Mathematics: Post your doubts here!

[Thought blocker]

from 4 cos sqr x
U kno cos sqr x= (1+ cos2x)/2
4(1+ cos2x)/2
2(1+ cos2x) this is wat they are saying

Solve full question for me.

 

[Lily9605]

from 4 cos sqr x
U kno cos sqr x= (1+ cos2x)/2
4(1+ cos2x)/2
2(1+ cos2x) this is wat they are saying

Thank you. So using that I got,
ln(y^3 +1 ) = 2x + sin2x + ln9
how do I find the y coordinates using this equation?

 

[mmmmmm]

why friction is added, I am refering to the first part?

i)
resolve the forces
R (normal reaction)= 5sin30

friction +5cos30= 6
friction=1.67 N

ii)
F= mew (R)
mew = F/R
=1.67/5sin30
= 0.68

 

[RoOkaYya G]

why friction is added, I am refering to the first part?

coz its in the same direction as the force

 

[mmmmmm]

coz its in the same direction as the force

The direction shuld be against the movement of the ring, but it is not.

 

[Thought blocker]

The direction shuld be against the movement of the ring, but it is not.

The ring is sliding down due to force of gravity (Rings weight pulling it down) hence friction would be up hence it is added.

 

[RoOkaYya G]

The direction shuld be against the movement of the ring, but it is not.

friction is upward w.r.t weight whch is downward.

 

[Thought blocker]

The ring is sliding down due to force of gravity (Rings weight pulling it down) hence friction would be up hence it is added.

 

[RoOkaYya G]

The direction shuld be against the movement of the ring, but it is not.

see..since the system is in equilibrium...means upward = downward na?
but since the weight is 6
the upward force shld be 6 too
so logically it shld be same on both side (same magnitude on both side to obtain system in equilibrium)
but since 5cos30 is less than 6....logically friction is upward
u can check it with ths logic if u want
got it?

 

[Andhikasm]

Hello people, can you please help me in statistics may/june 2007 P6 Question 3b; the normal distribution one?? PLEASE I've been trying to get this done for months!

 

[Thought blocker]

June 2014 Question Paper 31 (96Kb) how to tackle the following Qu.?? :
qu 4 /qu.5(ii)/qu.7(iii)/10(ii)
urgent plzzz!!!!!!

june 2014 p32 qu.7 (b). how to tackle this question????

I am not gonna solve 2014 paper 3 before April 2015
I am sorry.
Rutzaba help.

 

[Thought blocker]

Hello people, can you please help me in statistics may/june 2007 P6 Question 3b; the normal distribution one?? PLEASE I've been trying to get this done for months!

I am in hurry, cant edit post with symbols, hope you understand. :¬

Let the random variable X be normally distributed where with u(meu) n s(sigma) as mean n standard deviation respectively....
Nw we need to find
P((u+s)<X<(u-s))

Now standarising
we have
p{((u-s-u)/s)<Z<((u-s-u)/s)}
so, P(-1<Z<1)
so phi(1)-phi(-1)
=phi 1 -[1-phi(-1)]
={2phi(1)}-1
=(2*0.8413)-1
= .6826

therefore for 1 observation the p is .6862
so, for 800 it is .6826*800
=546 (Ans)

 

[Thought blocker]

thanks cause am taking my exams in november 2015. thanks for ur help .|.

Aha, you have a year time
All the best.
I am not solving it now coz I wanna solve it in April so I can know what were the twists in paper 3.
I hope Rutaba helps you.

 

[Rutzaba]

Guys sorry i dont have access to my pc...

 

[Thought blocker]

Guys sorry i dont have access to my pc...

He can wait for april to come. Relax.

 

[Thought blocker]

Help meeee.

In?

 

[lxelle]

T

In?

Hang on ha, tryna upload the file.

 

[lxelle]

This question! ii) onwards.

 

[Thought blocker]

This question! ii) onwards.

It takes time to open image. #SlowNet
Can you post the link?

 

[lxelle]

N

It takes time to open image. #SlowNet
Can you post the link?

No link though. It's my own ques.