Rational Area and Irrational Perimeter Question for a Rhombus

Look at Picture

Correct C.

Step-by-step explanation:

Rhombus with side √5 units has perimeter

P = 4 √5 units which is irrational number

Rhombus with diagonals with lengths √2 and √18 has area

A = (d₁ · d₂)/2 = (√2 · √18)/2 = (√36)/2 = 6/2 = 3 units²

A = 3 units² which is rational number

God is with you!!!

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! : )

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