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Is answer 28 ?
this is the answerIs answer 28 ?
d(1/cosx)/dy= -(-sinx)/cos^2(x)= sinx/cosx.1/cosx
Oops. I suck.this is the answer
|z| = √[(a)^2+(b)^2], and |z|^2 = a^2+b^2can 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)
No because i^2 = -1, so -(bi)^2 = -(-b^2)= +b^2ye a^2 - b^2 nahi hoga??
Which part is that?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 )
oo ryt sorry... i'm a little slow todayNo because i^2 = -1, so -(bi)^2 = -(-b^2)= +b^2
Neeche brackets me likha bhi hai
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...
I've solved this but the working is pretty lengthy.
i did this question sooooooo long ago i forgot how i did it am searching for the working . may take a whilein 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?
|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
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