FIND THE RATE CHANGE FOR FG AND OVER WHAT INTERVALS OF TIME IS MICHA TRAVELING THE FASTEST AND WHY ARE SOME LINE SEGMENTS ON THE GRAPH STEEPER THAN OTHERS

1.as we have differents slopes, it means he had differents speeds
2) rates of change
AB is 5
CD is 5
EF is -1.25
BC is 2.5
DE is 0
FG is -4.16666
3) AB and CD
Explanation
when you have a graph where
x-axis=time
y-axis = distance
the rate of change is the speed, then
so
[tex]\text{slope}=\frac{\Delta y}{\Delta x}=\frac{change\text{ in distance}}{\text{change in time}}=speed[/tex]Step 1
why are some line segments on the graph sleeper than others?
as we have differents slopes, it means he had differents speeds
Step 2
speed of each line segment
a)rate of change AB
A=(0,0) origin
B=(4,20)
find the speed( slope)
[tex]\text{slope}=\frac{\Delta y}{\Delta x}=\frac{change\text{ in distance}}{\text{change in time}}=\frac{y_2-y_1}{x_2-x_1}=\frac{20-0}{4-0}=\frac{20}{4}=5[/tex]it means, the rate of change in AB is 5
b)rate of change CD
C=(12,40)
D=(16,60)
find the slope
[tex]\text{slope}=\frac{\Delta y}{\Delta x}=\frac{change\text{ in distance}}{\text{change in time}}=\frac{y_2-y_1}{x_2-x_1}=\frac{60-40}{16-12}=\frac{20}{4}=5[/tex]it means, the rate of change in CD is 5
c)rate of change EF
E=(20,60)
F=(28,50)
find the slope
[tex]\text{slope}=\frac{\Delta y}{\Delta x}=\frac{change\text{ in distance}}{\text{change in time}}=\frac{y_2-y_1}{x_2-x_1}=\frac{50-60}{28-20}=\frac{-10}{8}=-\frac{5}{4}=-1.25[/tex]it means, the rate of change in EF is -1.25
d)rate of change BC
B=(4,20)
C=(12,40)
find the slope
[tex]\text{slope}=\frac{\Delta y}{\Delta x}=\frac{change\text{ in distance}}{\text{change in time}}=\frac{y_2-y_1}{x_2-x_1}=\frac{40-20}{12-4}=\frac{20}{8}=2.5[/tex]it means, the rate of change in BC is 2.5
e)Rate of change DE
D=(16,60)
E=(20,60)
find the slope
[tex]\text{slope}=\frac{\Delta y}{\Delta x}=\frac{change\text{ in distance}}{\text{change in time}}=\frac{60-60}{20-16}=\frac{0}{4}=0[/tex]it means, the rate of change in DE is 0
f)rate of change change FG
F=(28,50)
G=(40,0)
find the slope
[tex]\begin{gathered} \text{slope}=\frac{\Delta y}{\Delta x}=\frac{change\text{ in distance}}{\text{change in time}}=\frac{0-50}{40-28}=\frac{-50}{12}=-4.16666 \\ \end{gathered}[/tex]it means, the rate of change in FG is -4.16666
Step 3
what interval is the fastest?
the fastest interval is when you have the bigger slope , then
[tex]AB\text{ and CD}[/tex]because the speed is 5