It wants it to be in slope-intercept form.
y=mx+b
We have to first find the slope and plug it into point-slope form.
y-y1=m(x-x1)
Find the slope of the second line. (I did this one first on accident)
Rise/run= 3/1= 3 The slope is 3. Plug that in along with the point (0,3)
y-3=3(x-0)
y-3=3x
Add 3 to the other side.
y= 3x +3 <- <em>for the second line</em><em>
</em>
Now, the second.
rise/run= 1/2= .5 Use point (6,0)
y-0=.5(x-6)
y= .5x-3
y=.5x-3 <- for the first line
I hope this helps!
~kaiker
Answer:
Option C is right.
Step-by-step explanation:
Given is a graph with two triangles marked on it.
Triangle ABC is in the I quadrant with vertices (2,2) (2,10) and (8,12)
Triange A'B'C' is in the III quadrant with vertices (-1,-1), (-1,-5) and (-4,-6)
On comparison we find corresponding side of AB is A'B'
Length of AB = 8 and Length of A'B' = 4.
Hence A'B'C' is obtained by dilating ABC by a scale factor of 1/2.
Now since moved to III quadrant from I quadrant we find that there is a rotation of triangle ABC about the origin. The degree of rotation is 180 degrees.
Hence A'B'C' is obtained by dilating ABC by a scale factor of 1/2 and then rotating it about the origin by 180 degrees
Answer:
(3, -2) and (9, 0)
Step-by-step explanation:
x-3(-2)=9
x+6= 9
x=3
9-3y=9
-3y=0
y=0
Answer: D=Pie x C
Step-by-step explanation:
trust me and plz make me the brillienrttest
