Point s should be at 6,-2 and you have plotted at 6,2.
then, the line y=x goes from bottom left corner, through the origin to the top right corner. reflect your shape in this.

7 0

3 years ago

Hurry im despret i need help

svp [43]

The slope-point form of line:

$y-y_1=m(x-x_1)\\\\m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the points (-9, 7) and (6, 2). Substitute:

$m=\dfrac{2-7}{6-(-9)}=\dfrac{-5}{6+9}=-\dfrac{5}{15}=-\dfrac{1}{3}\\\\y-7=-\dfrac{1}{3}(x-(-9))\\\\\boxed{y-7=-\dfrac{1}{3}(x+9)}$

The slope-intercept form of line:

$y=mx+b$ .

We have the slope m:

$y=-\dfrac{1}{3}x+b$

Pu the coordinates of the point (6, 2) to the equation:

$2=-\dfrac{1}{3}(6)+b$

$2=-2+b$ add 2 to both sides

$4=b\to b=4$

$\boxed{y=-\dfrac{1}{3}x+4}$

6 0

3 years ago

What is the rule for the reflection

hram777 [196]

Step-by-step explanation:

We reflect We triangle across both axis so the x and y values will both switch.

Our triangle is originally in the second quadrant but then go to the 4the after we reflect it.

Second Quadrant are

^;8+1—](

3 0

3 years ago

Who can answer this Algebraic Expression for me? 3(m-5)+m

egoroff_w [7]

Simply. 4m - 5.

3(m-5)+m

First, use distributive property

3m-15+ m

Combine like terms

3m + m

The answer

4m - 15

3 0

2 years ago