Which sequence of transformations will map figure k onto figure k′?
two congruent kites, figure k and figure k prime, are drawn on a coordinate grid. figure k has vertices at 4, 3, at 6, 5, at 4, 8, and at 2, 5. figure k prime has vertices at 4, negative 8, at 6, negative 5, at 4, negative 3, and at 2, negative 5
reflection across x = 4, 180° rotation about the origin, and a translation of (x + 8, y)
reflection across x = 4, 180° rotation about the origin, and a translation of (x − 8, y)
reflection across y = 4, 180° rotation about the origin, and a translation of (x + 8, y)
reflection across y = 4, 180° rotation about the origin, and a translation of (x − 8, y)

a.   since the profit made in 2012 is equal to the profit made in 2014, the profit is symmetric about the line y = 120,000.

step-by-step explanation:

8y^50

step-by-step explanation:

54 km / hours

step-by-step explanation:

we need to convert seconds to hours

60 seconds = 1 minutes

60 minutes = 1hour

15 m         60 second   60 minutes

  * *         = 54000 meters/ hour

second     1 minute             1 hour

now we need to convert meters to km

1000 m = 1 km

54000 m       1 km

*   = 54 km / hours

hour               1000m

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