Given the points (-3,4), (-1,5), (5,10) and (7,11) of a parallelogram. what transformation would translate the parallelogram so the lower left vertex would be at the origin and all other points would fall in the first quadrant?

Consider this equation. |y + 6| = 2 what can be concluded of the equation? check all that apply. there will be one solution. there will be two solutions. the solution to –(y + 6) = 2 will be also be a solution to the given absolute value equation. the solution(s) will be the number(s) on the number line 2 units away from –6. the value of y must be positive since the variable is inside absolute value signs.

If two parallel lines are cut by a transversal, interior angles on the same side of the transversal are supplementary. a. always b. sometimes c. never

F(x) = x^2+x-2/x^2-3x-4 what is the domain and range, x-and y-intercepts, horizontal asymptotes, and vertical asymptotes?