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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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8eihsenag@
send meeting id
oh ok. Let me open teamview

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Other answers:

uhm... can not connect? O.o
is it the same ID you game me last time?
m14-003-342 ?
nope that was finished.. you need to start a fresh meeting.. click on Meeting, then you will see "presentation" icon on left pane
m09-542-741
what do I do now?
@ganeshie8 uhm... hello? :D
joining..
oh I see your name
It says you haven't started the presentation...
O.o how do I start it?
Let : \[f(x) = \sum\limits_{k=0}^{n}a_kx^k\] \[g(x) = \sum\limits_{k=0}^{n}a_k(x+jp)^k\] show that \(f(x)\equiv g(x) \pmod{p}\)
\((x+jp)^k = x^k + p*(some integer)\)
let me skip typing the sum, so it's x^i (jp)^(k-i), i from 0 to k ? oh, x^0 (jp)^(k-0) + x^1 (jp)^(k-1) + x^2 (jp)^(k-2) + ... + x^(k-1)(jp)^1 + x^k (jp)^0 yeah I see p as a factor of every term except the last term p*(some integer) + x^k yes
\((x+jp)^k = x^k + p*(some integer) \equiv x^k + 0 \pmod{p}\)

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