Find the coordinates of the orthocenter of a triangle with vertices at each set of points on a coordinate plane. A. ​(0,0), ​(12 ​,6​), ​(6 ​,48 ​) b. ​(2 ​,4​), ​(10 ​,12​), ​(3 ​,19​)

the answer is c

hope this !

y=3n + 1

step-by-step explanation:

this is wrong because it says the second number is one less and i added one instead

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

add a pic of the graph, then ill edit this "answer" and solve it for you : 3

step-by-step explanation: