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Question
secondary 4 | A Maths
3 Answers Below
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Date Posted: 5 years ago
Views: 214
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These are the solutions to all your qn.Hope they help. :)
Date Posted:
5 years ago
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Qn 36
Date Posted:
5 years ago
for part i how you suddenly get 27
The (3-2x) is fixed.Since you are finding the x coefficient,I need to times 3 and -2x to the terms in the power 9 expression I have found.This is where I got 27bx-2x.27 is the end product of multiplying 3 and 9 together.
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Qn 35
Date Posted:
5 years ago
how you get 16kx^2+hk^8=-9x^8
It's basically the same as stated in the earlier post.Your focus is on the terms with X variable when you times (3-2x) & (1+bx)^9 together.By the way,It's 16kx^8+hx^8=-9x^8.
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