Mathematics: Post your doubts here!

[Tkp]

Please anyone could explain to me mechanics /jun11 variant 43 no.4 and no.5 urgently ?

ok just see the diagram carefully.the object is in equilbrium.so the frictional force must be opposite to 6.1 N.
so to find the frictional force 6.1+5costheta(as the theta is in the x axis)=f,so f is 7.5
u=f/r(r is mg as the object is in equilbrium)
3rd 1 is same as 1st 1.now the 6.1 is replaced with 8.6n.so 8.6+5c0stheta-f=ma
and the accleration must be in left direction as the object is moving

 

[Tkp]

cnt help with no.4.i also need help in this part

Please anyone could explain to me mechanics /jun11 variant 43 no.4 and no.5 urgently ?

 

[GorgeousEyes]

ok just see the diagram carefully.the object is in equilbrium.so the frictional force must be opposite to 6.1 N.
so to find the frictional force 6.1+5costheta(as the theta is in the x axis)=f,so f is 7.5
u=f/r(r is mg as the object is in equilbrium)
3rd 1 is same as 1st 1.now the 6.1 is replaced with 8.6n.so 8.6+5c0stheta-f=ma
and the accleration must be in left direction as the object is moving

Thaaanks !

 

[GorgeousEyes]

cnt help with no.4.i also need help in this part

So we will have to w8 for someone to explain this to us

 

[Tkp]

So we will have to w8 for someone to explain this to us

haha.ur right

 

[Rahma Abdelrahman]

Ya rahmaaa thank youu .. hahaha bas msa shr7ty 7elw

bgadd? shokran

 

[Rahma Abdelrahman]

Thnx i got it

ok gd

 

[tanmaydube]

View attachment 22700

Please solve this!!

Thank you

Please help!

 

[tanmaydube]

View attachment 22703

Help again, stuck at the first part!!!

Please help!

 

[tanmaydube]

View attachment 22704

Cannot do B (ii)

Please help!

 

[leosco1995]

Just can't seem to get any of the answers right.. even though I think I know what I'm doing. Help would be appreciated with both parts.

 

[Dug]

View attachment 22703

Help again, stuck at the first part!!!

y = sin³2x cos³2x

A = ⌡sin³2x cos³2x dx

u = sin2x
du/dx = 2cos2x
dx = du/2cos2x

A = ⌡sin³2x cos³2x du/2cos2x
A = ⌡sin³2x (cos²2x)(cos2x) du/2cos2x
A = ⌡sin³2x cos²2x du/2
A = ½ ⌡sin³2x (1 - sin²2x) du
A = ½ ⌡u³ (1 - u²) du
A = ½ ⌡u³ - u^5 du
A = ½ (u^4/4 - u^6/6)

For limits, set y = 0
sin³2x cos³2x = 0

For this function to be zero,
sin³2x = 0 or cos³2x = 0
x = 0, 45

Put these values in u = sin2x and we get the limits 0 and 1.

A = ½ (1/4 - 1/6)
A = 1/24

 

[tanmaydube]

Dug thanks a lot! please help me with the other ones too! that would be great!

 

[A star]

25%?

how r u long time no see?

 

[sk8rchick]

hey guys am new to this so plzz help!
i need igcse o level 2010 papers any one know where i can get them.. plzz help me!!

 

[Nagaanusan]

hey guys am new to this so plzz help!
i need igcse o level 2010 papers any one know where i can get them.. plzz help me!!

http://papers.xtremepapers.com/CIE/Cambridge IGCSE/

 

[minie23]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w08_qp_3.pdf

Please help me for no. 8 and 10(ii) (iii) (iv)

Dug

 

[sk8rchick]

guy i need help with paper 1 2010 here's the paper i need help in question no. 13 any one plzz help ASAP

 

[sk8rchick]

thnaks but i want edexcel igce 2010 papers if u have plzzz help me with it.. thanks for this one it was very usefull

 

[InnocentAngel]

Please help me with this simple question in S1:
How does a sample differ from a population?