We will show that the reflexive closure of a transitive closure of a relation is equal to the transitive closure of its reflexive closure through a series of simpler results: [3 marks] let ribe a transitive relation on a set s, show that its reflexive closure rr is also transitive. [3 marks] let rą be a reflexive relation on a set s, show that its transitive closure tr, is also reflexive. [3 marks] using the previous results or otherwise, show that r(tr) = t(rr) for any relation r on a set. hint: you may find the fact that transitive (resp. reflexive) closures of r are the smallest transitive (resp. reflexive) relation containing r useful.

eve and marks ratios are equivalent

step-by-step explanation:

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

cretdit to source

$113.42

i hope that is right.

step-by-step explanation: