan u show me the question pls ?
btw, when ur done, please explain it to me too.. its bothering me now, i can't study anything elsecan u show me the question pls ?
sure!btw, when ur done, please explain it to me too.. its bothering me now, i can't study anything else
Farru here u go!btw, when ur done, please explain it to me too.. its bothering me now, i can't study anything else
but iKhaled what i learnt was to bring all the x values on one side all the y values on one side...u guys learnt a different method or is it a totally different topic?Farru here u go!
dy/dx = (y^2 +4)/xy
dx/dy = xy/(y^2 +4)
1/x dx/dy = y/(y^2+4)
1/x dx = y/(y^2+4) dy
∫1/x dx = ∫y/(y^2+4) dy
i think u know how to integeate..if not tell me and i will teach u too but here is the rest
ln x = 1/2ln(y^2+4) + c
sub the values of x and y and u will get c = -1/2ln4
ln x = 1/2ln(y^2+4) -1/2ln4
ln x + 1/2ln 4 = 1/2 ln(y^2+4)
2ln x + ln 4 = ln (y^2+4)
ln (4x^2) = ln (y^2+4)
4x^2 = y^2 + 4
y^2 = 4x^2-4
thats it!!
nope thats the method..u put x on one side and y on the other side thats the separate integrals and thats what i did see..but iKhaled what i learnt was to bring all the x values on one side all the y values on one side...u guys learnt a different method or is it a totally different topic?
oh sorry i just read this post after i replied..anyway u r welcome !Oh wait, i take that reply back.. iget what you're doing
Thanks a lot!! Jazak Allahu Khair brother!!!
and Farru i'm really sorry for being useless...
looked like french for a non p3 studentFarru here u go!
dy/dx = (y^2 +4)/xy
dx/dy = xy/(y^2 +4)
1/x dx/dy = y/(y^2+4)
1/x dx = y/(y^2+4) dy
∫1/x dx = ∫y/(y^2+4) dy
i think u know how to integeate..if not tell me and i will teach u too but here is the rest
ln x = 1/2ln(y^2+4) + c
sub the values of x and y and u will get c = -1/2ln4
ln x = 1/2ln(y^2+4) -1/2ln4
ln x + 1/2ln 4 = 1/2 ln(y^2+4)
2ln x + ln 4 = ln (y^2+4)
ln (4x^2) = ln (y^2+4)
4x^2 = y^2 + 4
y^2 = 4x^2-4
thats it!!
LOL!!looked like french for a non p3 student
Farru here u go!
dy/dx = (y^2 +4)/xy
dx/dy = xy/(y^2 +4)
1/x dx/dy = y/(y^2+4)
1/x dx = y/(y^2+4) dy
∫1/x dx = ∫y/(y^2+4) dy
i think u know how to integeate..if not tell me and i will teach u too but here is the rest
ln x = 1/2ln(y^2+4) + c
sub the values of x and y and u will get c = -1/2ln4
ln x = 1/2ln(y^2+4) -1/2ln4
ln x + 1/2ln 4 = 1/2 ln(y^2+4)
2ln x + ln 4 = ln (y^2+4)
ln (4x^2) = ln (y^2+4)
4x^2 = y^2 + 4
y^2 = 4x^2-4
thats it!!
Its ok sis!! Thank you anyway.Oh wait, i take that reply back.. iget what you're doing
Thanks a lot!! Jazak Allahu Khair brother!!!
and Farru i'm really sorry for being useless...
looked like french for a non p3 student
bother? you aren't bothering me
so first thing is writing the forces so that you wouldn't get confused by them, as the 5.5 N and the W N are pulling the particle upwards.
then draw a line in the middle of the right angle to make 2 letter Z (one from AP1X to the line and the other from CP2X to the line) and label the angles AP1X with a symbol and CP2X with another symbol, so now you have the angle AP1X equal to the left side of the line you drew and CP2X is the same as the right side of the line.
so lami says that 7.3/sin(90) = 5.5/sin(AP1X) = W/sin(CP2X)
so using cross-multiplication (5.5 x sin(90))/7.3 = 48.9 (angle CP2X)
angle AP1X = 90 - 48.9 = 41.1
so angle AP1X= 41.1
then to get W you do the same but you use the angle 41.1 so
(sin(41.1) x 7.3)/sin(90) = 4.8
so W= 4.8
did you get it now? if you want me to scan the paper I solved it on i don't mind
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