Mathematics: Post your doubts here!

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Ebby_SaLeem

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Can someone explain me union and intersection of the compliements of sets, A and B in this case.
PLEASE! With proper and clear diagrams... thanks!
 
soul

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can someone please explain to me the link between work done by driving force, work one by resistance, k.e, p.e
 
Pie-man

Pie-man

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soul said:
can someone please explain to me the link between work done by driving force, work one by resistance, k.e, p.e
Click to expand...
the link from the rule "Final Energy = Initial Energy + Driving force - Resistance force" ? or the one "Driving force - Resisting force = mass x acceleration"?

so you get the final and initial energy using "K.E. [1/2 mass x velocity(squared) ] + P.E. (mass x gravity [10] x height)" and substituting the mass, height and velocity as given at each point.

if given a driving force or resistant force you put them in the equation "F.E. = I.E. + DF - RF"
 
knowitall10

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Alright.. i don't know if it's right though.. like i said, i'm taking Edexcel, not Cambridge:)
I'm sorry i'm late, the internet stopped working :/
 

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knowitall10

knowitall10

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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:(
 
iKhaled

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knowitall10 said:
btw, when ur done, please explain it to me too.. its bothering me now, i can't study anything else
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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!!
 
knowitall10

knowitall10

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iKhaled said:
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!!
Click to expand...
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?
 
knowitall10

knowitall10

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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...:notworthy:
 
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