Mathematics: Post your doubts here!

[daredevil]

was ur previpus doubt solved? The one u tagged me in?

nope... cud u plz explain it like step by step??
i reallyy suck at vectors >.<

 

[Rutzaba]

nope... cud u plz explain it like step by step??
i reallyy suck at vectors >.<

ul have to wait

 

[daredevil]

ul have to wait

okaay no prob

 

[daredevil]

View attachment 42442

M/j 12 qp 33

2nd part

when u get an answer to this... tag me to it too...

 

[Rutzaba]

Put the line into parametric form.
If its (0,1,-1) +t(1,2,2) t im using in place of lambda
U will get (t) i +(1+2t) j + (-1+2t)k
Put the coefficient of i j k into the place of x y z in plane p
3(t) + 2( 1+2t) +4(-1+2t) = 13
Solve this uwill get t=1
Put this back into parametric form of the line
(t) i +(1+2t) j + (-1+2t)k u will get
1,3,1 these are coordinate of A
Put the coefficient of i j k into the place of x y z in plane q
-2(t) +(1+2t)+(-1+2t)=4
U will get t=2
Put it back into
(t) i +(1+2t) j + (-1+2t)k
U will get (2,5,3)
This is B
Now B -A =(1,2,2)
Its magnitude will be root of 1sqr + 2sqr+ 2sqr
Root 9=3
Sorry but im from cell islie der lagi

 

[Rutzaba]

Phew, that took a while to write clearly and illustrating properly sorry for the delay
View attachment 41014 View attachment 41015 View attachment 41016

when u get an answer to this... tag me to it too...

there u go

 

[A star]

when u get an answer to this... tag me to it too...

the one that is quoted in his question
the equationi formed in the first part is
y=e^15(e^(-2t))+ln70)
now we know the inital mass is 70 so
p=(the long equation quoted above/70)*100
as t==> infinity p==> 1 there u go

 

[sitooon]

How to simplify -e^-y = 0.5e^2x - 1.5 -----> y=ln(2/(3-e^2x ))
How to draw Re z =1 on argand diagram !

 

[daredevil]

did you get it cause i cannot get whats difficult and wat you are asking my friend. u need a psychiatrist

ho gaya the one about the min/max point.... thanks

 

[daredevil]

Put the line into parametric form.
If its (0,1,-1) +t(1,2,2) t im using in place of lambda
U will get (t) i +(1+2t) j + (-1+2t)k
Put the coefficient of i j k into the place of x y z in plane p
3(t) + 2( 1+2t) +4(-1+2t) = 13
Solve this uwill get t=1
Put this back into parametric form of the line
(t) i +(1+2t) j + (-1+2t)k u will get
1,3,1 these are coordinate of A
Put the coefficient of i j k into the place of x y z in plane q
-2(t) +(1+2t)+(-1+2t)=4
U will get t=2
Put it back into
(t) i +(1+2t) j + (-1+2t)k
U will get (2,5,3)
This is B
Now B -A =(1,2,2)
Its magnitude will be root of 1sqr + 2sqr+ 2sqr
Root 9=3
Sorry but im from cell islie der lagi

thankkss alot!!

I just hope i remember this procedure.. i ust cant seem to drum it into my head.
i understand it every time someone does it for me but then give me another question and its stuck there again.... i guess i'll do all vector questions from many papers now...

 

[A*(a*)]

anyone who can help me with this question?
Q. z3 - az + 10 = 0
one of the root is given = 1+ 2i
(i) find a.
(ii) find all other roots
PLEASEE ANYONE?

 

[Rutzaba]

thankkss alot!!

I just hope i remember this procedure.. i ust cant seem to drum it into my head.
i understand it every time someone does it for me but then give me another question and its stuck there again.... i guess i'll do all vector questions from many papers now...

tell me wen u nxt come online n i try to help u visually insha Allah

 

[A*(a*)]

tell me wen u nxt come online n i try to help u visually insha Allah

anyone who can help me with this question?
Q. z3 - az + 10 = 0
one of the root is given = 1+ 2i
(i) find a.
(ii) find all other roots
PLEASEE ANYONE?

 

[Rutzaba]

anyone who can help me with this question?
Q. z3 - az + 10 = 0
one of the root is given = 1+ 2i
(i) find a.
(ii) find all other roots
PLEASEE ANYONE?

i hate complex numbers dude...
but leme c

 

[Rutzaba]

anyone who can help me with this question?
Q. z3 - az + 10 = 0
one of the root is given = 1+ 2i
(i) find a.
(ii) find all other roots
PLEASE ANYONE?

OK you know tat if 1+2i is a factor of this equation that means if this equation is divided by this factor the remainder wud be zero
place the (1+2i in place of z in the eq
(1+2i)^3 -a(1+2i) +10=0
(1+4i+4i²)(1+2i) -a-2ai+10=0
1+4i+4i² +2i+ 8i² +8i³-a-2ai+10=0
now u know that i²= -1
1+4i+4(-1)+2i+ 8(-1) +8(-1)(i)-a-2ai+10 Notice that 8i³ was turned into 8i²(i) the i² became -1 and the i remained.
-1-2i-2ai-a =0 reduced form.
-(1+2i)= 2ai+a
-(1+2i)= a(1+2i)
a=-1

 

[salvatore]

anyone who can help me with this question?
Q. z3 - az + 10 = 0
one of the root is given = 1+ 2i
(i) find a.
(ii) find all other roots
PLEASEE ANYONE?

If 1+2i is a root, that means substituting this into the equation will give you 0.

(1+2i)^3 - a(1+2i) +10

The best way to solve (1+2i)^3 is to square it first, then multiply by (1+2i).
(1+2i)^2 = (1)^2 + 4i + (2i)^2
i^2 = -1
=> 4i - 3
(4i-3)(1+2i) = 4i + 8(i)^2 - 3 - 6i
again, i^2 = -1
=> -11 - 2i

-11 - 2i - a - 2ai + 10 = 0
-2i - a - 2ai = 1
-2i - a(1 + 2i) = 1
-a = (1 + 2i)/(1+2i)
a = -1

 

[daredevil]

tell me wen u nxt come online n i try to help u visually insha Allah

I am online... =)

 

[daredevil]

part i

i am getting
'1' instead of 3 .... -.-

 

[Rutzaba]

View attachment 42532

part i

i am getting
'1' instead of 3 .... -.-

a to rha hai three

 

[Rutzaba]

when u substitute u not only substitute x
but u aso find out a value for dx
and u also change the limits