The easiest way is to graph it based upon the slope (m) and y-intercept (b), in the standard slope-intercept form: y = m (x) + b.
The line above intercepts the y-axis at y = -2, which is b. The slope (m) = rise/run = (y2-y1)/(x2-x1 ); so for the point (-4, 2) to (-6, 4) is:
(4-2)/(-6--4) = 2/(-6+4) = 2/-2 = -1.
So one form of the equation would be:
y = -1x - 2
Now the other form of an equation is point-slope: y-k = m (x-h), where the point is at (h, k)
and if we pick -5 for x (bc 5 it listed in 3 of the answers), the y at x=-5 looks like around +3
so we get: y-k = -1 (x--5)...
y-3 = -(x+5)... therefore D) is the correct answer:
Answer:
i need points im really sorry
Step-by-step explanation:
J=-8c²-4cd-d²
k=-3c²+7cd-3d²
j+k=
-8c²-4cd-d²-3c²+7cd-3d²
-11c²-4d²+3cd
Answer:
B. Yes. ΔDEF can be mapped to ΔRPQ by a 180° rotation about the origin followed by a translation 2 units down.
Step-by-step explanation:
^^^just did it on edg and got it right.