Answer :
By graph of object A the rate of change is estimated at 17/6.
Object B's rate of change is 10/3=20/6.
Comparing object A to object B:
17/6 <20/6
Object B is faster by:
17/6x=20/6
(6/17)(17/6)x=(20/6)(6/17)
x=20/17≈1.18
B)
Object B by a factor of 1.19
Since I made an estimation from graph of object A , my solution is off by. 01 but it is the closest I can get. The solution B is correct.
Object B's rate of change is 10/3=20/6.
Comparing object A to object B:
17/6 <20/6
Object B is faster by:
17/6x=20/6
(6/17)(17/6)x=(20/6)(6/17)
x=20/17≈1.18
B)
Object B by a factor of 1.19
Since I made an estimation from graph of object A , my solution is off by. 01 but it is the closest I can get. The solution B is correct.
Answer:
B)Object B by a factor of 1.19
Step-by-step explanation:
Object A and Object B are two objects in motion. Given the graph of Object A and the equation y =
10
3
x of Object B, which of the two moving objects has greater speed and by what factor? (round to nearest hundredth)
A)
Object A by a factor of 1.19
B)
Object B by a factor of 1.19
C)
Object A by a factor of 1.53
D)
Object B by a factor of 1.53
the slope of A is 17/6 judging form the graph
Object B's slope is 10/3=20/6.(multiply the denominator and numerator by 2)
Comparing object A to object B, we know that
17/6 <20/6
dividing both sides by 17/6
(17/6)/(17/6)x=(20/6)/(17/6)
x=20/17≈1.18
B)Object B by a factor of 1.19