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WelcomeI didn't know this ... damn, tysm!!!
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WelcomeI didn't know this ... damn, tysm!!!
It has to be given. Otherwise anything works because sin x = cos (90-x).acosθ + bsinθ = rsin(θ+α) or rsin(θ-α) or rcos(θ+α) or rcos(θ-α)
How do I know when to use what? :O
anything works but it is simplest using a particular method ... you know, time management and allIt has to be given. Otherwise anything works because sin x = cos (90-x).
You need two vectors that lie in the plane. Then cross both of them to find the normal vector.Can someone please explain how to get a second vector for the question (ii) for the vector product ?
View attachment 59259
Use binomeal expansion.
lol i know that ... i didn't get the right answerUse binomeal expansion.
So first part you have 3/3-2x + -x-2/x^2+4
Ohh such a stupid mistake i made ... I'm soo sorry for making you do it Thank you! I realised my mistake ... instead of substituting value of A, i had substituted value of B ... I can only pray to God I don't make such mistakes in my exam.So first part you have 3/3-2x + -x-2/x^2+4
For binomial expansion as mentioned in data booklet we need the bracket like (1+x)^n
So change 3/3-2x ==> 3/3(1-(2x/3)) ==> (1-(2x/3))^-1
Similarly
- 1) And expand (1-(2x/3))^-1
-x-2/x^2+4 ==> -x-2/4((x^2/4)+1) ==> 4^-1(-x-2)((x^2/4)+1)^-1
Now add 1 and 2 to obtain full expansion of 3/3-2x + -x-2/x^2+4
- 2) Expand ((x^2/4)+1)^-1 and then multiply the expansion by 4^-1(-x-2)
Hahaha no problem!Ohh such a stupid mistake i made ... I'm soo sorry for making you do it Thank you! I realised my mistake ... instead of substituting value of A, i had substituted value of B ... I can only pray to God I don't make such mistakes in my exam.
Thank you.
Re: Maths help available here!!! Stuck somewhere?? Ask here!
Could you help me with a solution to the Chapter 11 review problems in new syllabus mathematics 3 by lee peng yee?
UPDATE: Link to Sequences Help by destined007 added!
View attachment 59296
(ii) & (iii) part
All questions? oe just the last two?Could anyone help me with these problems?Would be greatly appreciatedView attachment 59298
Can you explain the "cross product....."again.
Can you explain the "cross product....."again.
Thanks
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