we are given points A and B
A=(3,150)
B=(4,200)
we can find points
x1=3 , y1=150
x2=4 , y2=200
(a)
the change in y-values from Point A to Point B is
[tex]=y_2-y_1[/tex]
now, we can plug values
[tex]=200-150[/tex]
[tex]=50feet[/tex]...........Answer
(b)
the change in x-values from Point A to Point B is
[tex]=x_2-x_1[/tex]
now, we can plug values
[tex]=4-3[/tex]
[tex]=1second [/tex]...........Answer
(c)
the rate of change of the linear function in feet per second is
[tex]=\frac{y_2-y_1}{x_2-x_1}[/tex]
now, we can plug values
[tex]=\frac{50}{1}feet/sec[/tex]
[tex]=50feet/sec[/tex].................Answer
The straight line graph indicates that rate of change of the distance traveled with time by the red kangaroo is a constant
Reasons:
The y-values at point A = 150 feet
The y-values at point B = 200 feet
Change in y-values from point A to point B, Δy is given as follows;
Δy = y-values at point B - y-values at point A
Δy = 200 feet - 150 feet = 50 feet
Change in y-values from point A to point B, Δy = 50 feet
The x-values at point A = 3 seconds
The x-values at point B = 4 seconds
Change in x-values from point A to point B, Δx is given as follows;
Δx = x-values at point B - x-values at point A
Δx = 4 seconds - 3 seconds= 1 second
Change in x-values from point A to point B, Δx = 1 second.
The rate of change of the linear function is [tex]\frac{\Delta y}{\Delta x}[/tex], therefore, we have;
[tex]Rate \ of \ change \ of \ the \ linear \ function = \dfrac{\Delta y}{\Delta x} = \dfrac{50 \ feet}{1 \ second} = 50 \ feet \ per \ second[/tex]
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