C' ( 0, 3)
A' ( 2 , 1)
C'A' = CA
C'A' = √(2^2 + 2^2) = √8 = 2√2
C'A' = CA = 2√2
Hi there!
The linear equation would be y = 3x + 11. This is because the slope is 3, found by using the formula (y2-y1 / x2-x1). Then, we can plug in the slope and a point into y=mx+b form. Next, we can plug in 0 for x and 0 for y to figure out if the line goes through the point (0,0). Doing this, we discover that the line does not go through the point (0,0).
Hope this helps!! :)
If there's anything else that I can help you with, please let me know!
I think it’s C not sure tho
Step-by-step explanation:
8x + 14y = 4 ...1
-6x - 7y = -10 ...2
...2 × 2 in 2 sides of the equation
-12x -14y = -20 ...3
...1 + ...3
8x + 14y +(-12x -14y) = 4 +(-20)
8x + 14y - 12x -14y = 4 - 20
-4x = -16
x = (-16)/(-4)
= 4
replace x in ...1 or ...2 you will reserve y value
and answer in the term of (x,y)
1. Horizontal asymptotes: None
Vertical asymptote: x=-2
RD: 2
Domain: First picture
Range: first pic
2.HA: y=1
VA: x= -5 negative
RD: x= 5 no negative
Domain: second pic
Range second pic
3.HA: y=0
VA: x=-4,4
RD: none
D: 3rd pic
R: 3rd pic
4.HA: none
VA: x=3
RD: x=-2
D: 4th
R: 4th
I hope this helps!!!!!!!