Graph the set of points. Which model is most appropriate for the set?
1 answer:
Check the slope between the moving points
<span>slope between (–6, –1) & (–3, 2) = 1 </span>
<span>slope between (-3, 2) & (–1, 4) = 1 </span>
<span>slope between (-1, 4) & (2, 7) = 1 </span>
<span>the points are collinear and make an angle of 45 degrees with the x-axis </span>
<span>we can have an equation of a line passing through (-6,-1) and slope 1 as </span>
<span>(y + 1) = 1(x + 6) </span>
<span>y = x + 5 is your linear model mostly</span>
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Answer:
CD= 15.03
Step-by-step explanation:
x1=7, x2= -8
y1=-4, y2=-5
distance CD= √(x2- x1) 2+ (y2-y1)2
= √( -8-7)2 + [-5-(-4)]2
= √(15)2+ (-5+4)2
= √225+ (1)2
= √225+1
= √226
= 15.03
Answer:
you need diffrent ones child
Step-by-step explanation:
There's a strong negative association between the two variables.
50 since a triangle has to equal 180, trust!

by using the integration formula
we get,

now put the value of t=\sin\theta in the above equation
we get,

hence proved