Mathematics: Post your doubts here!

[daredevil]

plz solve this =`|

 

[Thought blocker]

View attachment 39335

plz solve this =`|

Is answer 28 ?

 

[Thought blocker]

 

[daredevil]

Is answer 28 ?

this is the answer

 

[asd]

View attachment 39331



View attachment 39333
Q6 and Q7 from this paper...

it's M/J 2013 p32
A star

d(1/cosx)/dy= -(-sinx)/cos^2(x)= sinx/cosx.1/cosx
=tanx.secx
ln(secx + tanx) = (secx.tanx + sec^2(x)) . 1/(secx + tanx)
simply, it should give you secx.

And for 7, Expand cos(x+45) and then add (-sqrt(2)sinx) to it. then you can express it in that form.

 

[Thought blocker]

this is the answer

Oops. I suck.

 

[sumeru]

View attachment 39331



View attachment 39333
Q6 and Q7 from this paper...

it's M/J 2013 p32
A star

 

[asd]

can someone please help me with the hence show part in part (ii)??View attachment 39317

|z| = √[(a)^2+(b)^2], and |z|^2 = a^2+b^2
z.z*= (a+ib)(a-ib)= a^2 +aib -aib -(bi)^2 = a^2 + b^2
(N.B i^2 = -1)

(z-ki) = a+ib-ki = a +(b-k)i
(z-ki)*= a -(b-k)i = a-bi +ki
Ooops, HAHAHA never saw you wrote part (ii)
can't do that one either, I guess Namehere can help you with this

 

[daredevil]

|z| = √[(a)^2+(b)^2], and |z|^2 = a^2+b^2
z.z*= (a+ib)(a-ib)= a^2 +aib -aib -bi^2 = a^2 + b^2
(N.B i^2 = -1)

ye a^2 - b^2 nahi hoga??

 

[asd]

ye a^2 - b^2 nahi hoga??

No because i^2 = -1, so -(bi)^2 = -(-b^2)= +b^2
Neeche brackets me likha bhi hai

 

[Heyyy]

Does the +12 suddenly turn into a -12? Anyways, I know how to do that part, but thanks anyways! ^^ I wanted you to do the Hence show part (the one with the = 4 )

Which part is that?

 

[daredevil]

oo

No because i^2 = -1, so -(bi)^2 = -(-b^2)= +b^2
Neeche brackets me likha bhi hai

oo ryt sorry... i'm a little slow today

 

[sumeru]

View attachment 39335

plz solve this =`|

You'll get confused, if u look at the mark scheme...the answer is properly given in the examiner's report....so look at the er...

 

[daredevil]

in question it is 3+x^2 but u have multiplied with x^2...?? shouldn't it be...

(3+[sqrt3 tan(thita)]^2)^-1/2

this was the question.. no?

 

[asd]

View attachment 39335

plz solve this =`|

I've solved this but the working is pretty lengthy.
Ill just tell you in short, you gotta break the 1/x^2(2x+1) into partial fractions, and then integrate. youre gonna get -2ln(x) -1/x + 2ln(2x+1)= lny +c
Then use (1,1) to find c. Then youre gonna get too many ln's. just put them together in a single ln function (divide the ones being subtracted, multiply the ones being added) and then cancel the ln from both sides. THEN, substitute 2 in x and youre gonna get 25e/36e^(1/2) = 25e^(1/2) / 36

 

[A star]

in question it is 3+x^2 but u have multiplied with x^2...?? shouldn't it be...

(3+[sqrt3 tan(thita)]^2)^-1/2

this was the question.. no?

i did this question sooooooo long ago i forgot how i did it am searching for the working . may take a while

 

[Namehere]

|z| = √[(a)^2+(b)^2], and |z|^2 = a^2+b^2
z.z*= (a+ib)(a-ib)= a^2 +aib -aib -(bi)^2 = a^2 + b^2
(N.B i^2 = -1)

(z-ki) = a+ib-ki = a +(b-k)i
(z-ki)*= a -(b-k)i = a-bi +ki
Ooops, HAHAHA never saw you wrote part (ii)
can't do that one either, I guess Namehere can help you with this

I´m actually looking for someone who can solve Q7(ii), the part of hence show that the modulus of z-2i = 4.
Here is the question once again:

 

[saif mahmoud]

please anyone where can i find from 1993 to 2000 AS biology past paper asap!! extremely important plz im running out of time

 

[asd]

I´m actually looking for someone who can solve Q7(ii), the part of hence show that the modulus of z-2i = 4.
Here is the question once again:

Well, with some help from the mark scheme, I can tell that the expression |z-2i| = zz* -2iz + 2iz + 4
Now if you look at it, its pretty similar to the expression already given in the question except that we have a positive 4, instead of -12.

 

[sumeru]

l

can someone please help me with the hence show part in part (ii)??View attachment 39317

For q(2).....