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Domain and range of a relation, Consider R be a relation from A to B, that ...

Consider R be a relation from A to B, that is, take R A Χ B. Then Domain R = {a: a € A, (a, b) € R for any b € B} i.e. domain of R is the set of all the first components of

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integral 0 to pi e^cosx cos (sinx) dx, Let u = sin(x). Then du = cos(x) dx...

Let u = sin(x). Then du = cos(x) dx. So you can now antidifferentiate e^u du. This is e^u + C = e^sin(x) + C.  Then substitute your range 0 to pi. e^sin (pi)-e^sin(0) =0-0 =0

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Derive for the filter from z=a and poles at z=b andz=c, where a, b, c are the real constants the corresponding difference equation. For what values of parameters a, b, and c the fi

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