Use DE←→ and FG←→ to answer the question. DE←→ contains the points D(1,−2) and E(3,4). FG←→ contains the points F(−1,2) and G(4,0). Is DE←→ perpendicular to FG←→? Why or why not? A. No, because the product of the slopes is not −1. B. Yes, because the product of the slopes is −1. C. No, because the product of the slopes is not 1. D. Yes, because the product of the slopes is 1.

A. No, because the product of the slopes is not -1

Step-by-step explanation:

The slope of a line that passes through points (x1, y1) and (x2, y2) is computed as follows:

m = (y2 - y1)/(x2 - x1)

Therefore:

Two lines are perpendicular if the product of their slopes is equal to minus one. In this case, m1*m2 = 3*(-0.4) = -1.2 ≠ -1

guess at question:

a vertical line intersects the graph of g(x,y) = 0 at two points. what conclusion follows about the function f(x) implicitly defined by g(x,f(x)) = 0?

the domain of f(x) cannot include the x coordinate of any such vertical line.

informally, f(x) is not a function, and

g(x,y)=0 cannot be solved for y uniquely.

the relation r = {(x,y) such that g(x,y) = 0} is a function if

for all (a,b) ∊ r and (c,d) ∊ r, a=c implies b=d.

step-by-step explanation:

answer: b

step-by-step explanation: