Find the circumference and area of a circle with a radius of 3
1 answer:
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Answer:
Step-by-step explanation:
There is an error in the question. The table does not show two linear functions. y₁ is a linear function, but y₂ is not a straight line. It makes a bend at (-6,1).
Line 1 goes through (-12,-3) and (0,5).
slope = (5-(-3))/(0-(-12)) = 2/3
y-intercept = 5
y₁ = (2/3)x + 5
Line 2 goes through (-12,-2) and (-6,1).
slope = (1-(-2))/(-6-(-12)) = 1/2
y₂ = (1/2)x + 4
(2/3)x + 5 = (1/2)x + 4
x = -6
y = (2/3)x + 5 = 1
Solution: (-6,1)
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!!!!!!!
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!!!!!!!