Question

pls help me in qns:
10 , 11 ,17 and 20

Answer 1

man at least type them out one at a time... i hate it when ppl just post their homework sets.

Answer 2

So #10. What's an arbitrary term of
\[ \left( \frac{3x}{2} - \frac{1}{3x} \right)^6 \]

Well, you can write down the terms using the binomial theorem: it is
\[ \sum_{i=0}^6 {6 \choose i} \left(\frac{3x}{2}\right)^i \left( \frac{-1}{3x} \right)^{n-i} \]

Answer 3

oh and that's not even hw, that's a take home exam

Answer 4

hey its not homework these are model qns i hav xam tmrw..i got sm doubts thats y i posted .bahrom

Answer 5

ok

Answer 6

didnt u see the heading of document?

Answer 7

james wat abt the other 3?

Answer 8

17)
\[f(x)= \frac{x^{100}}{100}+\frac{x^{99}}{99}+\frac{x^{98}}{98}+\frac{x^{97}}{97}+......+\frac{x^2}{2}+\frac{x^1}{1}+1\]
now let's find f'(x)
\[f'(x)=x^{99}+x^{98}+x^{97}+....+x^2+x^1+1\]
if x=1
we get
\[f'(1)=1 + 1+ 1....1 (100 times)\]
so
\[f'(1)=100\]
if x=0
\[f'(0)=0+1\]
so
\[f'(1)=100\times f'(0)\]

Answer 9

Aravind: (a) please at least acknowledge that I've helped you with one question before going on to the next. You do this all the time, and I have to tell you in Western culture, it's very rude
(b) I haven't even finished that first question yet; what about the second part of it. Can you see how to do it? If so, say so. If not, engage in conversation.
(c) I'm with Bahrom. Much better to post each of these questions one at a time.

Answer 10

hey mertsyj can you help me with a problem?

Answer 11

oh srry ..thx i knw the second part

Answer 12

thx ash