Given triangle ABC with coordinates A(−6, 4), B(−6, 1), and C(−8, 0), and its image A′B′C′ with A′(−2, 0), B′(−5, 0), and C′(−6,
Zinaida [17]
Answer:
The line of reflection is at y = x+6.
Step-by-step explanation:
The perpendicular bisector of AA' is a line with slope 1 through the midpoint of AA', which is (-4, 2). In point-slope form, the equation is ...
y = 1(x +4) +2
y = x + 6 . . . . . . . line of reflection
His name is in the name of the name in the name of the name in the name of the name
<h2><u>
Answer With Explanation:</u></h2>
<u>Firstly, let's start with <XOZ: =55°</u>
We know that <ZOQ is 70° and angles on a line add up to 180° so we do 180-70=110 110 divided by 2 = 55 so the 2 angles (XOZ & XOP are 55)
<u>Secondly, <OMN, <MON & <ONM = All are 60°</u>
These 2 angles are joined to create an equilateral triangle which always adds up to 180°
So, there are 3 points to this triangle, therefore we divide 180 by 3 which is 60. The angles are 60°
<u>Thirdly, <QON: =55°</u>
This angle lies on the line XON which needs to add up to 180°
As we worked out before, <XOZ was 55°
So, <ZOQ was already given as 70°
We then do 55+70=125 then 180-125=55°
<QON is 55°
(I'm only in Grade 9 LOL)
Answer:
6cm
Step-by-step explanation:
= 6
