Please show all steps and explain: Prove or disprove that a positive integer congruent to 1 modulo 4 cannot have a prime factor congruent to 3 modulo 4.

.5 lb

step-by-step explanation:

12 in / 6 in =2 in

.25 x 2 = .5 lb

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

i believe your answer is 99.7% of 7-year-old children are between 46.6 inches and 49.5 inches tall.