orientation steps

course Mth 158

Checking disks:Task: `q001. This is for Mth 158 students only.

Disks for your course should have been packaged with your textbook. You should follow the instructions given by the publishers to run your disks.

Please verify below that your disks work and that you have been able to access the material. If not, briefly describe the problem.

Your response (insert your response beginning in the next line):

I downloaded apple quicktime 7 but it still says that windows cannot open this file. I am able to see the chapter listings, but when I click on a number for a specific problem, it can’t show the tutorial.

Step 6

When you submit information via a form, the form will request your name, email access and access code.

As previously instructed, your access code should be copied from a conveniently located file, not typed in. This is because any error in entering your access code will almost certainly result in an invalid invalid access code, which can cause serious problems with the system.

The email address you include should be your VCCS email address.

Please indicate these requirements in your own words:

Your response (insert your response beginning in the next line):

Submitted info will include my name, email and access which should be copied instead of typed in to prevent errors

Task: `q002.

Other than the name and email address you enter when requested by the form, you should not include any other identifying information on the form. This will ensure that, even in the unlikely event that someone stumbles on your access page, there will be nothing on that page to identify you.

Please indicate this in your own words.

Your response (insert your response beginning in the next line):

No other info should be included on forms to ensure confidentiality.

Task: `q003.

If you wish to maintain your privacy, then after accessing your information you should delete the history of your browser and close the browser in order to maintain the security of your access page. In Internet Explorer you do this by clicking on Tools > Internet Options > General and choosing to Clear History and Delete Temporary Internet Files.

You should safeguard your code. However if despite your best efforts you have reason to believe your access code has been compromised you may email the instructor from your VCCS account and request a new access code.

Please indicate your understanding of this advice in the box below:

Your response (insert your response beginning in the next line):

Keep privacy by deleting history and email the instructor if I feel the need to get a new access code.

Task: `q004. Everything you submit should be submitted through a form. If your work is formatted (e.g., for a lab report, because it contains graphs, or for some other reason) you may also 'back it up' by emailing the formatted document, but the document should first be copied and pasted into the Submit Work form and submitted. If everything has been done according to instructions, it is seldom necessary for the instructor to look at your formatted file.

Your response (insert your response beginning in the next line):

All documents should be submitted through a form and a back up can be emailed in the case of graphs or such.

Task: `q005.

Keep the Thread:

When sending messages don't assume that your instructor remembers the last message you sent. If your present message is related to the previous message, or answers a question posed by the instructor in a previous message, be sure you don't remove all that information when you send your new message. Your instructor will have less time to spend on your work if it is often necessary to go back and research previous correspondence to figure out what you are talking about. It is your responsibility to give the instructor everything required for a useful response in one message.

For example if the instructor poses the question 'now what do you get when you divide $400 by $10 / hour', don't send back a message that reads

Dear Instructor,

the answer is 40

Your student.

with nothing more in the message. Include the question, include your original work on the problem, and state your answer in a format that makes it clear you understand what you are doing.

State this in your own words.

Your response (insert your response beginning in the next line):

If a response email is needed to answer a question asked by the instructor, send the original question, my work and my answer so that the instructor knows what I’m doing.

Task: `q006.

Make it Self-Contained: When submitting word-processed documents make your message self-contained, so that it isn't necessary for the reader to consult other documents, and be sure it can be read efficiently by scrolling down the page, with no need to scroll right or left.

Don't spread things over multiple documents (e.g., a word processed document with an explanation and a spreadsheet document with some tables and graphs) but copy and paste everything into one document so the reader doesn't have to keep flipping from one document to the other. Also if you choose to use a spreadsheet as your document be sure you leave ample room for the instructor to insert comments, and be sure the document does not require right-left scrolling. Note that word-processor documents with copied-and-pasted spreadsheet information are generally more flexible than spreadsheets because of the ease of inserting comments, but that if these guidelines are followed either is acceptable.

Explain your understanding of these requirements.

Your response (insert your response beginning in the next line):

Keep submitted work in one document to prevent having to consult multiple documents by the instructor.

Task: `q007.

Please understand that to make it possible to work individually with a large number of students, your instructor works with you mainly by responding to your assignments and questions. Your instructor will generally not initiate contact with you. It is your responsibility to initiate contact with the instructor by submitting assignments and questions, and you can generally expect a reply to be posted by the end of the day after it is sent, seven days a week.

If a reply is not posted within this time you should resubmit your work, but be sure to indicate at the beginning that the work has already been submitted.

Please restate this in your own words.

Your response (insert your response beginning in the next line):

Contact with the instructor should be limited to responses to questions, etc and will be in a timely fashion. If we don’t hear back by the end of the day, we should resubmit with an indicator that it has already been submitted.

Task: `q008.

You will normally use web forms (e.g., the Submit Work form) to communicate information. However some documents are formatted in such a way that the text-only format available on the form is not sufficient. In this case you should still submit a text version of the document using the form, but you may also send a backup copy using your VCCS student email.

Any email you send should contain access information in the precise format prescribed in the message you received with your Access Code. This syntax should be copied and pasted, as indicated in that message, rather than being typed in.

Any email should also contain your course (e.g., Mth 173 or Phy 121, etc). in its subject line.

Emails must be sent from your VCCS email account. Messages sent from other accounts may or may not receive responses, but it will not be possible for the instructor to search such messages, and information sent from other accounts risks being screened out by our network's SPAM filter.

Please indicate how you will include your course and access information when and if you send email messages:

Your response (insert your response beginning in the next line):

Our course should be in the subject line and access info should be copies and pasted, and should only be sent through VCCS email.

Task: `q009.

Any document you submit to the instructor must be a word-processed document. It is not possible for the instructor to efficiently edit comments into Excel documents, scanned documents or documents submitted on paper or by fax.

Tables and graphs from Excel and other programs can be copied and pasted into word-processed documents.

• Note that for most purposes graphs need to be described in words. the important skill is not the use of a computer program to construct a graph, but your understanding of the essential properties and the meaning of the graph.

The only exception to this policy is tests, which are completed on paper and in your handwriting.

Please summarize your understanding of this policy.

Your response (insert your response beginning in the next line):

Submitted work should be in word processed format. Graphs should when possible, described in words. Tests are the only exception to the word processing requirement.

Task: `q010.

It is usually much more efficient and effective for the instructor to work with you through forms and your Access Page than face-to-face.

The instructor enjoys working face-to-face with students, but it is simply not possible to accommodate the desire of a class full of asynchronous distance students for individual meetings.

For this reason the instructor's office hours are conducted by responding to information received through forms, and face-to-face meetings are normally restricted to matters that have already been addressed to the fullest possible extent via electronic communication (e.g., access page and forms) and have proven impossible to resolve efficiently by this means.

This policy makes it possible for the instructor to respond fully to your submitted work and to do so in a timely manner. It also ensures that all students, local as well as non-local, get equal attention from the instructor.

Please explain this briefly in your own words.

Your response (insert your response beginning in the next line):

Face to face meetings with the instructor are difficult and therefore restricted to those absolutely necessary that can’t be handled through written formats.

Task: `q011.

When you take a test you must submit the Test Taken Form (at the General Information page, just below the Submit Work Form) to alert the instructor to be on the lookout for your test. If a test has gone astray (rare but tests can be misrouted) we can more easily trace it if we know promptly that it's missing.

Please state this in your own words:

Your response (insert your response beginning in the next line):

Once a test is completed, a test taken form should be sent so the instructor can promptly track it down it if ends up in cyberspace.

Task: `q012. Zipped files are acumbersome to deal with and the instructor does not accept them. Be sure files are sent in their original format. Note that some email programs (most notable AOL's) zip files without telling the user. You should either override this automatic process, if possible, or send files one at a time to ensure that they don't get Zipped. Please explain this policy briefly in your own words.

Your response (insert your response beginning in the next line):

No zipped files are acceptable.

Task: `q013.

Save all communications from the instructor; you're responsible for doing so. You are also responsible for keeping track of what you have sent the instructor, who can better serve students if time isn't wasted rereading something you re-sent because you weren't sure you sent in the first place. Don't delete your records of sent or received items and if you're not sure whether you've sent something check your record of received items. Once your work appears at your Access site (explained when you get your access code), however, you may assume that it is safely stored in at least two places.

State this in your own words.

Your response (insert your response beginning in the next line):

Save all communication between you and the instructor, don’t delete items at least until it appears at the access site.

Step 7

Task: `q001.

If you were in a course that meets in a classroom you would be attending regularly. In an asychronous distance course, while you have the course materials and access to a great deal of instruction, you do not have the benefit of regular meetings, and it can be difficult to find the time to work on the course.

You cannot allow this course to become something you need to 'find time' to do. You need to schedule a regular time to work on this course, and you need to schedule a sufficient number of hours to do this work.

Though there are exceptions both ways, most people manage to establish a regular schedule are successful in these courses, and most people who fail to establish a regular schedule are not successful.

Please explain in your own words why it is important to establish a schedule for this course.

Your response (insert your response beginning in the next line):

In order to give the course a suitable amount of time to learn it, I need to set up a specific schedule to allow for the work.

Task: `q002.

Write your work out on paper. Don't try to do multi-step problems on your computer keyboard. It's quicker to write them out then transcribe your work on the keyboard, and the act of writing things down has a number of advantages. It allows you to organize your thoughts, to make multiple representations of the situation, and to save your work for reference. Writing, sketching, etc. also reinforce the learning process.

Make sketches to represent the things you are thinking about and try to organize your thoughts as you proceed.

You should in any case make notes for future reference as you work. Also, in some cases a single complex problem or situation will be broken down into a series of questions, and it will be important to maintain the thread of the problem, and keeping at least brief notes will allow you to do so.

Your response (insert your response beginning in the next line):

Use paper and pencil to take notes, work problems out and organize thoughts, then transfer info to the computer.

Task: `q003.

On anything you send the instructor, including but not limited to q_a_ assignments, queries and tests, sufficient documentation is required to allow the instructor to follow your thinking and the details of your solution.

An example of good documentation to the question 'How long will it take to make $400 at $10 per hour?':

'At $10 / hour it will take $400 / ($10 / hour) = 40 hours to make $400.'

A poor answer to the same question:

'4000'

This is a poor answer first because it's undocumented and second because it's wrong.

It would be fairly easy for the instructor to figure out where the 4000 came from--most likely you multiplied when you should have divided, though you may have just been really careless with your 0's--so it might be possible to help you see what you did wrong here. However this is usually not the case with undocumented answers on more complicated problems.

The more usual case is that your instructor has no clue about what you did wrong and no reasonable way to 'reverse-engineer' your solution and address your error.

On a test the bad thing about such an answer is that even if you thought correctly through several steps and made only one minor error in your arithmetic, you didn't document the process and there would be no way to give you any partial credit.

Note also that if a question can be answered with 'true' or 'false' it doesn't matter whether you put down the right answer or not, if all you put down is 'true' or 'false' it is impossible to tell whether you got the answer by a correct process or by a coin flip, and in this course credit is not give for coin flips.

As another example, if a test problem asks for the graph of an expression it is not sufficient to copy the output of your graphing calculator; unless the problem specifically tells you to use the graphing calculator you must document how the characteristics of the graph result from the given expression. Document your answers, show the instructor that you know why the answer is what it is, or you risk getting no credit for the question.

Explain why it's important for you to document your work.

Your response (insert your response beginning in the next line):

The instructor can only tell if you understand where your answer came from if you explain it in words. Don’t just give an answer or you risk not getting credit for it.

Task: `q004.

Don't waste your time misrepresenting what you know.

• Some students copy the homework of other students or receive inappropriate help on homework.

• Some students go through a q_a_ or query program and look at the answers, then essentially copy the answers on the next run.

The instructor notices this pattern but doesn't penalize it, and some students get the 10% or 15% of their grade that's based on homework and daily assignments in this manner. However students who use this strategy tend not to learn the material well and almost never succeed on the tests that make up the vast majority of their grade.

Please state these ideas in your own words.

Your response (insert your response beginning in the next line):

If you spend more time searching easy solutions to the homework assigned (like copying another student’s notes), it will look good at first but will bite you in the end when you don’t know what you are doing on tests which account for the majority of your grade.

Task: `q005.

Also it isn't strictly necessary to do all the homework and daily assignments, since test (and for physics students lab) grades are the dominant factors in your final grade. Some students do indeed succeed without submitting much work other than tests (and for physics courses labs). However for most students it is necessary to go through the process and learn the material by submitting the assignments and getting instructor feedback.

Please state this in your own words.

Your response (insert your response beginning in the next line):

You w ill benefit from turning in assignments by not only doing the practice but also from getting feedback from your instructor.

Task: `q006.

When documenting tests items you need to use the methods appropriate to your course. Just because you can get the right answer in one way or another does not mean that you are using a procedure on which you can build further understanding. It's not a matter of 'my way' vs. 'your way'. The structure of the subject dictates the things you need to understand. If you are taking a test on material which requires you to write and solve equations, for example, then using trial and error to arrive at even a correct solution is not valid and would not receive credit.

State this policy in your own words.

Your response (insert your response beginning in the next line):

Before taking tests, be sure to know the material and how to logically solve the questions rather than doing trial and error to find answers.

Task: `q007.

It is also not valid to justify a solution by copying a picture or a solution from a calculator (unless of course the problem specifies that the calculator is to be used in this manner). It is fine to use a calculator to do your arithmetic, but you must show the steps in solving and equation, and if the solution includes a graph you must explain the behavior of that graph rather than just copying calculator output. The key is that while a calculator can be very useful, operations like entering a function or an equation and copying output from a calculator is not a college-level skill. If the process is part of the course, you have to show the steps of the process. State this policy in your own words.

Your response (insert your response beginning in the next line):

All work needs to be explained to allow the instructor to see that you know how solutions were reached. A calculator won’t always be readily available and I need to know how to solve problems without using it as a source.

001. typewriter notation

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Question: `q001. Explain the difference between x - 2 / x + 4 and (x - 2) / (x + 4). Then evaluate each expression for x = 2.

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Your solution:

Using the order of operations, the first statement indicates that you perform the division portion first, the second statement has both of the operations in parenthesis done first, then the division. The first statement solution would be 2 – 2 divided by 2 + 4 and indicates that you divide 2 by 2 first. That would lead to the next step being 2 – 1 + 4 which has an answer of 5. The second statement would be solved with parenthesis first so it would be (2 – 2) divided by (2 + 4). The next step is 0 divided by 6 and the answer is 0 because 0 divided by anything is still zero.

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Given Solution:

`aThe order of operations dictates that grouped expressions must be evaluated first, that exponentiation must be done before multiplication or division, which must be done before addition or subtraction.

It makes a big difference whether you subtract the 2 from the x or divide the -2 by 4 first. If there are no parentheses you have to divide before you subtract. Substituting 2 for x we get

2 - 2 / 2 + 4

= 2 - 1 + 4 (do multiplications and divisions before additions and subtractions)

= 5 (add and subtract in indicated order)

If there are parentheses you evaluate the grouped expressions first:

(x - 2) / (x + 4) = (2 - 2) / ( 2 + 4 ) = 0 / 6 = 0.

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Self-critique (if necessary):

I got the same answer.

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Question: `q002. Explain the difference between 2 ^ x + 4 and 2 ^ (x + 4). Then evaluate each expression for x = 2.

Note that a ^ b means to raise a to the b power. This process is called exponentiation, and the ^ symbol is used on most calculators, and in most computer algebra systems, to represent exponentiation.

Your solution

The first statement means that you would do the power (for this problem to the second power) before adding 4 while the second statement has you adding the x (2) to the 4 and then raising the first number (2) to that power (6).

The first statement is 2 to the 2nd power which is 2 time 2 which is 4, then adding 4 giving the total 8.

The second statement is doing parenthesis first so 2 + 4 which is 6, then starting over and doing 2 raised to the 6th power—2 * 2 * 2 * 2 * 2 * 2 which is 64.

Confidence Assessment:sure of my answer

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Given Solution:

`a2 ^ x + 4 indicates that you are to raise 2 to the x power before adding the 4.

2 ^ (x + 4) indicates that you are to first evaluate x + 4, then raise 2 to this power.

If x = 2, then

2 ^ x + 4 = 2 ^ 2 + 4 = 2 * 2 + 4 = 4 + 4 = 8.

and

2 ^ (x + 4) = 2 ^ (2 + 4) = 2 ^ 6 = 2*2*2*2*2*2 = 64.

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Self-critique (if necessary):

I got the same

Self-critique Rating:

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Question: `q003. What is the numerator of the fraction in the expression x - 3 / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x? What is the denominator? What do you get when you evaluate the expression for x = 2?

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Your solution:

The numerator is 3 and the denominator is [(2x – 5)^2 * 3x + 1

If I replace x with 2, it becomes 2 minus (3 divided by [(2 times 2 – 5)squared times 3 times 2 + 1] and then subtract 2 and add (2 + 7 times 2)

2 – 3/7 – 2 + 14 which equals 13 4/7 (13 and 4 7ths)

Confidence Assessment: a little shaky about this one. I see why we have to practice this, it’s confusing seeing it with computer generated symbols

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Given Solution:

`aThe numerator is 3. x isn't part of the fraction. / indicates division, which must always precede subtraction. Only the 3 is divided by [ (2x-5)^2 * 3x + 1 ] and only [ (2x-5)^2 * 3x + 1 ] divides 3.

If we mean (x - 3) / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x we have to write it that way.

The preceding comments show that the denominator is [ (2x-5)^2 * 3x + 1 ]

Evaluating the expression for x = 2:

- 3 / [ (2 * 2 - 5)^2 * 3(2) + 1 ] - 2 + 7*2 =

2 - 3 / [ (4 - 5)^2 * 6 + 1 ] - 2 + 14 = evaluate in parenthese; do multiplications outside parentheses

2 - 3 / [ (-1)^2 * 6 + 1 ] -2 + 14 = add inside parentheses

2 - 3 / [ 1 * 6 + 1 ] - 2 + 14 = exponentiate in bracketed term;

2 - 3 / 7 - 2 + 14 = evaluate in brackets

13 4/7 or 95/7 or about 13.57 add and subtract in order.

The details of the calculation 2 - 3 / 7 - 2 + 14:

Since multiplication precedes addition or subtraction the 3/7 must be done first, making 3/7 a fraction. Changing the order of the terms we have

2 - 2 + 14 - 3 / 7 = 14 - 3/7 = 98/7 - 3/7 = 95/7.

COMMON STUDENT QUESTION: ok, I dont understand why x isnt part of the fraction? And I dont understand why only the brackets are divided by 3..why not the rest of the equation?

INSTRUCTOR RESPONSE: Different situations give us different algebraic expressions; the situation dictates the form of the expression.

If the above expression was was written otherwise it would be a completely different expression and most likely give you a different result when you substitute.

If we intended the numerator to be x - 3 then the expression would be written (x - 3) / [(2x-5)^2 * 3x + 1 ] - 2 + 7x, with the x - 3 grouped.

If we intended the numerator to be the entire expression after the / the expression would be written x - 3 / [(2x-5)^2 * 3x + 1 - 2 + 7x ].

STUDENT COMMENT: I wasn't sure if the numerator would be 3 or -3. or is the subtraction sign just that a sign in this case?

INSTRUCTOR RESPONSE: In this case you would regard the - sign as an operation to be performed between the value of x and the value of the fraction, rather than as part of the numerator. That is, you would regard x - 3 / [ (2x-5)^2 * 3x + 1 ] as a subtraction of the fraction 3 / [ (2x-5)^2 * 3x + 1 ] from the term x.

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Self-critique (if necessary):

This will require some extra thinking on my part.

Self-critique Rating:

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Question: `q004. Explain, step by step, how you evaluate the expression (x - 5) ^ 2x-1 + 3 / x-2 for x = 4.

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Your solution:

(4 – 5)^2 * 4 – 1 + ¾ - 2

-1^2 times 4 – 1 + ¾ - 2

1 times 4 – 1 + ¾ - 2

4 – 1 + ¾ 2

3 + ¾ - 2

3 – 2 + ¾

1 and 3/4

Confidence Assessment:

Fairly confident

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Given Solution:

`aWe get

(4-5)^2 * 4 - 1 + 3 / 4 - 2

= (-1)^2 * 4 - 1 + 3 / 4 - 2 evaluating the term in parentheses

= 1 * 4 - 1 + 3 / 4 - 2 exponentiating (2 is the exponent, which is applied to -1 rather than multiplying the 2 by 4

= 4 - 1 + 3/4 - 2 noting that 3/4 is a fraction and adding and subtracting in order we get

= 1 3/4 = 7 /4 (Note that we could group the expression as 4 - 1 - 2 + 3/4 = 1 + 3/4 = 1 3/4 = 7/4).

COMMON ERROR:

(4 - 5) ^ 2*4 - 1 + 3 / 4 - 2 =

-1 ^ 2*4 - 1 + 3 / 4-2 =

-1 ^ 8 -1 + 3 / 4 - 2.

INSTRUCTOR COMMENTS:

There are two errors here. In the second step you can't multiply 2 * 4 because you have (-1)^2, which must be done first. Exponentiation precedes multiplication.

Also it isn't quite correct to write -1^2*4 at the beginning of the second step. If you were supposed to multiply 2 * 4 the expression would be (-1)^(2 * 4).

Note also that the -1 needs to be grouped because the entire expression (-1) is taken to the power. -1^8 would be -1 because you would raise 1 to the power 8 before applying the - sign, which is effectively a multiplication by -1.

STUDENT QUESTION: if it's read (-1)^8 it would be 1 or would you apply the sign afterward even if it is grouped and it be a -1?

INSTRUCTOR RESPONSE: The 8th power won't occur in this problem, of course, but you ask a good question.

-1^8 would require raising 1 to the 8th power, then applying the negative sign, and the result would be -1.

(-1)^8 would be the 8th power of -1, which as you see would be 1.

STUDENT COMMENT: I think it would be easier to visualize what your trying to raise to the exponent if you actually put parenthesis around the 2, that part seems to get tricky on the computer.

INSTRUCTOR RESPONSE: The expression was intentionally written to be misleading and make the point that, to avoid ambiguity, order of operations apply strictly, no matter what the expression looks like.

Normally, for clarity, the parentheses would be included. They aren't necessary, but when helpful it's a good idea to include them. You can, of course, have too many parentheses in an expression, making it harder than necessary to sort out. In practice we try to strike a balance.

The original expression was

(x - 5) ^ 2x-1 + 3 / x-2

White spaces make no difference in how an expression is evaluated, but they can help show the structure; e.g.,

(x - 5)^2 * x - 1 + 3 / x -2

is a visual improvement over the original. The * between the 2 and the x is not strictly necessary, but is also helpful.

((((x - 5) ^ 2)) * x) - 1 + (3 / x) - 2

verges on having too many parentheses at the beginning; it does help clarify the 3 / x.

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Self-critique (if necessary):

I know I have to write it out first

Self-critique Rating: ok

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Question: `q005. At the link

http://www.vhcc.edu/dsmith/genInfo/introductory problems/typewriter_notation_examples_with_links.htm

(copy this path into the Address box of your Internet browser; alternatively use the path

http://vhmthphy.vhcc.edu/ > General Information > Startup and Orientation (either scroll to bottom of page or click on Links to Supplemental Sites) > typewriter notation examples

and you will find a page containing a number of additional exercises and/or examples of typewriter notation.Locate this site, click on a few of the links, and describe what you see there.

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Your solution:

These links show more examples of how the math is written in typewriter format and will be very helpful as I attempt to use the keyboard for my answers.

Confidence Assessment:

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Given Solution:

`aYou should see a brief set of instructions and over 30 numbered examples. If you click on the word Picture you will see the standard-notation format of the expression. The link entitled Examples and Pictures, located in the initial instructions, shows all the examples and pictures without requiring you to click on the links. There is also a file which includes explanations.

The instructions include a note indicating that Liberal Arts Mathematics students don't need a deep understanding of the notation, Mth 173-4 and University Physics students need a very good understanding,

while students in other courses should understand the notation and should understand the more basic simplifications.

There is also a link to a page with pictures only, to provide the opportunity to translated standard notation into typewriter notation.

end program

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Self-critique (if necessary):

Self-critique Rating:

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Question: `q006 Standard mathematics notation is easier to look at; it's easier to see the meaning of the expressions.

However it's very important to understand order of operations, and students do get used to this way of doing it.

You should of course write everything out in standard notation when you work it on paper.

It is likely that you will at some point use a computer algebra system, and when you do you will probably have to enter expressions using a keyboard, so it is well worth the trouble to get used to this notation.

Indicate your understanding of why it is important to understand this notation.

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Your solution:

At some point I will probably have to write out expressions and answer problems using a keyboard so it’s a good idea to get used to this type of math presentation.

002. Describing Graphs

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Question: `q001. You will frequently need to describe the graphs you have constructed in this course. This exercise is designed to get you used to some of the terminology we use to describe graphs. Please complete this exercise and email your work to the instructor. Note that you should do these graphs on paper without using a calculator. None of the arithmetic involved here should require a calculator, and you should not require the graphing capabilities of your calculator to answer these questions.

Problem 1. We make a table for y = 2x + 7 as follows: We construct two columns, and label the first column 'x' and the second 'y'. Put the numbers -3, -2, -1, -, 1, 2, 3 in the 'x' column. We substitute -3 into the expression and get y = 2(-3) + 7 = 1. We substitute -2 and get y = 2(-2) + 7 = 3. Substituting the remaining numbers we get y values 5, 7, 9, 11 and 13. These numbers go into the second column, each next to the x value from which it was obtained. We then graph these points on a set of x-y coordinate axes. Noting that these points lie on a straight line, we then construct the line through the points.

Now make a table for and graph the function y = 3x - 4.

Identify the intercepts of the graph, i.e., the points where the graph goes through the x and the y axes.

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Your solution:

The x intercept is 4/3 and the point if (4/3, 0) and the y intercept is -4 so the point is (0, -4)

The point where the graph goes through the y axis is at -4 and the point where it goes through the x axis is 4/3.

Confidence Assessment: good

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Given Solution:

`aThe graph goes through the x axis when y = 0 and through the y axis when x = 0.

The x-intercept is therefore when 0 = 3x - 4, so 4 = 3x and x = 4/3.

The y-intercept is when y = 3 * 0 - 4 = -4. Thus the x intercept is at (4/3, 0) and the y intercept is at (0, -4).

Your graph should confirm this.

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Self-critique (if necessary):

Self-critique Rating:

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Question: `q002. Does the steepness of the graph in the preceding exercise (of the function y = 3x - 4) change? If so describe how it changes.

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Your solution: the line is straight so the steepness doesn’t change.

Confidence Assessment:

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Given Solution:

`aThe graph forms a straight line with no change in steepness.

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Self-critique (if necessary):

Self-critique Rating:

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Question: `q003. What is the slope of the graph of the preceding two exercises (the function ia y = 3x - 4;slope is rise / run between two points of the graph)?

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Your solution:

The rise is 18 and the run is 6 which makes the slope 3.

Confidence Assessment: struggling with this

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Given Solution:

`aBetween any two points of the graph rise / run = 3.

For example, when x = 2 we have y = 3 * 2 - 4 = 2 and when x = 8 we have y = 3 * 8 - 4 = 20. Between these points the rise is 20 - 2 = 18 and the run is 8 - 2 = 6 so the slope is rise / run = 18 / 6 = 3.

Note that 3 is the coefficient of x in y = 3x - 4.

Note the following for reference in subsequent problems: The graph of this function is a straight line. The graph increases as we move from left to right. We therefore say that the graph is increasing, and that it is increasing at constant rate because the steepness of a straight line doesn't change.

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Self-critique (if necessary):

I need to investigate the rise and run more.

Self-critique Rating:

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Question: `q004. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = 0 and x = 3.

Would you say that the graph is increasing or decreasing?

Does the steepness of the graph change and if so, how?

Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?

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Your solution:

The graph is increasing at an increasing rate. The graphing points are (0,0), (1,1), (2,4) (3,9) showing an increased speed with the y coordinates.

Confidence Assessment: good

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Given Solution:

`aGraph points include (0,0), (1,1), (2,4) and (3,9). The y values are 0, 1, 4 and 9, which increase as we move from left to right.

The increases between these points are 1, 3 and 5, so the graph not only increases, it increases at an increasing rate

STUDENT QUESTION: I understand increasing...im just not sure at what rate...how do you determine increasing at an increasing rate or a constant rate?

INSTRUCTOR RESPONSE: Does the y value increase by the same amount, by a greater amount or by a lesser amount every time x increases by 1?

In this case the increases get greater and greater. So the graph increases, and at an increasing rate. *&*&.

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Self-critique (if necessary):

Self-critique Rating:

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Question: `q005. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = -3 and x = 0.

Would you say that the graph is increasing or decreasing?

Does the steepness of the graph change and if so, how?

Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?

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Your solution:

This graph is decreasing at a decreasing rate, coordinates are (-3,9) (-2,4) (-1,1) (0,0) that shows the y points moving slower..

Confidence Assessment:

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Given Solution:

`aFrom left to right the graph is decreasing (points (-3,9), (-2,4), (-1,1), (0,0) show y values 9, 4, 1, 0 as we move from left to right ). The magnitudes of the changes in x from 9 to 4 to 1 to 0 decrease, so the steepness is decreasing.

Thus the graph is decreasing, but more and more slowly. We therefore say that the graph is decreasing at a decreasing rate.

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Self-critique (if necessary):

Self-critique Rating:

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Question: `q006. Make a table of y vs. x for y = `sqrt(x). [note: `sqrt(x) means 'the square root of x']. Graph y = `sqrt(x) between x = 0 and x = 3.

Would you say that the graph is increasing or decreasing?

Does the steepness of the graph change and if so, how?

Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?

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Your solution:

This graph is increasing at an decreasing rate, points being (0,0) (1,1) (2,1.4142) (3,1.732).

Confidence Assessment:

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Given Solution:

`aIf you use x values 0, 1, 2, 3, 4 you will obtain graph points (0,0), (1,1), (2,1.414), (3. 1.732), (4,2). The y value changes by less and less for every succeeding x value. Thus the steepness of the graph is decreasing.

The graph would be increasing at a decreasing rate.

If the graph respresents the profile of a hill, the hill starts out very steep but gets easier and easier to climb. You are still climbing but you go up by less with each step, so the rate of increase is decreasing.

If your graph doesn't look like this then you probably are not using a consistent scale for at least one of the axes. If your graph isn't as desribed take another look at your plot and make a note in your response indicating any difficulties.

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Self-critique (if necessary): the explanation you gave that compares it to a hill is very helpful

Self-critique Rating:

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Question: `q007. Make a table of y vs. x for y = 5 * 2^(-x). Graph y = 5 * 2^(-x) between x = 0 and x = 3.

Would you say that the graph is increasing or decreasing?

Does the steepness of the graph change and if so, how?

Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?

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Your solution:

This graph is decreasing at a decreasing rate with points being (0,5/0) (1,5/2)(2,5/4)(3,5/8) . Since the y coordinates are getting less each time, the graph is decreasing more slowly.

Confidence Assessment:2

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Given Solution:

`a** From basic algebra recall that a^(-b) = 1 / (a^b).

So, for example:

2^-2 = 1 / (2^2) = 1/4, so 5 * 2^-2 = 5 * 1/4 = 5/4.

5* 2^-3 = 5 * (1 / 2^3) = 5 * 1/8 = 5/8. Etc.

The decimal equivalents of the values for x = 0 to x = 3 will be 5, 2.5, 1.25, .625. These values decrease, but by less and less each time.

The graph is therefore decreasing at a decreasing rate. **

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Self-critique (if necessary):

Self-critique Rating:

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Question: `q008. Suppose you stand still in front of a driveway. A car starts out next to you and moves away from you, traveling faster and faster.

If y represents the distance from you to the car and t represents the time in seconds since the car started out, would a graph of y vs. t be increasing or decreasing?

Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?

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Your solution:

The graph would be at an increasing rate and would be further each second so would increase at an increasing rate.

Confidence Assessment: 2

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Given Solution:

`a** The speed of the car increases so it goes further each second. On a graph of distance vs. clock time there would be a greater change in distance with each second, which would cause a greater slope with each subsequent second. The graph would therefore be increasing at an increasing rate. **

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Self-critique (if necessary):

Self-critique Rating:

003. PC1 questions

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Question: `q001 A straight line connects the points (3, 5) and (7, 17), while another straight line continues on from (7, 17) to the point (10, 29). Which line is steeper and on what basis to you claim your result?

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Your solution:

The second line is steeper, with more slope because there is the same rise (12) but less run (3). The rise and run for the first line is 12 and 4, giving a ratio of 3 while the second line has the ratio of 4.

Confidence Assessment: 2

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Given Solution:

`aThe point (3,5) has x coordinate 3 and y coordinate 5. The point (7, 17) has x coordinate 7 and y coordinate 17. To move from (3,5) to (7, 17) we must therefore move 4 units in the x direction and 12 units in the y direction.

Thus between (3,5) and (7,17) the rise is 12 and the run is 4, so the rise/run ratio is 12/4 = 3.

Between (7,10) and (10,29) the rise is also 12 but the run is only 3--same rise for less run, therefore more slope. The rise/run ratio here is 12/3 = 4.

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Self-critique (if necessary):

I need more practice on rise, run and slope so this will be easier to remember.

Self-critique Rating:2

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Question: `q002. The expression (x-2) * (2x+5) is zero when x = 2 and when x = -2.5. Without using a calculator verify this, and explain why these two values of x, and only these two values of x, can make the expression zero.

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Your solution:

This can be solved by finding x – 2 = 0 and 2x + 5 = 0 since these 2 parts are multiplied and anything times 0 is 0.

X – 2=0 x=0+2 so x=2

2x+5=0 2x=0-5 2x=-5 x=-5/2 or -2.5

Confidence Assessment:3

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Given Solution:

`aIf x = 2 then x-2 = 2 - 2 = 0, which makes the product (x -2) * (2x + 5) zero.

If x = -2.5 then 2x + 5 = 2 (-2.5) + 5 = -5 + 5 = 0.which makes the product (x -2) * (2x + 5) zero.

The only way to product (x-2)(2x+5) can be zero is if either (x -2) or (2x + 5) is zero.

Note that (x-2)(2x+5) can be expanded using the Distributive Law to get

x(2x+5) - 2(2x+5). Then again using the distributive law we get

2x^2 + 5x - 4x - 10 which simplifies to

2x^2 + x - 10.

However this doesn't help us find the x values which make the expression zero. We are better off to look at the factored form.

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Self-critique (if necessary):comfortable

Self-critique Rating:

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Question: `q003. For what x values will the expression (3x - 6) * (x + 4) * (x^2 - 4) be zero?

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Your solution:

X will be either 2 or -4. Since the 3 parenthesis are multiplied together, any of those could be equal to 0. 3x-6=0 3x=6 x=2, x+4=0 so x=-4 x^2-4=0, x^2=4 x=2 or -2

Confidence Assessment: 3

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Given Solution:

`aIn order for the expression to be zero we must have 3x-6 = 0 or x+4=0 or x^2-4=0.

3x-6 = 0 is rearranged to 3x = 6 then to x = 6 / 3 = 2. So when x=2, 3x-6 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero.

x+4 = 0 gives us x = -4. So when x=-4, x+4 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero.

x^2-4 = 0 is rearranged to x^2 = 4 which has solutions x = + - `sqrt(4) or + - 2. So when x=2 or when x = -2, x^2 - 4 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero.

We therefore see that (3x - 6) * (x + 4) * (x^2 - 4) = 0 when x = 2, or -4, or -2. These are the only values of x which can yield zero.**

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Self-critique (if necessary):

Self-critique Rating:

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Question: `q004. 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:

Although the height of the 2 are closer than the width of the 2 trapezoids just more than double, the difference in the width is more than ten times that so that leads to the conclusion that the second trapezoid has the larger area.

Confidence Assessment:2

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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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Question: `q005. Sketch graphs of y = x^2, y = 1/x and y = `sqrt(x) [note: `sqrt(x) means 'the square root of x'] for x > 0. We say that a graph increases if it gets higher as we move toward the right, and if a graph is increasing it has a positive slope. Explain which of the following descriptions is correct for each graph:

As we move from left to right the graph increases as its slope increases.

As we move from left to right the graph decreases as its slope increases.

As we move from left to right the graph increases as its slope decreases.

As we move from left to right the graph decreases as its slope decreases.

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Your solution:

The first graph increases at an increasing rate with a positive slope.

The second graph decreases at a slower rate and its slope decreases.

The third graph increases at a slower rate so it’s slope decreases.

Confidence Assessment:1

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Given Solution:

`aFor x = 1, 2, 3, 4:

The function y = x^2 takes values 1, 4, 9 and 16, increasing more and more for each unit increase in x. This graph therefore increases, as you say, but at an increasing rate.

The function y = 1/x takes values 1, 1/2, 1/3 and 1/4, with decimal equivalents 1, .5, .33..., and .25. These values are decreasing, but less and less each time. The decreasing values ensure that the slopes are negative. However, the more gradual the decrease the closer the slope is to zero. The slopes are therefore negative numbers which approach zero.

Negative numbers which approach zero are increasing. So the slopes are increasing, and we say that the graph decreases as the slope increases.

We could also say that the graph decreases but by less and less each time. So the graph is decreasing at a decreasing rate.

For y = `sqrt(x) we get approximate values 1, 1.414, 1.732 and 2. This graph increases but at a decreasing rate.

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Self-critique (if necessary):

I still get confused about the slope increase or decrease.

Self-critique Rating:

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Question: `q006. If the population of the frogs in your frog pond increased by 10% each month, starting with an initial population of 20 frogs, then how many frogs would you have at the end of each of the first three months (you can count fractional frogs, even if it doesn't appear to you to make sense)? Can you think of a strategy that would allow you to calculate the number of frogs after 300 months (according to this model, which probably wouldn't be valid for that long) without having to do at least 300 calculations?

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Your solution:

I would have 26.62 frogs in 3 months.

I know there’s a formula for this but can’t remember w/o looking at your answer…

Confidence Assessment:3 for answer, 1 for solution

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Given Solution:

`aAt the end of the first month, the number of frogs in the pond would be (20 * .1) + 20 = 22 frogs. At the end of the second month there would be (22 * .1) + 22 = 24.2 frogs while at the end of the third month there would be (24.2 * .1) + 24.2 = 26.62 frogs.

The key to extending the strategy is to notice that multiplying a number by .1 and adding it to the number is really the same as simply multiplying the number by 1.1. 10 * 1.1 = 22; 22 * 1.1 = 24.2; etc.. So after 300 months you will have multiplied by 1.1 a total of 300 times. This would give you 20 * 1.1^300, whatever that equals (a calculator will easily do the arithmetic).

A common error is to say that 300 months at 10% per month gives 3,000 percent, so there would be 30 * 20 = 600 frogs after 30 months. That doesn't work because the 10% increase is applied to a greater number of frogs each time. 3000% would just be applied to the initial number, so it doesn't give a big enough answer.

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Self-critique (if necessary):

Another learning item

&#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).

&#

Self-critique Rating:

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Question: `q007. Calculate 1/x for x = 1, .1, .01 and .001. Describe the pattern you obtain. Why do we say that the values of x are approaching zero? What numbers might we use for x to continue approaching zero? What happens to the values of 1/x as we continue to approach zero? What do you think the graph of y = 1/x vs. x looks for x values between 0 and 1?

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Your solution:

The value of x is getting smaller each time but will never reach zero. The x as we continue will just keep getting smaller infinitely. I would assume the graph will get steeper successively but never get on the line (reaching 0)

Confidence Assessment:2

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Given Solution:

`aIf x = .1, for example, 1 / x = 1 / .1 = 10 (note that .1 goes into 1 ten times, since we can count to 1 by .1, getting.1, .2, .3, .4, ... .9, 10. This makes it clear that it takes ten .1's to make 1.

So if x = .01, 1/x = 100 Ithink again of counting to 1, this time by .01). If x = .001 then 1/x = 1000, etc..

Note also that we cannot find a number which is equal to 1 / 0. Deceive why this is true, try counting to 1 by 0's. You can count as long as you want and you'll ever get anywhere.

The values of 1/x don't just increase, they increase without bound. If we think of x approaching 0 through the values .1, .01, .001, .0001, ..., there is no limit to how big the reciprocals 10, 100, 1000, 10000 etc. can become.

The graph becomes steeper and steeper as it approaches the y axis, continuing to do so without bound but never touching the y axis.

This is what it means to say that the y axis is a vertical asymptote for the graph .

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Self-critique (if necessary):

Self-critique Rating:

&#This also requires a self-critique.

&#

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Question: `q008. At clock time t the velocity of a certain automobile is v = 3 t + 9. At velocity v its energy of motion is E = 800 v^2. What is the energy of the automobile at clock time t = 5?

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Your solution:

3*5+9=24 so E=800 * 24^2. This is 46080

Confidence Assessment:2

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Given Solution:

`aFor t=5, v = 3 t + 9 = (3*5) + 9 = 24. Therefore E = 800 * 24^2 = 460800.

•

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Question: `q009. Continuing the preceding problem, can you give an expression for E in terms of t?

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Your solution:

The energy of motion is 800 times the velocity which is found by multiplying clock time by 3 and adding 9.

Confidence Assessment:2

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Given Solution:

`aSince v = 3 t + 9 the expression would be E = 800 v^2 = 800 ( 3t + 9) ^2. This is the only answer really required here.

For further reference, though, note that this expression could also be expanded by applying the Distributive Law:.

Since (3t + 9 ) ^ 2 = (3 t + 9 ) * ( 3 t + 9 ) = 3t ( 3t + 9 ) + 9 * (3 t + 9) = 9 t^2 + 27 t + 27 t + 81 = 9 t^2 + 54 t + 81, we get

E = 800 ( 9 t^2 + 54 t + 81) = 7200 t^2 + 43320 t + 64800 (check my multiplication because I did that in my head, which isn't always reliable).

001. Rates

Note that there are 10 questions in this assignment. The questions are of increasing difficulty--the first questions are fairly easy but later questions are very tricky. The main purposes of these exercises are to refine your thinking about rates, and to see how you process challenging information. Most students in most courses would not be expected to answer all these questions correctly; all that's required is that you do your best and follows the recommended procedures for answering and self-critiquing your work.

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Question: If you make $50 in 5 hr, then at what rate are you earning money?

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Your solution:

$50 divided by 5 = hourly rate which is $10 per hour.

Confidence Assessment:3

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Given Solution:

The rate at which you are earning money is the number of dollars per hour you are earning. You are earning money at the rate of 50 dollars / (5 hours) = 10 dollars / hour. It is very likely that you immediately came up with the $10 / hour because almosteveryone is familiar with the concept of the pay rate, the number of dollars per hour. Note carefully that the pay rate is found by dividing the quantity earned by the time required to earn it. Time rates in general are found by dividing an accumulated quantity by the time required to accumulate it.

You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.

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Self-critique (if necessary):

Self-critique Rating:

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Question: `q003.If you make $60,000 per year then how much do you make per month?

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Your solution:

$60,000 per year is divided into 12 months. 60,000/12=$5,000 per month.

Confidence Assessment: 3

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Given Solution:

Most people will very quickly see that we need to divide $60,000 by 12 months, giving us 60,000 dollars / (12 months) = 5000 dollars / month. Note that again we have found a time rate, dividing the accumulated quantity by the time required to accumulate it.

You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.

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Self-critique (if necessary):

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Question: `q004. Suppose that the $60,000 is made in a year by a small business. Would be more appropriate to say that the business makes $5000 per month, or that the business makes an average of $5000 per month?

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Your solution:

I would expect that the $5000 would be an average instead of flat amount.

Confidence Assessment:3

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Given Solution:

Small businesses do not usually make the same amount of money every month. The amount made depends on the demand for the services or commodities provided by the business, and there are often seasonal fluctuations in addition to other market fluctuations. It is almost certain that a small business making $60,000 per year will make more than $5000 in some months and less than $5000 in others. Therefore it is much more appropriate to say that the business makes and average of $5000 per month.

You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.

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Self-critique (if necessary):

Self-critique Rating:

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Question: `q005. If you travel 300 miles in 6 hours, at what average rate are you covering distance, and why do we say average rate instead of just plain rate?

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Your solution:

300 divided by 6 hours gives an average rate of 50 mph. This should be an average because speed will fluctuate.

Confidence Assessment: 3

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Given Solution:

The average rate is 50 miles per hour, or 50 miles / hour. This is obtained by dividing the accumulated quantity, the 300 miles, by the time required to accumulate it, obtaining ave rate = 300 miles / ( 6 hours) = 50 miles / hour. Note that the rate at which distance is covered is called speed. The car has an average speed of 50 miles/hour. We say 'average rate' in this case because it is almost certain that slight changes in pressure on the accelerator, traffic conditions and other factors ensure that the speed will sometimes be greater than 50 miles/hour and sometimes less than 50 miles/hour; the 50 miles/hour we obtain from the given information is clearly and overall average of the velocities.

You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.

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Self-critique (if necessary):

Self-critique Rating:

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Question: `q006. If you use 60 gallons of gasoline on a 1200 mile trip, then at what average rate are you using gasoline, with respect to miles traveled?

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Your solution:

60/1200 = .05. This is how much gas you are using per mile.

Confidence Assessment: 3

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Given Solution:

The rate of change of one quantity with respect to another is the change in the first quantity, divided by the change in the second. As in previous examples, we found the rate at which money was made with respect to time by dividing the amount of money made by the time required to make it.

By analogy, the rate at which we use fuel with respect to miles traveled is the change in the amount of fuel divided by the number of miles traveled. In this case we use 60 gallons of fuel in 1200 miles, so the average rate it 60 gal / (1200 miles) = .05 gallons / mile.

Note that this question didn't ask for miles per gallon. Miles per gallon is an appropriate and common calculation, but it measures the rate at which miles are covered with respect to the amount of fuel used. Be sure you see the difference.

Note that in this problem we again have here an example of a rate, but unlike previous instances this rate is not calculated with respect to time. This rate is calculated with respect to the amount of fuel used. We divide the accumulated quantity, in this case miles, by the amount of fuel required to cover t miles. Note that again we call the result of this problem an average rate because there are always at least subtle differences in driving conditions that result in more or fewer miles covered with a certain amount of fuel.

It's very important to understand the phrase 'with respect to'. Whether the calculation makes sense or not, it is defined by the order of the terms.

In this case gallons / mile tells you how many gallons you are burning, on the average, per mile. This concept is not as familiar as miles / gallon, but except for familiarity it's technically no more difficult.

You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as

Self-critique Rating:

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Question: `q007. The word 'average' generally connotes something like adding two quantities and dividing by 2, or adding several quantities and dividing by the number of quantities we added. Why is it that we are calculating average rates but we aren't adding anything?

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Your solution:

It has already been combined before the question is asked.

Confidence Assessment:3

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Given Solution:

The word 'average' in the context of the dollars / month, miles / gallon types of questions we have been answering was used because we expect that in different months different amounts were earned, or that over different parts of the trip the gas mileage might have varied, but that if we knew all the individual quantities (e.g., the dollars earned each month, the number of gallons used with each mile) and averaged them in the usual manner, we would get the .05 gallons / mile, or the $5000 / month. In a sense we have already added up all the dollars earned in each month, or the miles traveled on each gallon, and we have obtained the total $60,000 or 1200 miles. Thus when we divide by the number of months or the number of gallons, we are in fact calculating an average rate.

You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.

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Question: `q008. In a study of how lifting strength is influenced by various ways of training, a study group was divided into 2 subgroups of equally matched individuals. The first group did 10 pushups per day for a year and the second group did 50 pushups per day for year. At the end of the year to lifting strength of the first group averaged 147 pounds, while that of the second group averaged 162 pounds. At what average rate did lifting strength increase per daily pushup?

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Your solution:

Proportions would be used 10/147

An increase of 15 pounds of strength at the end of the year and a difference of 40 pushups per day. Other than that, I don’t know where to go.

Confidence Assessment: 0

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Given Solution:

The second group had 15 pounds more lifting strength as a result of doing 40 more daily pushups than the first. The desired rate is therefore 15 pounds / 40 pushups = .375 pounds / pushup.

You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.

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Self-critique (if necessary):

This type of question requires a lot of concentration

Self-critique Rating:

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Question: `q009. In another part of the study, participants all did 30 pushups per day, but one group did pushups with a 10-pound weight on their shoulders while the other used a 30-pound weight. At the end of the study, the first group had an average lifting strength of 171 pounds, while the second had an average lifting strength of 188 pounds. At what average rate did lifting strength increase with respect to the added shoulder weight?

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Your solution:

17 pounds improvement by using an extra 20 pounds of weights daily.

Confidence Assessment: 0

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Given Solution:

The difference in lifting strength was 17 pounds, as a result of a 20 pound difference in added weight. The average rate at which strength increases with respect added weight would therefore be 17 lifting pounds / (20 added pounds) = .85 lifting pounds / added pound. The strength advantage was .85 lifting pounds per pound of added weight, on the average.

You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.

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Self-critique (if necessary):still very confusing

Self-critique Rating:

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Question: `q010. During a race, a runner passes the 100-meter mark 12 seconds after the start and the 200-meter mark 22 seconds after the start. At what average rate was the runner covering distance between those two positions?

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Your solution:

He was covering 100 meters in 10 seconds or 10 meters per second.

Confidence Assessment: 2

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Given Solution:

The runner traveled 100 meters between the two positions, and required 10 seconds to do so. The average rate at which the runner was covering distance was therefore 100 meters / (10 seconds) = 10 meters / second. Again this is an average rate; at different positions in his stride the runner would clearly be traveling at slightly different speeds.

You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.

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Self-critique (if necessary):

I kept trying to figure out the catch with this one because I expected it’s difficulty to be like the previous 2.

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Question: `q011. During a race, a runner passes the 100-meter mark moving at 10 meters / second, and the 200-meter mark moving at 9 meters / second. What is your best estimate of how long it takes the runner to cover the 100 meter distance?

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Your solution:

I will average the first 20 meters to estimate of 9.5 seconds per 10 meter so he will cover the 100 meter distance in 95 seconds.

Confidence Assessment:2

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Given Solution:

At 10 meters/sec, the runner would require 10 seconds to travel 100 meters. However the runner seems to be slowing, and will therefore require more than 10 seconds to travel the 100 meters. We don't know what the runner's average speed is, we only know that it goes from 10 m/s to 9 m/s. The simplest estimate we could make would be that the average speed is the average of 10 m/s and 9 m/s, or (10 m/s + 9 m/s ) / 2 = 9.5 m/s. Taking this approximation as the average rate, the time required to travel 100 meters will be (100 meters) / (9.5 m/s) = 10.5 sec, approx.. Note that simply averaging the 10 m/s and the 9 m/s might not be the best way to approximate the average rate--for example we if we knew enough about the situation we might expect that this runner would maintain the 10 m/s for most of the remaining 100 meters, and simply tire during the last few seconds. However we were not given this information, and we don't add extraneous assumptions without good cause. So the approximation we used here is pretty close to the best we can do with the given information.

You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.

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Self-critique (if necessary):

Self-critique Rating:

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Question: `q012. We just averaged two quantities, adding them and dividing by 2, to find an average rate. We didn't do that before. Why we do it now?

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Your solution:

We did it now because it fit with the situation since we know that car speed and runner speed does fluctuate.

Confidence Assessment: 2

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Given Solution:

In previous examples the quantities weren't rates. We were given the amount of change of some accumulating quantity, and the change in time or in some other quantity on which the first was dependent (e.g., dollars and months, miles and gallons). Here we are given 2 rates, 10 m/s and 9 m/s, in a situation where we need an average rate in order to answer a question. Within this context, averaging the 2 rates was an appropriate tactic.

You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.

STUDENT QUESTION:

I thought the change of an accumulating quantity was the rate?

INSTRUCTOR RESPONSE:

Quick response: The rate is not just the change in the accumulating quantity; if we're talking about a 'time rate' it's the change in the accumulating quantity divided by the time interval (or in calculus the limiting value of this ratio as the time interval approaches zero).

More detailed response: If quantity A changes with respect to quantity B, then the average rate of change of A with respect to B (i.e., change in A / change in B) is 'the rate'. If the B quantity is clock time, then 'the rate' tells you 'how fast' the A quantity accumulates. However the rate is not just the change in the quantity A (i.e., the change in the accumulating quantity), but change in A / change in B.

For students having had at least a semester of calculus at some level: Of course the above generalizes into the definition of the derivative. y ' (x) is the instantaneous rate at which the y quantity changes with respect to x. y ' (x) is the rate at which y accumulates with respect to x.

001. Areas

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Question: `q001. There are 11 questions and 7 summary questions in this assignment.

What is the area of a rectangle whose dimensions are 4 m by 3 meters.

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Your solution:

Area of a rectangle is l * w so the area of this is 4 * 3 = 12m^2

Confidence Assessment: 3

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Given Solution:

`aA 4 m by 3 m rectangle can be divided into 3 rows of 4 squares, each 1 meter on a side. This makes 3 * 4 = 12 such squares. Each 1 meter square has an area of 1 square meter, or 1 m^2. The total area of the rectangle is therefore 12 square meters, or 12 m^2.

The formula for the area of a rectangle is A = L * W, where L is the length and W the width of the rectangle. Applying this formula to the present problem we obtain area A = L * W = 4 m * 3 m = (4 * 3) ( m * m ) = 12 m^2.

Note the use of the unit m, standing for meters, in the entire calculation. Note that m * m = m^2.

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Question: `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?

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Your solution:

A of triangle is bh/2 so the area of this is 4.0 * 3.0/2=6 m^2

Confidence Assessment:

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Given Solution:

`aA right triangle can be joined along its hypotenuse with another identical right triangle to form a rectangle. In this case the rectangle would have dimensions 4.0 meters by 3.0 meters, and would be divided by any diagonal into two identical right triangles with legs of 4.0 meters and 3.0 meters.

The rectangle will have area A = L * W = 4.0 m * 3.0 m = 12 m^2, as explained in the preceding problem. Each of the two right triangles, since they are identical, will therefore have half this area, or 1/2 * 12 m^2 = 6.0 m^2.

The formula for the area of a right triangle with base b and altitude h is A = 1/2 * b * h.

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Question: `q003. What is the area of a parallelogram whose base is 5.0 meters and whose altitude is 2.0 meters?

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Your solution:

A of parallelogram is base times height so the area of this is 5.0 meters times 2.0 meters which equals 10 m^2

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Given Solution:

`aA parallelogram is easily rearranged into a rectangle by 'cutting off' the protruding end, turning that portion upside down and joining it to the other end. Hopefully you are familiar with this construction. In any case the resulting rectangle has sides equal to the base and the altitude so its area is A = b * h.

The present rectangle has area A = 5.0 m * 2.0 m = 10 m^2.

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Question: `q004. What is the area of a triangle whose base is 5.0 cm and whose altitude is 2.0 cm?

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Your solution:

Area of triangle is base times altitude divided by 2 so this area is 5.0 * 2.0/2=5 cm^2

Confidence Assessment: 3

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Given Solution:

`aIt is possible to join any triangle with an identical copy of itself to construct a parallelogram whose base and altitude are equal to the base and altitude of the triangle. The area of the parallelogram is A = b * h, so the area of each of the two identical triangles formed by 'cutting' the parallelogram about the approriate diagonal is A = 1/2 * b * h. The area of the present triangle is therefore A = 1/2 * 5.0 cm * 2.0 cm = 1/2 * 10 cm^2 = 5.0 cm^2.

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Question: `q005. What is the area of a trapezoid with a width of 4.0 km and average altitude of 5.0 km?

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Your solution:

Area of trapezoid is width time altitude so this is 4.0*5.0=20km^2

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Given Solution:

`aAny trapezoid can be reconstructed to form a rectangle whose width is equal to that of the trapezoid and whose altitude is equal to the average of the two altitudes of the trapezoid. The area of the rectangle, and therefore the trapezoid, is therefore A = base * average altitude. In the present case this area is A = 4.0 km * 5.0 km = 20 km^2.

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Question: `q006. What is the area of a trapezoid whose width is 4 cm in whose altitudes are 3.0 cm and 8.0 cm?

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Your solution:

Area of this is 4 *[ (3+8)/2] which is 4*5.5=22cm^2

Confidence Assessment:

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Given Solution:

`aThe area is equal to the product of the width and the average altitude. Average altitude is (3 cm + 8 cm) / 2 = 5.5 cm so the area of the trapezoid is A = 4 cm * 5.5 cm = 22 cm^2.

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Question: `q007. What is the area of a circle whose radius is 3.00 cm?

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Your solution:

A of a circle is pi times radius^2 so this is 3.14*3^2 which equals 28.26cm^2

Confidence Assessment: 3

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Given Solution:

`aThe area of a circle is A = pi * r^2, where r is the radius. Thus

A = pi * (3 cm)^2 = 9 pi cm^2.

Note that the units are cm^2, since the cm unit is part r, which is squared.

The expression 9 pi cm^2 is exact. Any decimal equivalent is an approximation. Using the 3-significant-figure approximation pi = 3.14 we find that the approximate area is A = 9 pi cm^2 = 9 * 3.14 cm^2 = 28.26 cm^2, which we round to 28.3 cm^2 to match the number of significant figures in the given radius.

Be careful not to confuse the formula A = pi r^2, which gives area in square units, with the formula C = 2 pi r for the circumference. The latter gives a result which is in units of radius, rather than square units. Area is measured in square units; if you get an answer which is not in square units this tips you off to the fact that you've made an error somewhere.

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Question: `q008. What is the circumference of a circle whose radius is exactly 3 cm?

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Your solution:

C=2pi r so the circumference is 2*3.14*3=18.84 cm.

Confidence Assessment: 3

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Given Solution:

`aThe circumference of this circle is

C = 2 pi r = 2 pi * 3 cm = 6 pi cm.

This is the exact area. An approximation to 3 significant figures is 6 * 3.14 cm = 18.84 cm.

Note that circumference is measured in the same units as radius, in this case cm, and not in cm^2. If your calculation gives you cm^2 then you know you've done something wrong.

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Question: `q009. What is the area of a circle whose diameter is exactly 12 meters?

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Your solution:

Area of circle is pi r^2 so the diameter has to be divided by 2 in order to have the radius. After that the equation will be A=3.14*6^2

This equals 3.14 * 36 =113.04m^2

onfidence Assessment: 3

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Given Solution:

`aThe area of a circle is A = pi r^2, where r is the radius. The radius of this circle is half the 12 m diameter, or 6 m. So the area is

A = pi ( 6 m )^2 = 36 pi m^2.

This result can be approximated to any desired accuracy by using a sufficient number of significant figures in our approximation of pi. For example using the 5-significant-figure approximation pi = 3.1416 we obtain A = 36 m^2 * 3.1416 = 113.09 m^2.

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Question: `q010. What is the area of a circle whose circumference is 14 `pi meters?

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Your solution:

Area needs to know the radius of a circle so if the circumference is given, the first step is to solve C=2pi r

14=2pi r

R=7

So A=pi * 7^2 and 49*pi m^2

Confidence Assessment: a little shaky on this one

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Given Solution:

`aWe know that A = pi r^2. We can find the area if we know the radius r. We therefore attempt to use the given information to find r.

We know that circumference and radius are related by C = 2 pi r. Solving for r we obtain r = C / (2 pi). In this case we find that

r = 14 pi m / (2 pi) = (14/2) * (pi/pi) m = 7 * 1 m = 7 m.

We use this to find the area

A = pi * (7 m)^2 = 49 pi m^2.

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Question: `q011. What is the radius of circle whose area is 78 square meters?

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Your solution:

A=pi r^2 so if area is 78, it means 78=pi f^2

idence Assessment: you kind of lose me with not using 3.14 as pi

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Given Solution:

`aKnowing that A = pi r^2 we solve for r. We first divide both sides by pi to obtain A / pi = r^2. We then reverse the sides and take the square root of both sides, obtaining r = sqrt( A / pi ).

Note that strictly speaking the solution to r^2 = A / pi is r = +-sqrt( A / pi ), meaning + sqrt( A / pi) or - sqrt(A / pi). However knowing that r and A are both positive quantities, we can reject the negative solution.

Now we substitute A = 78 m^2 to obtain

r = sqrt( 78 m^2 / pi) = sqrt(78 / pi) m.{}

Approximating this quantity to 2 significant figures we obtain r = 5.0 m.

Self-critique Rating:

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Question: `q012. Summary Question 1: How do we visualize the area of a rectangle?

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Your solution:

It is the interior of the figure and can be divided into blocks of even sizes

Confidence Assessment: 2

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Given Solution:

`aWe visualize the rectangle being covered by rows of 1-unit squares. We multiply the number of squares in a row by the number of rows. So the area is A = L * W.

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Question: `q013. Summary Question 2: How do we visualize the area of a right triangle?

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Your solution:

We visualize it in a rectangle in rows and then cut the rectangle in half to make it the triangle.

fidence Assessment: 3

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Given Solution:

`aWe visualize two identical right triangles being joined along their common hypotenuse to form a rectangle whose length is equal to the base of the triangle and whose width is equal to the altitude of the triangle. The area of the rectangle is b * h, so the area of each triangle is 1/2 * b * h.

Self-critique Rating:

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Question: `q014. Summary Question 3: How do we calculate the area of a parallelogram?

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Your solution:

We multiply the base times the height as if the parallelogram was straightened into a rectangle.

Confidence Assessment: 3

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Given Solution:

`aThe area of a parallelogram is equal to the product of its base and its altitude. The altitude is measured perpendicular to the base.

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Question: `q015. Summary Question 4: How do we calculate the area of a trapezoid?

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Your solution:

Trapezoid heights are averaged together, then multiplied by base for the area.

Confidence Assessment: 2

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Given Solution:

`aWe think of the trapezoid being oriented so that its two parallel sides are vertical, and we multiply the average altitude by the width.

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Question: `q016. Summary Question 5: How do we calculate the area of a circle?

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Your solution:

Area of a circle is found by multiplying pi (3.14 is generally used) times the radius squared.

Confidence Assessment: 3

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Given Solution:

`aWe use the formula A = pi r^2, where r is the radius of the circle.

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Question: `q017. Summary Question 6: How do we calculate the circumference of a circle? How can we easily avoid confusing this formula with that for the area of the circle?

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Your solution:

Circumference of a circle is either diameter times pi or radius times 2 times pi.

Confidence Assessment:

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Given Solution:

`aWe use the formula C = 2 pi r. The formula for the area involves r^2, which will give us squared units of the radius. Circumference is not measured in squared units.

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Question: `q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.

I pictured each drawing and put each formula on the paper to see which I needed to use.

002. Volumes

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Question: `q001. There are 9 questions and 4 summary questions in this assignment.

What is the volume of a rectangular solid whose dimensions are exactly 3 cm by 5 cm by 7 cm?

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Your solution:

Volume is base area times height so this would be 3*5*7=105 cm^3

Confidence Assessment:3

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Given Solution:

`aIf we orient this object so that its 3 cm dimension is its 'height', then it will be 'resting' on a rectangular base whose dimension are 5 cm by 7 cm. This base can be divided into 5 rows each consisting of 7 squares, each 1 meter by 1 meter. There will therefore be 5 * 7 = 35 such squares, showing us that the area of the base is 35 m^2.

Above each of these base squares the object rises to a distance of 3 meters, forming a small rectangular tower. Each such tower can be divided into 3 cubical blocks, each having dimension 1 meter by 1 meter by 1 meter. The volume of each 1-meter cube is 1 m * 1 m * 1 m = 1 m^3, also expressed as 1 cubic meter. So each small 'tower' has volume 3 m^3.

The object can be divided into 35 such 'towers'. So the total volume is 35 * 3 m^3 = 105 m^3.

This construction shows us why the volume of a rectangular solid is equal to the area of the base (in this example the 35 m^2 of the base) and the altitude (in this case 3 meters). The volume of any rectangular solid is therefore

V = A * h,

where A is the area of the base and h the altitude.

This is sometimes expressed as V = L * W * h, where L and W are the length and width of the base. However the relationship V = A * h applies to a much broader class of objects than just rectangular solids, and V = A * h is a more powerful idea than V = L * W * h. Remember both, but remember also that V = A * h is the more imp

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Question: `q002. What is the volume of a rectangular solid whose base area is 48 square meters and whose altitude is 2 meters?

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Your solution:

Same as above, formula is Base area times altitude so 48*2=96m^3

Confidence Assessment:3

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Given Solution:

`aUsing the idea that V = A * h we find that the volume of this solid is

V = A * h = 48 m^2 * 2 m = 96 m^3.

Note that m * m^2 means m * (m * m) = m * m * m = m^2.

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Question: `q003. What is the volume of a uniform cylinder whose base area is 20 square meters and whose altitude is 40 meters?

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Your solution:

Cylinder volume is A*2 h so this would be 20m^2 times 40 =800m6#

Confidence Assessment:2

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Given Solution:

`aV = A * h applies to uniform cylinders as well as to rectangular solids. We are given the altitude h and the base area A so we conclude that

V = A * h = 20 m^2 * 40 m = 800 m^3.

The relationship V = A * h applies to any solid object whose cross-sectional area A is constant. This is the case for uniform cylinders and uniform prisms.

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Question: `q004. What is the volume of a uniform cylinder whose base has radius 5 cm and whose altitude is 30 cm?

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Your solution:

V=pi r^2 h which is pi * 5^2 * 30

=pi 25*30

=750 pi cm^3

onfidence Assessment:3

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Given Solution:

`aThe cylinder is uniform, which means that its cross-sectional area is constant. So the relationship V = A * h applies.

The cross-sectional area A is the area of a circle of radius 5 cm, so we see that A = pi r^2 = pi ( 5 cm)^2 = 25 pi cm^2.

Since the altitude is 30 cm the volume is therefore

V = A * h = 25 pi cm^2 * 30 cm = 750 pi cm^3.

Note that the common formula for the volume of a uniform cylinder is V = pi r^2 h. However this is just an instance of the formula V = A * h, since the cross-sectional area A of the uniform cylinder is pi r^2. Rather than having to carry around the formula V = pi r^2 h, it's more efficient to remember V = A * h and to apply the well-known formula A = pi r^2 for the area of a circle.

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Question: `q005. Estimate the dimensions of a metal can containing food. What is its volume, as indicated by your estimates?

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Your solution:

A can is 4 inches in diameter and 8 inches in height.

This would equal V=pi 4^2*8 or V=pi 128 in^3 or about 401.92 in^3 if I used 3.14 for pi

Confidence Assessment:3

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Given Solution:

`aPeople will commonly estimate the dimensions of a can of food in centimeters or in inches, though other units of measure are possible (e.g., millimeters, feet, meters, miles, km). Different cans have different dimensions, and your estimate will depend a lot on what can you are using.

A typical can might have a circular cross-section with diameter 3 inches and altitude 5 inches. This can would have volume V = A * h, where A is the area of the cross-section. The diameter of the cross-section is 3 inches so its radius will be 3/2 in.. The cross-sectional area is therefore A = pi r^2 = pi * (3/2 in)^2 = 9 pi / 4 in^2 and its volume is

V = A * h = (9 pi / 4) in^2 * 5 in = 45 pi / 4 in^3.

Approximating, this comes out to around 35 in^3.

Another can around the same size might have diameter 8 cm and height 14 cm, giving it cross-sectional area A = pi ( 4 cm)^2 = 16 pi cm^2 and volume V = A * h = 16 pi cm^2 * 14 cm = 224 pi cm^2.

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Question: `q006. What is the volume of a pyramid whose base area is 50 square cm and whose altitude is 60 cm?

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Your solution:

V=base area times height and then divide by 3.

V=50*60/3

=3000/3

=1000 cm^3

Confidence Assessment:3

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Given Solution:

`aWe can't use the V = A * h idea for a pyramid because the thing doesn't have a constant cross-sectional area--from base to apex the cross-sections get smaller and smaller. It turns out that there is a way to cut up and reassemble a pyramid to show that its volume is exactly 1/3 that of a rectangular solid with base area A and altitude h. Think of putting the pyramid in a box having the same altitude as the pyramid, with the base of the pyramid just covering the bottom of the box. The apex (the point) of the pyramid will just touch the top of the box. The pyramid occupies exactly 1/3 the volume of that box.

So the volume of the pyramid is V = 1/3 * A * h. The base area A is 30 cm^2 and the altitude is 60 cm so we have

V = 1/3 * 50 cm^2 * 60 cm = 1000 cm^3

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Question: `q007. What is the volume of a cone whose base area is 20 square meters and whose altitude is 9 meters?

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Your solution:

Cone=base area * altitude divided by 3

=20 * 9/3

=180/3

=60 m^3

onfidence Assessment:3

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Given Solution:

`aJust as the volume of a pyramid is 1/3 the volume of the 'box' that contains it, the volume of a cone is 1/3 the volume of the cylinder that contains it. Specifically, the cylinder that contains the cone has the base of the cone as its base and matches the altitude of the cone. So the volume of the cone is 1/3 A * h, where A is the area of the base and h is the altitude of the cone.

In this case the base area and altitude are given, so the volume of the cone is

V = 1/3 A * h = 1/3 * 20 m^2 * 9 m = 60 m^3.

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Question: `q008. What is a volume of a sphere whose radius is 4 meters?

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Your solution:

Vsphere=4/3 pi r^3

=4/3 pi * 4^3

=4/3 pi * 64

=pi 256/3 m^3

Confidence Assessment:3

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Given Solution:

`aThe volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere. In this case r = 4 m so

V = 4/3 pi * (4 m)^3 = 4/3 pi * 4^3 m^3 = 256/3 pi m^3.

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Question: `q009. What is the volume of a planet whose diameter is 14,000 km?

Your solution

V=4/3pi*7,000^3

Confidence Assessment: I know the formula but calculator isn’t that high

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Given Solution:

`aThe planet is presumably a sphere, so to the extent that this is so the volume of this planet is V = 4/3 pi r^3, where r is the radius of the planet. The diameter of the planet is 14,000 km so the radius is half this, or 7,000 km. It follows that the volume of the planet is

V = 4/3 pi r^3 = 4/3 pi * (7,000 km)^3 = 4/3 pi * 343,000,000,000 km^3 = 1,372,000,000,000 / 3 * pi km^3.

This result can be approximated to an appropriate number of significant figures.

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Question: `q010. Summary Question 1: What basic principle do we apply to find the volume of a uniform cylinder of known dimensions?

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Your solution:

The volume is determined by pi r^2 h when dimensions are known

Confidence Assessment:2

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Given Solution:

`aThe principle is that when the cross-section of an object is constant, its volume is V = A * h, where A is the cross-sectional area and h the altitude. Altitude is measure perpendicular to the cross-section.

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Question: `q011. Summary Question 2: What basic principle do we apply to find the volume of a pyramid or a cone?

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Your solution:

Volume of pyramid or cone uses area and height, and then divides by 3.

Confidence Assessment:3

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Given Solution:

`aThe volumes of these solids are each 1/3 the volume of the enclosing figure. Each volume can be expressed as V = 1/3 A * h, where A is the area of the base and h the altitude as measured perpendicular to the base.

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Question: `q012. Summary Question 3: What is the formula for the volume of a sphere?

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Your solution:

A sphere uses the radius and pi, cubing radius and then multiplying all by 4/3

Confidence Assessment:2

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Given Solution:

`aThe volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere.

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Question: `q013. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.

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Self-critique (if necessary):

In order to answer these, I had to have the formulas written down and accessible.

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