Mathematics: Post your doubts here!

[PhyZac]

Assalamu Alikum Wa Rahmatullahi Wa Barakatoho.. littlecloud11
May June 2008
and 9 (i)
Sorry for not posting link ....but xtremepapers isnt working good.

 

[Dug]

Can somebody explain 2 (ii)?

n = 15
Σx = 1572

Σ(x - 104.8)^2 = Σx^2 - (2)(104.8)Σx + Σ(104.8)^2 = Σx^2 - (2)(104.8)(1572) + (15)(104.8^2) --- i

Σ(x - 100)^2 = 499.2
Expanding LHS:
Σx^2 - 2(100)Σx + Σ(100^2) = 499.2
Σx^2 - (2)(100)(1572) + 150000 = 499.2
Σx^2 = 164899.2 --- ii

Put (ii) in (i):
Σ(x - 104.8)^2 = 153.6

 

[littlecloud11]

Assalamu Alikum Wa Rahmatullahi Wa Barakatoho.. littlecloud11
May June 2008
and 9 (i)
Sorry for not posting link ....but xtremepapers isnt working good.

I'm guessing this is for the P3 paper.

9i) y = e^(-x/2) √(1+2x)
dy/dx = -1/2 e^(-x/2) * √(1+2x) + 1/2* 2 * (1+2x)^(-1/2) * e^(-x/2)
dy/dx = [-e^(-x/2) * √(1+2x)]/2 + e^(-x/2)/ √(1+2x)

For M, dy/dx = 0
[-e^(-x/2) * √(1+2x)]/2 + e^(-x/2)/ √(1+2x) = 0
Take LCM
[-e^(-x/2) * √(1+2x)* √(1+2x) + e^(-x/2) *2] / 2√(1+2x) =0
cross multiply, so the denominator becomes 0
-e^(-x/2) * (1+2x) +2e^(-x/2) = 0
take -e^(-x/2) common-
-e^(-x/2) [(1+2x) -2] =0
so,
-e^(-x/2) = 0 (NA) or -(1+2x) -1 = 0
1+2x -2 = 0
2x= 1
x =1/2

 

[Syed Abdul Haseeb]

Salam to one and all, i need some help in P1(Hugh Neill and Douglas Quadling) Ch:13 Vectors Miscellaneous 13 Q 9, 13, 14 and 15 thankz in advance

 

[zunair]

It's neither of those two. The answer for the question would be -π/3 ...

f(x)=x/2 + π/6
g(x)=Cos(x)
g(f(x)) = Cos((x/2+π/6))
Cos((x/2+π/6)) = 1
x/2+π/6= Cos^-1(1) (Cos Inverse of 1 = 0)
x/2+π/6=0
x/2=-π/6
x=-2π/6
x=-π/3

The working done here is perfect. There is an error in the marking scheme. I always check my answers against the Examiner's Report.

Cheers

Thank you very much but there's this thinngy confusing me that inverse of cos 1 is 0 as well as 2pi so when put equal to 2 pi the answer turns out to be 5pi/3 as well

 

[zunair]

Thank you very much but there's this thinngy confusing me that inverse of cos 1 is 0 as well as 2pi so when put equal to 2 pi the answer turns out to be 5pi/3 as well

where can i get examiner reports

 

[daredevil]

HELLPP Q.1 and Q.2 of M/J 12 .. Paper 62
please solve that for me... thanx

 

[syed1995]

where can i get examiner reports

Read the question again. Limits are -Pi/2 to Pi/2. so you will ignore any values greater than Pi/2.


The examiner reports are all available on xtremepapers... here:

 

[daredevil]

can some1 pleassee tell me how to make a box and whiskers diagram??

 

[Dug]

can some1 pleassee tell me how to make a box and whiskers diagram??

Here

 

[Rutzaba]

Hi guys.... Please help me with these questions... would be appreciated... Thanks
Q5,Q6ii, Q7iii, Q8i, Q9ii

Q 6 part two a and b.
a)A conjugate in simple words mean that the sign ( that is negative or positive) of the imaginary number (the coefficient of i ) would change.
z= √3 + i
z*= √3 - i
1....things to remember = i = √-1 and thus i^2= -1
2.... if there is a term in i in the denominator and we have to remove it... we multiply the whole term with its conjugate and divide the whole term with its conjugate too.
we need to find
a) 2z +z* = 2(√3 +i) + √3- i
2 √3 +2i + √3 -i
3 √3 + i ...... ans.

b)because the format of the fraction isnt good here i have made u a pic to understand from

http://i1275.photobucket.com/albums/y444/Rutzaba/xpc_zpsd8fbcce5.png

 

[daredevil]

Here

can play the video becuz youtube is not working in pakistan... do u have anything else??

 

[Dug]

can play the video becuz youtube is not working in pakistan... do u have anything else??

Here

 

[parthrocks]

can someone plz help me with this..I need the solution asap!plz
Maths expert
Need it asap.plz attach a scan if possible showing all the steps.ok

 

[parthrocks]

and this one too plz help

 

[daredevil]

Here

thannkuu

 

[daredevil]

can someone plz help me with this..I need the solution asap!plz
Maths expert
Need it asap.plz attach a scan if possible showing all the steps.ok

c u can do it like this :
separate the 4 and the 10s :-

4^4 x (10^-1 x 10^-2 x 10^-3 x 10^1)
= 256 x 10^-1-2-3+1
= 256 x 10^-5
= 0.00256
=0.002560

don't be confused because of the last zero because it is a trick. it does not actually matter how many zeros we put at the end the answer is still 0.002560

get it??

 

[zunair]

Thanks a lot

 

[yubakkk]

 

[yubakkk]

plz answer will working.