B3 calculus

#$&*

course PHY-242

005. Calculus

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Question: `q001. There are 12 questions in this document.

The graph of a certain function is a smooth curve passing through the points

(3, 5), (7, 17) and (10, 29).

Between which two points do you think the graph is steeper, on the average?

Why do we say 'on the average'?

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Your solution:

Steepness of (3, 5) and (7, 17)

(17 - 5)/(7 - 3) = 12/4 = 3

Steepness of (7, 17) and (10, 29)

(29 - 17)/(10 - 7) = 12/3 = 4

Steepness of (3, 5) and (10, 29)

(29 - 5)/(10 - 3) = 24/7 = 3.42

The graph is steepest between (7, 17) and (10, 29) having the average slope of

4.

We say on average since we assume graph is a straight line. In reality, the

graph probably has a steeper slope at one point and smoother slope at another

point.

confidence rating #$&*: 3

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Given Solution:

`aSlope = rise / run.

Between points (7, 17) and (10, 29) we get rise / run = (29 - 17) / (10 - 7)

=12 / 3 = 4.

The slope between points (3, 5) and (7, 17) is 3 / 1. (17 - 5) / (7 -3) = 12 /

4 = 3.

The segment with slope 4 is the steeper. The graph being a smooth curve,

slopes may vary from point to point. The slope obtained over the interval is a

specific type of average of the slopes of all points between the endpoints.

2. Answer without using a calculator: As x takes the values 2.1, 2.01, 2.001

and 2.0001, what values are taken by the expression 1 / (x - 2)?

1. As the process continues, with x getting closer and closer to 2, what

happens to the values of 1 / (x-2)?

2. Will the value ever exceed a billion? Will it ever exceed one trillion

billions?

3. Will it ever exceed the number of particles in the known universe?

4. Is there any number it will never exceed?

5. What does the graph of y = 1 / (x-2) look like in the vicinity of x = 2?

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Your solution:

2. Answer without using a calculator: As x takes the values 2.1, 2.01, 2.001

and 2.0001, what values are taken by the expression 1 / (x - 2)?

As x takes the following values, the expression approaches infinity

1. As the process continues, with x getting closer and closer to 2, what

happens to the values of 1 / (x-2)?

The values reach infinity

2. Will the value ever exceed a billion? Will it ever exceed one trillion

billions?

Yes, it will, since infinity is greater than a billion or a trillion.

3. Will it ever exceed the number of particles in the known universe?

Yes, since infinity is greater

4. Is there any number it will never exceed?

Infinity is greater than any Real Number, so it will exceed all Real numbers

5. What does the graph of y = 1 / (x-2) look like in the vicinity of x = 2?

Vertical Assemtope

confidence rating #$&*: 3

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Given Solution:

`aFor x = 2.1, 2.01, 2.001, 2.0001 we see that x -2 = .1, .01, .001, .0001.

Thus 1/(x -2) takes respective values 10, 100, 1000, 10,000.

It is important to note that x is changing by smaller and smaller increments

as it approaches 2, while the value of the function is changing by greater and

greater amounts.

As x gets closer in closer to 2, it will reach the values 2.00001, 2.0000001,

etc.. Since we can put as many zeros as we want in .000...001 the reciprocal

100...000 can be as large as we desire. Given any number, we can exceed it.

Note that the function is simply not defined for x = 2. We cannot divide 1 by

0 (try counting to 1 by 0's..You never get anywhere. It can't be done. You can

count to 1 by .1's--.1, .2, .3, ..., .9, 1. You get 10. You can do similar

thing for .01, .001, etc., but you just can't do it for 0).

As x approaches 2 the graph approaches the vertical line x = 2; the graph

itself is never vertical. That is, the graph will have a vertical asymptote at

the line x = 2. As x approaches 2, therefore, 1 / (x-2) will exceed all

bounds.

Note that if x approaches 2 through the values 1.9, 1.99, ..., the function

gives us -10, -100, etc.. So we can see that on one side of x = 2 the graph

will approach +infinity, on the other it will be negative and approach -

infinity.

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Self-critique (if necessary):

ok

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Self-critique Rating: ok

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Question: `q003. One straight line segment connects the points (3,5) and (7,9)

while another connects the points (10,2) and (50,4). From each of the four

points a line segment is drawn directly down to the x axis, forming two

trapezoids. Which trapezoid has the greater area? Try to justify your answer

with something more precise than, for example, 'from a sketch I can see that

this one is much bigger so it must have the greater area'.

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Your solution:

Area of trapezoid: 0.5*h(b1+b2)

the points (3,5) and (7,9)

0.5*h(b1+b2)

0.5*(7-3)(5+9)

0.5*(4)(14)

0.5*(4)(14)

0.5*56

28

the points (10,2) and (50,4)

0.5*h(b1+b2)

0.5*(50-10)(2+4)

0.5*(40)(6)

0.5*240

120

By finding the area of the trapezoid, the trapezoid with the points (10,2) and

(50,4) has the greater area.

confidence rating #$&*: 3

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Given Solution:

`aYour sketch should show that while the first trapezoid averages a little

more than double the altitude of the second, the second is clearly much more

than twice as wide and hence has the greater area.

To justify this a little more precisely, the first trapezoid, which runs from

x = 3 to x = 7, is 4 units wide while the second runs from x = 10 and to x =

50 and hence has a width of 40 units. The altitudes of the first trapezoid are

5 and 9,so the average altitude of the first is 7. The average altitude of the

second is the average of the altitudes 2 and 4, or 3. So the first trapezoid

is over twice as high, on the average, as the first. However the second is 10

times as wide, so the second trapezoid must have the greater area.

This is all the reasoning we need to answer the question. We could of course

multiply average altitude by width for each trapezoid, obtaining area 7 * 4 =

28 for the first and 3 * 40 = 120 for the second. However if all we need to

know is which trapezoid has a greater area, we need not bother with this step.

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Self-critique (if necessary):

ok

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Self-critique Rating: ok

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Question: `q004. If f(x) = x^2 (meaning 'x raised to the power 2') then which

is steeper, the line segment connecting the x = 2 and x = 5 points on the

graph of f(x), or the line segment connecting the x = -1 and x = 7 points on

the same graph? Explain the basis of your reasoning.

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Your solution:

f(x) = x^2

x, y

-1, 1

2, 4

5, 25

7, 49

line segment connecting the x = 2 and x = 5

x, y

2, 4

5, 25

average slope (25-4)/(5-2) = 21/3 = 7

line segment connecting the x = -1 and x = 7

x, y

-1, 1

7, 49

average slope (49-1)/(7-(-1)) = 48/(7+1) = 48/8 = 6

Conclusion:

line segment connecting the x = 2 and x = 5 is steeper (due to the slope).

confidence rating #$&*: 3

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Given Solution:

`aThe line segment connecting x = 2 and the x = 5 points is steeper: Since f

(x) = x^2, x = 2 gives y = 4 and x = 5 gives y = 25. The slope between the

points is rise / run = (25 - 4) / (5 - 2) = 21 / 3 = 7.

The line segment connecting the x = -1 point (-1,1) and the x = 7 point (7,49)

has a slope of (49 - 1) / (7 - -1) = 48 / 8 = 6.

The slope of the first segment is greater.

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Self-critique (if necessary):

ok

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Self-critique Rating: ok

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Question: `q005. Suppose that every week of the current millennium you go to

the jeweler and obtain a certain number of grams of pure gold, which you then

place in an old sock and bury in your backyard. Assume that buried gold lasts

a long, long time ( this is so), that the the gold remains undisturbed (maybe,

maybe not so), that no other source adds gold to your backyard (probably so),

and that there was no gold in your yard before..

1. If you construct a graph of y = the number of grams of gold in your

backyard vs. t = the number of weeks since Jan. 1, 2000, with the y axis

pointing up and the t axis pointing to the right, will the points on your

graph lie on a level straight line, a rising straight line, a falling straight

line, a line which rises faster and faster, a line which rises but more and

more slowly, a line which falls faster and faster, or a line which falls but

more and more slowly?

2. Answer the same question assuming that every week you bury 1 more gram than

you did the previous week.

3. Answer the same question assuming that every week you bury half the amount

you did the previous week.

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Your solution:

1. It's hard to tell since I don't know the exact amount of gold that you

purchase every week. If I assume that the quantity of gold purchased is the

same every week, the it would form a positive slope linear line.

2. Rising line which rises faster and faster

3. a line which rises but more and more slowly

confidence rating #$&*: 3

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Given Solution:

`a1. If it's the same amount each week it would be a straight line.

2. Buying gold every week, the amount of gold will always increase. Since you

buy more each week the rate of increase will keep increasing. So the graph

will increase, and at an increasing rate.

3. Buying gold every week, the amount of gold won't ever decrease. Since you

buy less each week the rate of increase will just keep falling. So the graph

will increase, but at a decreasing rate. This graph will in fact approach a

horizontal asymptote, since we have a geometric progression which implies an

exponential function.

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Self-critique (if necessary):

ok

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Self-critique Rating: ok

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Question: `q006. Suppose that every week you go to the jeweler and obtain a

certain number of grams of pure gold, which you then place in an old sock and

bury in your backyard. Assume that buried gold lasts a long, long time, that

the the gold remains undisturbed, and that no other source adds gold to your

backyard.

1. If you graph the rate at which gold is accumulating from week to week vs.

the number of weeks since Jan 1, 2000, will the points on your graph lie on a

level straight line, a rising straight line, a falling straight line, a line

which rises faster and faster, a line which rises but more and more slowly, a

line which falls faster and faster, or a line which falls but more and more

slowly?

2. Answer the same question assuming that every week you bury 1 more gram than

you did the previous week.

3. Answer the same question assuming that every week you bury half the amount

you did the previous week.

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Your solution:

1. a rising straight line

2. a line which rises faster and faster

3. a line which rises but more and more slowly

confidence rating #$&*: 3

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Given Solution:

`aThis set of questions is different from the preceding set. This question now

asks about a graph of rate vs. time, whereas the last was about the graph of

quantity vs. time.

Question 1: This question concerns the graph of the rate at which gold

accumulates, which in this case, since you buy the same amount eact week, is

constant. The graph would be a horizontal straight line.

Question 2: Each week you buy one more gram than the week before, so the rate

goes up each week by 1 gram per week. You thus get a risingstraight line

because the increase in the rate is the same from one week to the next.

Question 3. Since half the previous amount will be half of a declining amount,

the rate will decrease while remaining positive, so the graph remains positive

as it decreases more and more slowly. The rate approaches but never reaches

zero.

STUDENT COMMENT: I feel like I am having trouble visualizing these graphs

because every time for the first one I picture an increasing straight line

INSTRUCTOR RESPONSE: The first graph depicts the amount of gold you have in

your back yard. The second depicts the rate at which the gold is accumulating,

which is related to, but certainly not the same as, the amount of gold.

For example, as long as gold is being added to the back yard, the amount will

be increasing (though not necessarily on a straight line). However if less and

less gold is being added every year, the rate will be decreasing (perhaps

along a straight line, perhaps not).

FREQUENT STUDENT RESPONSE

This is the same as the problem before it. No self-critique is required.

INSTRUCTOR RESPONSE

This question is very different that the preceding, and in a very significant

and important way. You should have

self-critiqued; you should go back and insert a self-critique on this very

important question and indicate your insertion by

preceding it with ####. The extra effort will be more than worth your

trouble.

These two problems go to the heart of the Fundamental Theorem of Calculus,

which is the heart of this course, and the extra effort will be well worth it

in the long run. The same is true of the last question in this document.

STUDENT COMMENT

Aha! Well you had me tricked. I apparently misread the question. Please don�t

do this on a test!

INSTRUCTOR RESPONSE

I don't usually try to trick people, and wasn't really trying to do so here,

but I was aware when writing these two problems that most students would be

tricked.

My real goal: The distinction between these two problems is key to

understanding what calculus is all about. I want to at least draw your

attention to it early in the course.

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Self-critique (if necessary):

ok

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Self-critique Rating: ok

``q007. If the depth of water in a container is given, in centimeters, by 100

- 2 t + .01 t^2, where t is clock time in seconds, then what are the depths at

clock times t = 30, t = 40 and t = 60? On the average is depth changing more

rapidly during the first time interval or the second?

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Your solution:

100 - 2 t + .01 t^2

100 - 2 (30) + .01 (30)^2

100 - 60 + .01 * 900

100 - 60 + 9

40 + 9

49

100 - 2 (40) + .01 (40)^2

100 - 80 + .01 * 1600

100 - 80 + 16

20 + 16

36

100 - 2 (60) + .01 (60)^2

100 - 120 + .01 * 3600

100 - 120 + 36

-20 + 36

16

First Interval = 49-36 = 13

Second Interval = 36-16 = 20

The depth is changing more during the second interval.

confidence rating #$&*: 3

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Given Solution:

`aAt t = 30 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 30 + .01 * 30^2 =

49.

At t = 40 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 40 + .01 * 40^2 = 36.

At t = 60 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 60 + .01 * 60^2 = 16.

49 cm - 36 cm = 13 cm change in 10 sec or 1.3 cm/s on the average.

36 cm - 16 cm = 20 cm change in 20 sec or 1.0 cm/s on the average.

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Self-critique (if necessary):

ok

@&

&#Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the parts of the given solution on which your solution didn't agree, and if necessary asking specific questions (to which I will respond).

&#

*@

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Self-critique Rating: ok

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Question: `q008. If the rate at which water descends in a container is given,

in cm/s, by 10 - .1 t, where t is clock time in seconds, then at what rate is

water descending when t = 10, and at what rate is it descending when t = 20?

How much would you therefore expect the water level to change during this 10-

second interval?

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Your solution:

at 0 sec:

10 - .1 t

10 - .1 (0)

10

at 10 sec:

10 - .1 t

10 - .1 (10)

10 - 1

9

at 20 sec:

10 - .1 t

10 - .1 (20)

10 - 2

8

Every 10 seconds, the rate of water decreases by 1 interval.

confidence rating #$&*: 3

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Given Solution:

`aAt t = 10 sec the rate function gives us 10 - .1 * 10 = 10 - 1 = 9, meaning

a rate of 9 cm / sec.

At t = 20 sec the rate function gives us 10 - .1 * 20 = 10 - 2 = 8, meaning a

rate of 8 cm / sec.

The rate never goes below 8 cm/s, so in 10 sec the change wouldn't be less

than 80 cm.

The rate never goes above 9 cm/s, so in 10 sec the change wouldn't be greater

than 90 cm.

Any answer that isn't between 80 cm and 90 cm doesn't fit the given

conditions..

The rate change is a linear function of t. Therefore the average rate is the

average of the two rates, or 8.5 cm/s.

The average of the rates is 8.5 cm/sec. In 10 sec that would imply a change of

85 cm.

STUDENT RESPONSES

The following, or some variation on them, are very common in student comments.

They are both very good questions. Because of the importance of the required

to answer this question correctly, the instructor will typically request for a

revision in response to either student response:

I don't understand how the answer isn't 1 cm/s. That's the difference

between 8 cm/s and 9 cm/s.

I don't understand how the answer isn't 8.5 cm/s. That's the average of

the 8 cm/s and the 9 cm/s.

INSTRUCTOR RESPONSE

A self-critique should include a full statement of what you do and do not

understand about the given solution. A phrase-by-phrase analysis of the

solution is not unreasonable (and would be a good idea on this very important

question), though it wouldn't be necessary in most situations.

An important part of any self-critique is a good question, and you have asked

one. However a self-critique should if possible go further. I'm asking that

you go back and insert a self-critique on this very important question and

indicate your insertion by preceding it with ####, before submitting it. The

extra effort will be more than worth your trouble.

This problem, along with questions 5 and 6 of this document, go to the heart

of the Fundamental Theorem of Calculus, which is the heart of this course, and

the extra effort will be well worth it in the long run.

You should review the instructions for self-critique, provided at the link

given at the beginning of this document.

STUDENT COMMENT

The question is worded very confusingly. I took a stab and answered correctly.

When answering, """"How much would you

therefore expect the water level to change during this 10-second interval?""""

It is hard to tell whether you are asking for

what is the expected change in rate during this interval and what is the

changing """"water level."""" But now, after looking at

it, with your comments, it is clearer that I should be looking for the later.

Thanks!

INSTRUCTOR RESPONSE

'Water level' is clearly not a rate. I don't think there's any ambiguity in

what's being asked in the stated question.

The intent is to draw the very important distinction between the rate at which

a quantity changes, and the change in the quantity.

It seems clear that as a result of this question you understand this and will

be more likely to make such distinctions in your subsequent work.

This distinction is at the heart of the calculus and its applications. It is

in fact the distinction between a derivative and an integral.

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Self-critique (if necessary):

ok

@&

&#This also requires a self-critique.

&#

*@

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Self-critique Rating: ok

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Question: `q009. Sketch the line segment connecting the points (2, -4) and (6,

4), and the line segment connecting the points (2, 4) and (6, 1). The first of

these lines if the graph of the function f(x), the second is the graph of the

function g(x). Both functions are defined on the interval 2 <= x <= 6.

Let h(x) be the function whose value at x is the product of the values of

these two functions. For example, when x = 2 the value of the first function

is -4 and the value of the second is 4, so when x = 2 the value of h(x) is -4

* 4 = -16.

Answer the following based just on the characteristics of the graphs you have

sketched. (e.g., you could answer the following questions by first finding

the formulas for f(x) and g(x), then combining them to get a formula for h(x);

that's a good skill but that is not the intent of the present set of

questions).

What is the value of h(x) when x = 6?

Is the value of h(x) ever greater than its value at x = 6?

What is your best description of the graph of h(x)?

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Your solution:

What is the value of h(x) when x = 6?

4 * 1 = 4

Is the value of h(x) ever greater than its value at x = 6?

f(x) = (2, -4) (3, -2) (4, 0) (5, 2) (6, 4)

g(x) = (2, 4) (3, 3.25) (4, 2.5) (5, 1.75) (6, 1)

h(x) = (2, -16) (3, -6.5) (4, 0) (5, 3.5) (6, 4)

On the interval 2 <= x <= 6, the greatest value will be at x=6. The answer to the equation is 'no'

What is your best description of the graph of h(x)?

It looks somewhat like a cubic equation

confidence rating #$&*: 3

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Question: `q010. A straight line segment connects the points (3,5) and (7,9),

while the points (3, 9) and (7, 5) are connected by a curve which decreases at

an increasing rate. From each of the four points a line segment is drawn

directly down to the x axis, so that the first line segment is the top of a

trapezoid and the second a similar to a trapezoid but with a curved 'top'.

Which trapezoid has the greater area?

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Your solution:

Area of trapezoid: 0.5*h(b1+b2)

the points (3,5) and (7,9)

0.5*h(b1+b2)

0.5*(7-3)(5+9)

0.5*(4)(14)

0.5*(4)(14)

0.5*56

28

the points (3,9) and (7,5)

0.5*h(b1+b2)

0.5*(7-3)(5+9)

0.5*(4)(14)

0.5*56

28

By finding the area of the trapezoid, the two trapezoids seem to have the same area. However, since the points (3, 9) and (7, 5) are connected by a curve which decreases at an increasing rate, then the trapezoid with points (3, 9) and (7, 5) has the greater area.

confidence rating #$&*: 3

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Question: `q011. Describe the graph of the position of a car vs. clock time,

given each of the following conditions:

The car coasts down a straight incline, gaining the same amount of speed

every second

The car coasts down a hill which gets steeper and steeper, gaining more

speed every second

The car coasts down a straight incline, but due to increasing air

resistance gaining less speed with every passing second

Describe the graph of the rate of change of the position of a car vs. clock

time, given each of the above conditions.

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Your solution:

The car coasts down a straight incline, gaining the same amount of speed

every second

negative linear line

The car coasts down a hill which gets steeper and steeper, gaining more

speed every second

decreases at an increasing rate

The car coasts down a straight incline, but due to increasing air

resistance gaining less speed with every passing second

decreases at a decreasing rate

confidence rating #$&*: 3

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Question: `q012. If at t = 100 seconds water is flowing out of a container at

the rate of 1.4 liters / second, and at t = 150 second the rate is 1.0 liters

/ second, then what is your best estimate of how much water flowed out during

the 50-second interval?

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Your solution:

Average: (1.4 liters / second + 1.0 liters / second) / 2 = 1.2 liters / second

1.2 liters / second * 50 seconds = 60 liters

confidence rating #$&*:3

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Self-critique Rating: 3"

&#This looks good. Let me know if you have any questions. &#