Mathematics: Post your doubts here!

[PhyZac]

solve this too plz phyzac

I solved this before I am sure!
I will look for it

 

[PhyZac]

applepie1996
We have to find the point p where the perpendicular distance to the two planes is same.

first we will for an equation the way the paper gave.
therefore
for plane m
|x+2y-2z-1|/sqrt((1^2) + (2^2) + (-2^2) = |x+2y-2z-1|/3 [P.S, sqrt is square root]
for plane n
|2x-2y+z-7|/sqrt((2^2)+(-2^2)+(1^2) = |2x-2y+z-7|/3


since you are finding a point where both distance is same , therefore.
|x+2y-2z-1|/3 = |2x-2y+z-7|/3 (3 can be cancelled both sides)
=
|x+2y-2z-1| = |2x-2y+z-7|

now we will sub the line values in the equation.
the x component (1+2t) [P.S, t is lamda ]
the y comp. (1+t)
the z comp. (-1+2t)

therefore.
| (1+2t) +2 ( 1+t) -2(-1+2t) -1 | = |2(1+2t) - 2(1+t) + (-1+2t) - 7 |
simplifying it you get
|4| = |-8+4t|
now to solve modulus, we do squaring method
16 = 64-64t+16t^2
simplify
2 = 8 - 8t +2t^2
2t^2 - 8t + 6 = 0
solve it and get
t = 3 or t= 1
when t=3 the position of point is [P.S, u do this by sub t value in line equation)
(7, 4, 5) => OA
when t = 1 the position of point is
(3, 2, 1) => OB

BA (or AB, same thing) = OA - OB
= (4 , 2, 4)

now find the mod
sqrt(4^2 + 2^2 + 4^2)
= 6

I dint do every step in detail, (very long) if u dont get any step, just ask. [Pray for me please.]

 

[Rutzaba]

applepie1996
We have to find the point p where the perpendicular distance to the two planes is same.

first we will for an equation the way the paper gave.
therefore
for plane m
|x+2y-2z-1|/sqrt((1^2) + (2^2) + (-2^2) = |x+2y-2z-1|/3 [P.S, sqrt is square root]
for plane n
|2x-2y+z-7|/sqrt((2^2)+(-2^2)+(1^2) = |2x-2y+z-7|/3


since you are finding a point where both distance is same , therefore.
|x+2y-2z-1|/3 = |2x-2y+z-7|/3 (3 can be cancelled both sides)
=
|x+2y-2z-1| = |2x-2y+z-7|

now we will sub the line values in the equation.
the x component (1+2t) [P.S, t is lamda ]
the y comp. (1+t)
the z comp. (-1+2t)

therefore.
| (1+2t) +2 ( 1+t) -2(-1+2t) -1 | = |2(1+2t) - 2(1+t) + (-1+2t) - 7 |
simplifying it you get
|4| = |-8+4t|
now to solve modulus, we do squaring method
16 = 64-64t+16t^2
simplify
2 = 8 - 8t +2t^2
2t^2 - 8t + 6 = 0
solve it and get
t = 3 or t= 1
when t=3 the position of point is [P.S, u do this by sub t value in line equation)
(7, 4, 5) => OA
when t = 1 the position of point is
(3, 2, 1) => OB

BA (or AB, same thing) = OA - OB
= (4 , 2, 4)

now find the mod
sqrt(4^2 + 2^2 + 4^2)
= 6

I dint do every step in detail, (very long) if u dont get any step, just ask. [Pray for me please.]

i will
and pray for me too pleez im not well

 

[xhizors]

I just wanted to make sure the answer is correct.

So
they want
Σ(x − x~)^2
so for me this was very hard, since we dont have individual numbers, so i went to the formula book and found the first equation helpful.
If you check it ( i cant write it down) you will see the numerator is similar to the wanted question, so basically I made it the subject of formula, and i got
Σx^2 - n(x~)^2 as the formula

Now it is easy
we have Σx^2 value
and n = 150
and mean is Σx/n

which book??

 

[applepie1996]

applepie1996
We have to find the point p where the perpendicular distance to the two planes is same.

first we will for an equation the way the paper gave.
therefore
for plane m
|x+2y-2z-1|/sqrt((1^2) + (2^2) + (-2^2) = |x+2y-2z-1|/3 [P.S, sqrt is square root]
for plane n
|2x-2y+z-7|/sqrt((2^2)+(-2^2)+(1^2) = |2x-2y+z-7|/3


since you are finding a point where both distance is same , therefore.
|x+2y-2z-1|/3 = |2x-2y+z-7|/3 (3 can be cancelled both sides)
=
|x+2y-2z-1| = |2x-2y+z-7|

now we will sub the line values in the equation.
the x component (1+2t) [P.S, t is lamda ]
the y comp. (1+t)
the z comp. (-1+2t)

therefore.
| (1+2t) +2 ( 1+t) -2(-1+2t) -1 | = |2(1+2t) - 2(1+t) + (-1+2t) - 7 |
simplifying it you get
|4| = |-8+4t|
now to solve modulus, we do squaring method
16 = 64-64t+16t^2
simplify
2 = 8 - 8t +2t^2
2t^2 - 8t + 6 = 0
solve it and get
t = 3 or t= 1
when t=3 the position of point is [P.S, u do this by sub t value in line equation)
(7, 4, 5) => OA
when t = 1 the position of point is
(3, 2, 1) => OB

BA (or AB, same thing) = OA - OB
= (4 , 2, 4)

now find the mod
sqrt(4^2 + 2^2 + 4^2)
= 6

I dint do every step in detail, (very long) if u dont get any step, just ask. [Pray for me please.]

JazakAllah
InshaAllah i will pray for you
and May Allah fulfill all your wishes

 

[PhyZac]

which book??

Sorry??

I solved that from my head, not any book.

 

[PhyZac]

Anyone having notes for Pure 3 integration??? Please, please, please.

 

[xhizors]

I just wanted to make sure the answer is correct.

So
they want
Σ(x − x~)^2
so for me this was very hard, since we dont have individual numbers, so i went to the formula book and found the first equation helpful.
If you check it ( i cant write it down) you will see the numerator is similar to the wanted question, so basically I made it the subject of formula, and i got
Σx^2 - n(x~)^2 as the formula

Now it is easy
we have Σx^2 value
and n = 150
and mean is Σx/n

I made it the subject of formula, and i got
Σx^2 - n(x~)^2 as the formula
sorry but how?

 

[PhyZac]

I made it the subject of formula, and i got
Σx^2 - n(x~)^2 as the formula
sorry but how?

Yes, that is correct, exactly like i told you..
now sub the value of Σx^2 = 8287.5
and n = 150
and the mean is 645/150 = 4.3
then take its square and multiply with 150
subtract from 8287.5

 

[Esme]

Anyone having notes for Pure 3 integration??? Please, please, please.

I don't have any website link but I have them in hard copy. Do you want me to scan them ??

 

[PhyZac]

I don't have any website link but I have them in hard copy. Do you want me to scan them ??

Well, if it is okay for you, and wont bother you then yes, very much wanted..!! Jazaki Allah khairan!

 

[Jiyad Ahsan]

I don't have any website link but I have them in hard copy. Do you want me to scan them ??

seriously i need p3 integration notes too !! i haven't practiced it for so long now.. its all gone to some recycle bin in my head

 

[Esme]

Well, if it is okay for you, and wont bother you then yes, very much wanted..!! Jazaki Allah khairan!

seriously i need p3 integration notes too !! i haven't practiced it for so long now.. its all gone to some recycle bin in my head

Okay give me around half an hour, I'll upload them

 

[Jiyad Ahsan]

PhyZac

could you help me understand how to know what loci to draw on argand diagrams?i cant seem to grasp the concept..

 

[Esme]

PhyZac Jiyad Ahsan
Here... They are photocopied notes so they might not be very neat sorry.

 

[Jiyad Ahsan]

PhyZac Jiyad Ahsan
Here... They are photocopied notes so they might not be very neat sorry.

thanx alot !!

 

[PhyZac]

PhyZac Jiyad Ahsan
Here... They are photocopied notes so they might not be very neat sorry.

Thank you so much so much so much!!!!!!!!!!!!!!!!!!! Jazaki Allah Khairan!!! Thanks ALOOOOOOOOOT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! May Allah S.w.t grant you and your family with Jannatul firdous, and shower His blessing on you. Aameeeeeeeeeeeeeeeeennn!!!!!!!!!! Thank you SO much!
May Allah S.w.t grant you with best results Aaameeen!

 

[PhyZac]

PhyZac

could you help me understand how to know what loci to draw on argand diagrams?i cant seem to grasp the concept..

I learned it from this file

 

[Esme]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w09_qp_31.pdf
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w09_ms_31.pdf

Q7iv) How do we show the shading for this: 0 < arg(z − u) < 1/4 π ?

 

[Esme]

Thank you so much so much so much!!!!!!!!!!!!!!!!!!! Jazaki Allah Khairan!!! Thanks ALOOOOOOOOOT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! May Allah S.w.t grant you and your family with Jannatul firdous, and shower His blessing on you. Aameeeeeeeeeeeeeeeeennn!!!!!!!!!! Thank you SO much!
May Allah S.w.t grant you with best results Aaameeen!

You're welcome.. I'm glad I could help after all the help you've given me
And thank you for all the duas..
May Allah grant you success too and shower his blessings on you Ameen