Acoordinate plane from negative 8 to 8 along the x-axis and from negative 7 to 7 along the y-axis. which steps should be used to graph the equation y – 4 = y minus 4 equals startfraction one-third endfraction left-parenthesis x plus 2 right-parenthesis.(x + 2)? 1. plot the point (2, 4). 2. from that point, count left 3 units and down 1 unit and plot a second point. 3. draw a line through the two points. 1. plot the point (2, 4). 2. from that point, count left 1 unit and down 3 units and plot a second point. 3. draw a line through the two points. 1. plot the point (–2,4). 2. from that point, count left 3 units and down 1 unit and plot a second point. 3. draw a line through the two points. 1. plot the point (–2,4). 2. from that point, count left 1 unit and down 3 units and plot a second point. 3. draw a line through the two points.

see the explanation

Step-by-step explanation:

we have

This is the equation of a line in point slope form

where

the point is (-2,4)

the slope is m=1/3

Remember that the formula of slope is "rise over run", where the "rise" (means change in y, up or down) and the "run" (means change in x, left or right)

so

To graph the line

1. Plot the point (–2,4).

2. From that point, count left 3 units and down 1 unit and plot a second point.

3. Draw a line through the two points

answer: this doesnt make any sense especially the photo.

step-by-step explanation:

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.