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yup i did that, in a pretty messy way thou. lol ... Thanku thanku!Here's what i've done:
argu = arg( (1+2i)/(1-3i) )
argu = arg(1+2i) - arg(1-3i)
arg(-1/2 + i/2) = tan^-1(2) - tan^-1(-3)
tan^-1(-1) = tan^-1(2) + tan^(3)
3π/4 = tan^-1(2) + tan^-1(3)
Hope you got it
Aoa wr wb
Can someone please help me, for qn 10 ii of this paper, while integrating, i don't get where the -ve sign comes before the 3/2
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_qp_3.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_ms_3.pdf
Thank you
oh right! Thank you so much for thatWa laikum as salaam
S denotes integration
S (9-h)^-1/3
Now to integrate this you multiply by 3/2 and divide by -1. This gives -3/2
If you're wondering why we divide by -1, it's the coefficient of h. And when we integrate we divide by the coefficient of the variable.
why do we divide by 3/2 ? i hope i dont sound dumb?!Wa laikum as salaam
S denotes integration
S (9-h)^-1/3
Now to integrate this you multiply by 3/2 and divide by -1. This gives -3/2
If you're wondering why we divide by -1, it's the coefficient of h. And when we integrate we divide by the coefficient of the variable.
why do we divide by 3/2 ? i hope i dont sound dumb?!
Why CIE why !?
OMG!!! i feel so dumbo! u know what i read (9-h)^-1 is divided by '3'...hahaha ! i know that thing thou...lol thanks anyway.... c
Can someone please explain to me q10 (iii)... i've been stuck on it for a while now... please help guys :/
Also please explain q7(iii) and q8(i).. in q8(i) i keep getting -10 instead of just 10u.. Just explain that point where im making the mistake.. thanks alot
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_32.pdf
Marking scheme: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_ms_32.pdf
Here's what i've done:
argu = arg( (1+2i)/(1-3i) )
argu = arg(1+2i) - arg(1-3i)
arg(-1/2 + i/2) = tan^-1(2) - tan^-1(-3)
tan^-1(-1) = tan^-1(2) + tan^(3)
3π/4 = tan^-1(2) + tan^-1(3)
Hope you got it
Can someone please explain to me q10 (iii)... i've been stuck on it for a while now... please help guys :/
Also please explain q7(iii) and q8(i).. in q8(i) i keep getting -10 instead of just 10u.. Just explain that point where im making the mistake.. thanks alot
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_32.pdf
Marking scheme: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_ms_32.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_33.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_ms_33.pdf
Q7iii) In such cases where they haven't given the initial value for the iteration formula, we just assume it to be 0 ?
MustafaMotani Told ya, I'd annoy you
Also, in complex numbers what is the square root of i ?
in this case i would have used somewhat similar values like 1 or 2 ...but there was one question in which i used pi/4 as initial value to get the answer .. (it ws a trig question) ..
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