Mathematics: Post your doubts here!

[Dark Destination]

Does anyone have any S1 notes? Have an exam tomorrow.

 

[Rizwan Javed]

Does anyone have any S1 notes? Have an exam tomorrow.

Check these links. They might help.
https://www.xtremepapers.com/community/attachments/s1-compilation-doc.10482/
https://www.xtremepapers.com/community/threads/binomial-and-normal-distribution-notes-here.16132/

 

[***amd***]

if y = (x - 150)/10
we are given mean y and standard deviation (SD) y

I know
mean x = 10 * mean y
but can we find SDx by the information given above?

 

[Rizwan Javed]

if y = (x - 150)/10
we are given mean y and standard deviation (SD) y

I know
mean x = 10 * mean y
but can we find SDx by the information given above?

I think we can find SDx.

SDy = SDx/10
then
SDx = 10SDy

 

[***amd***]

I think we can find SDx.

SDy = SDx/10
then
SDx = 10SDy

One of my frnds said that too, but see this...

 

[Rizwan Javed]

One of my frnds said that too, but see this...

I solved the data you showed you me to find the SD(z) and SD (x).

SD(z) comes out to be 16.6
SD(x) comes out to be 16608.9

according to the coding equation:
z = (x-150000)/1000

SD(z) = SD(x)/1000 ---(1)

if you substitute the values, both sides of the equation get equal; this shows that the relationship (1) holds.

 

[***amd***]

I solved the gave you showed you me to find the SD(z) and SD (x).

SD(z) comes out to be 16.6
SD(x) comes out to be 16608.9

according to the coding equation:
z = (x-150000)/1000

SD(z) = SD(x)/1000 ---(1)

if you substitute the values, both sides of the equation get equal; this shows that the relationship (1) holds.

oh..... sorry. I was taking variance as SD. my bad.
thanks btw

 

[Anum96]

Thank you so much! I got it! Actually I was totally ignoring some of the possible arrangements; my bad Anyway thanks once again.

You're welcome!

 

[Anum96]

plzz help ASAP..!!

Mass = 1100
Fd = 1800
Fr = 700N
Distance = x
Speed = v
i) K.E = 1/2(1100)v^2 = 550v^2
P.E = (1100)(10)(160) = 1760000 Now we have to find it in terms of x which is distance. So it will become 1760000/1760 = 1000x

kv^2 = x
1800x = 700x+ 550v^2+ 1000x
^This is the WD equation. I just substituted everything in terms of x and v
Solve it further you will get
100x = 550v^2
x = 550/100v^2
x = 5.5v^2
Hence k = 5.5 Ans.

ii) 5.5v^2 = 1760
v^2 = 320

1800(x-1760) = 700(x-1760)+ 550(v^2-320) [we subtracted 320 because it was our initial velocity and so to find gain in K.E we already have v^2(initial velocity)
1800x-3168000 = 700x - 1232000+ 550v^2 - 176000
1800x-700-3168000 = -1408000+ 550v^2
1100x-3168000+ 1408000 = 550v^2
1100x+ 1760000 = 550v^2
(1100x+ 176000)/550 = v^2
v^2 = 2x - 3200 (shown)

 

[Copy Cat]

I am a bit confused like i reached till the step
Int of X/(4-X^2) dx = Int of 1/4 dt

Int of X/(4-X^2) dx should be [ln(4-X^2)]/-2 right?

 

[Anum96]

I am a bit confused like i reached till the step
Int of X/(4-X^2) dx = Int of 1/4 dt

Int of X/(4-X^2) dx should be [ln(4-X^2)]/-2 right?

dx/dt = 1/x - x/4
Take L.C.M
dx/dt = (4 - x^2)/4x
Cross multiply
dx*4x = 4dt - x^2dt
dx*4x = dt(4-x^2)
(dx*4x)/(4-x^2) = dt
Switch sides and integrate.
t = -2ln(4-x^2) + c

t = o
x = 1

>> 0=-2ln(4-1) + c
0 = -2ln3
c = 2ln3
Substitute.
t = -2ln(4-x^2) + 2ln3
t - 2ln3 = -2ln(4-x^2)
-t + 2ln3 = 2ln(4-x^2)
-t = 2ln(4-x^2) - 2ln3
-t = 2ln[(4-x^2)/3]
-t/2 = ln[(4-x^2)/3]
exp(-t/2) = (4 - x^2)/3
3exp(-t/2) = 4 - x^2
3exp(-t/2) - 4 = -x^2
switch sides and change sign
x^2 = 4 - 3exp(-t/2)

I hope I'm right tho :3

 

[***amd***]

I am a bit confused like i reached till the step
Int of X/(4-X^2) dx = Int of 1/4 dt

Int of X/(4-X^2) dx should be [ln(4-X^2)]/-2 right?

u forgot to add a constant in the integrated equation, i guess
like...
[{ln (4 - x^2)} / -2 ] + ln c = t/4
c comes out to be 9

 

[Mr.Physics]

https://googledrive.com/host/0B1ZiqBksUHNYc0tWVkExdEc4Nlk/January 2014 (IAL) QP - Unit 1 Edexcel Physics.pdf
Question 18 (a) (ii)
Help me guys I can't figure it out
zahra azam My Name Anum96 ***amd***

 

[***amd***]

https://googledrive.com/host/0B1ZiqBksUHNYc0tWVkExdEc4Nlk/January 2014 (IAL) QP - Unit 1 Edexcel Physics.pdf
Question 18 (a) (ii)
Help me guys I can't figure it out
zahra azam My Name Anum96 ***amd***

i tried 2 ways, and i am getting T = 170820 N
i guess u just have to round it off to 200000, cuz it says 'about 2 * 10^5'

 

[Mr.Physics]

i tried 2 ways, and i am getting T = 170820 N
i guess u just have to round it off to 200000

K. But can you explain how did you get that ? :/

 

[***amd***]

K. But can you explain how did you get that ? :/

as the iceberg has a constant speed, it has no resultant force
no resultant force, then u can...
1. make a triangle of forces of 1 drag force and two tensions in the ropes, and use sine rule (triangle will be isosceles as two forces i.e. tensions are equal) OR
2. apply Lami's theorem [ 330000 / sin 30 = T / sin 165 ] OR
3. find horizontal components of both T's, and equate them with 330000, like, 2 * T cos 15 = 330000

 

[Mr.Physics]

as the iceberg has a constant speed, it has no resultant force
no resultant force, then u can...
1. make a triangle of forces of 1 drag force and two tensions in the ropes, and use sine rule (triangle will be isosceles as two forces i.e. tensions are equal) OR
2. apply Lami's theorem [ 330000 / sin 30 = T / sin 165 ] OR
3. find horizontal components of both T's, and equate them with 330000, like, 2 * T cos 15 = 330000

Thank you so much

 

[Copy Cat]

dx/dt = 1/x - x/4
Take L.C.M
dx/dt = (4 - x^2)/4x
Cross multiply
dx*4x = 4dt - x^2dt
dx*4x = dt(4-x^2)
(dx*4x)/(4-x^2) = dt
Switch sides and integrate.
t = -2ln(4-x^2) + c

t = o
x = 1

>> 0=-2ln(4-1) + c
0 = -2ln3
c = 2ln3
Substitute.
t = -2ln(4-x^2) + 2ln3
t - 2ln3 = -2ln(4-x^2)
-t + 2ln3 = 2ln(4-x^2)
-t = 2ln(4-x^2) - 2ln3
-t = 2ln[(4-x^2)/3]
-t/2 = ln[(4-x^2)/3]
exp(-t/2) = (4 - x^2)/3
3exp(-t/2) = 4 - x^2
3exp(-t/2) - 4 = -x^2
switch sides and change sign
x^2 = 4 - 3exp(-t/2)

I hope I'm right tho :3

Thanks a bunch....

 

[Copy Cat]

last part

 

[DaniyalK]

View attachment 58034
last part

take lcm
dh/dt = (100-h^2)/20h^2
(20h^2/(100-h^2)) dh = dt
Use result from part 2.
-20 + 2000/(10-h)(10+h) dh = dt
use partial fractions
-20 + 100/(10-h) + 100/(10+h) dh = dt
then integrate