# precalculus

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

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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~

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1. 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= 0
c) x= 1
d x= 3

sorry mistake on #2

its B?

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2. Problem 1.
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.

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3. Problem 2.
The answer is not B, (x=0).

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4. For Number 1, the answer is A

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)

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5. Problem 1
f(4)- f(1)
=11 - (-1)
= +12

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Problem 3.
C looks correct

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6. Why is problem 1 not A? I plugged in 4 and 1 into each of the equations given and A gave me negative avg rate T.T

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7. 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.

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8. Problem 1.
Look at my calculation, the average rate is +12,not a negative number.

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9. This is the correct question, sorry I forgot to type in the one part of the question.

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)

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10. Then the answer must be c)

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

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11. C is the answer I got.

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12. For problem 2, it is A? x= -1

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13. For problem 2, I got A, x=-1.

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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 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

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15. With the values given, I did not find any of the answers on the graph.

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.

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16. I believe its between (10,6) and (12,6) because we know that the starting point is at 6..

I don't know after that

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