Use desmos. com to graph and list the coordinates on here. asap

Please, see the answer in the attached files.

Step-by-step explanation:

1.) Suppose:

A=(-3,3)=(xa,ya)→xa=-3, ya=3

B=(0,6)=(xb,yb)→xb=0, yb=6

C=(2,4)=(xc,yc)→xc=2, yc=4

Centered at P=(2,2)=(xp,yp)→xp=2, yp=2

Scale factor: f=2

xa'=xp + f (xa-xp)=2+2(-3-2)=2+2(-5)=2-10→xa'=-8

ya'=yp + f (ya-yp)=2+2(3-2)=2+2(1)=2+2→ya'=4

A'=(xa',ya')→A'=(-8,4)

xb'=xp + f (xb-xp)=2+2(0-2)=2+2(-2)=2-4→xb'=-2

yb'=yp + f (yb-yp)=2+2(6-2)=2+2(4)=2+8→yb'=10

B'=(xb',ya')→B'=(-2,10)

xc'=xp + f (xc-xp)=2+2(2-2)=2+2(0)=2+0→xc'=2

yc'=yp + f (yc-yp)=2+2(4-2)=2+2(2)=2+4→yc'=6

C'=(xc',yc')→C'=(2,6)

Please, see the graph in the attached files.

2.) Suppose:

A=(-5,4)=(xa,ya)→xa=-5, ya=4

B=(3,8)=(xb,yb)→xb=3, yb=8

C=(5,4)=(xc,yc)→xc=5, yc=4

Centered at P=(-3,0)=(xp,yp)→xp=-3, yp=0

Scale factor: f=1/2

xa'=xp + f (xa-xp)=-3+(1/2)(-5-(-3))=-3+(1/2)(-5+3)=-3+(1/2)(-2)=-3-1→xa'=-4

ya'=yp + f (ya-yp)=0+(1/2)(4-0)=0+(1/2)(4)=0+2→ya'=2

A'=(xa',ya')→A'=(-4,2)

xb'=xp + f (xb-xp)=-3+(1/2)(3-(-3))=-3+(1/2)(3+3)=-3+(1/2)(6)=-3+3→xb'=0

yb'=yp + f (yb-yp)=0+(1/2)(8-0)=0+(1/2)(8)=0+4→yb'=4

B'=(xb',yb')→B'=(0,4)

xc'=xp + f (xc-xp)=-3+(1/2)(5-(-3))=-3+(1/2)(5+3)=-3+(1/2)(8)=-3+4→xc'=1

yc'=yp + f (yc-yp)=0+(1/2)(4-0)=0+(1/2)(4)=0+2→yc'=2

C'=(xc',yc')→C'=(1,2)

Please, see the graph in the attached files.

maybe - definition of supplementary angles. im not sure. sorry

step-by-step explanation:

15 units use socratic

step-by-step explanation: 15 units