Step-by-step explanation:
(- 4xy - 2) - 5(xy - 2) - 1
-4xy - 2 - 5xy + 10 - 1
Solving like terms
-9xy + 7
It depend on what question an algebric can have as many but depends???
Since it's a reflection over the x axis, the x coordinates will stay the same while the y coordinates will be opposite of what they are.
After reflected: A'(-2,-3), B'(0,-3), C'(-1,1)
For parts A, B, C, and D you most likely created a line. What question E is asking is for you to create a line that is perpendicular to the line you already created that also passes through the point (1,1). What is important to understand here is that the slope of the perpendicular line is the negative reciprocal of the original line's slope... if the original slope is (-4/3) than the perpendicular slope is (3/4)... then you should just plug that new slope into point-slope form or slope-intercept form to get your equation... y-y1 = m(x-x1) ... y-1= (3/4)(x-1) ... so it would be y=(3/4)x + 1/4 then for part f just convert into standard form which is just manipulating the variables... look up standard form equation on Google and manipulate the variables from there.
