Can someone verify if my answer is right?
1. Which function has a negative average rate of change on the interval 1<x<4
a) f(x) = x^2 - x - 1
b) g(x) = 1.6x - 2
c) h(x) = -x^2 + 9
d j(x) = -3
I picked d, because I subsitute all the numbers for all letters and the only negative one was D because there is no x value to add in.
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2. For which value of x is the instantaneous rate of change of h(x) = 0.5x^2 + x - 2 closest to 0?
a) x= -1
b) x= -1
c) x= 0
d x= 1
I chose B
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3. A student is walking in a straight line in front of a motion sensor. What does he need to do to produce a horizontal segment on the distance versus time graph?
a) move quickly toward the sensor
b) move slowly toward the sensor
c) stand in place
d) move quickly away from the sensor
The answer is c.
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4. A student is walking in a straight line in front of a sensor. The sensor begins collecting data when the student is 6m away. The student walks toward the sensor for 4 s at a rate of 0.5 m/s. Which of the points on the graph of the distance versus time graph for this walk?
a) (6, 0)
b) (8,2)
c) (10,6)
d) (12,6)
I chose D.
Help appreciated~
16 answers
a) x= -1
b) x= 0
c) x= 1
d x= 3
sorry mistake on #2
its B?
The answer is NOT D.
The average rate of change is:
for x=1, f(1)=-3
for x=4, f(4)=-3
Average rate of change is 0, not a negative number. One of the answers does give a negative average rate of change.
The answer is not B, (x=0).
wow, totally miscalculated.
What about other questions am I right?
For number 3
4. A student is walking in a straight line in front of a sensor. The sensor begins collecting data when the student is 6m away. The student walks toward the sensor for 4 s at a rate of 1 m/s.
Then she walks away from the sensor for 8 s at a rate of 0.5 m/s.Which of the points on the graph of the distance versus time graph for this walk?
a) (6, 0)
b) (8,2)
c) (10,6)
d) (12,6)
*Sorry, forgot one part of the question T.T I chose D, (12,6)
A is not the answer.
f(4)- f(1)
=11 - (-1)
= +12
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Problem 3.
C looks correct
at t=6s, dist=5m
at t=8s, dist=6m
at t=10s, dist=7m
at t=12s, dist=8m
I don't see any of those as choices.
Look at my calculation, the average rate is +12,not a negative number.
4. A student is walking in a straight line in front of a sensor. The sensor begins collecting data when the student is 6m away. The student walks toward the sensor for 4 s at a rate of 1 m/s.
Then she walks away from the sensor for 8 s at a rate of 0.5 m/s.Which of the points on the graph of the distance versus time graph for this walk?
a) (6, 0)
b) (8,2)
c) (10,6)
d) (12,6)
h(x) = -x^2 + 9/4-1
= -(4)^2 + 9/4-1
= -7/4-1
=-x(1)^2 + 9
= 8
= -7 - 8/3
=-5
there c is negative
4. A student is walking in a straight line in front of a sensor. The sensor begins collecting data when the student is 6m away. The student walks toward the sensor for 4 s at a rate of 1 m/s.
Then she walks away from the sensor for 8 s at a rate of 0.5 m/s.Which of the points on the graph of the distance versus time graph for this walk?
a) (6, 0)
b) (8,2)
c) (10,6)
d) (12,6)
I got D
For a distance vs time graph. X axis is time and Y axis is distance.
at t=6s, dist=5m
at t=8s, dist=6m
at t=10s, dist=7m
at t=12s, dist=8m
I don't see any of those as choices.
I don't know after that