course phy201 ЦOᆮ€èªÍ¸¥éªá¯qøíC®†cÝassignment #001
......!!!!!!!!...................................
21:24:32 `q001. Explain the difference between x - 2 / x + 4 and (x - 2) / (x + 4). The evaluate each expression for x = 2.
......!!!!!!!!...................................
RESPONSE --> In this equation because of the grouping the differences would be with x=2 that the first equation x - 2 / x + 4 would be equal to 5 and equation2 (x - 2) / (x + 4) would be equal to 0. Thatmakes grouping very important. confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:25:19 `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.
......!!!!!!!!...................................
RESPONSE --> I agree with the response and conclude mine to be a briefer explanation. confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:27:09 2 ^ 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.
......!!!!!!!!...................................
RESPONSE --> This aagian shows that grouping matter in the answer. self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:31:26 `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?
......!!!!!!!!...................................
RESPONSE --> The numerator for the farction 3 / [ (2x-5)^2* 3x + 1 ] is 3 where as the denominator is [ (2x-5)^2 * 3x + 1 ] and when x =2 would be 95/7 or 13 4/7. confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:32:02 The 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 ].
......!!!!!!!!...................................
RESPONSE --> I was right with my answer in 95/7. self critique assessment: 2
.................................................
......!!!!!!!!...................................
21:35:29 `q004. Explain, step by step, how you evaluate the expression (x - 5) ^ 2x-1 + 3 / x-2 for x = 4.
......!!!!!!!!...................................
RESPONSE --> (x - 5) ^ 2x-1 + 3 / x-2 for x = 4. (4 - 5) ^ 2(4)-1 + 3 / 4-2 (-1 ^2)4 -1 +3/4 -2 4 -3 +3/4 1+3/4 1 3/4 7/4 confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:36:05 We get (4-5)^2 * 4 - 1 + 3 / 1 - 4 = (-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.
......!!!!!!!!...................................
RESPONSE --> I recieved the correct answer on this problem. self critique assessment:
.................................................
......!!!!!!!!...................................
21:37:04 *&*& Standard mathematics notation is easier to see. On the other hand 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 have to enter expressions through a typewriter, so it is well worth the trouble to get used to this notation. Indicate your understanding of the necessity to understand this notation.
......!!!!!!!!...................................
RESPONSE --> I understand that standard math notation though hard to put in computer is easier to work. self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:39:24 `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.
......!!!!!!!!...................................
RESPONSE --> 2 equalviant equations one in standard and one in typewriter format. confidence assessment: 2
.................................................
......!!!!!!!!...................................
21:40:01 You 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,
......!!!!!!!!...................................
RESPONSE --> I understand the differnet formats self critique assessment: 2
.................................................
......!!!!!!!!...................................
21:40:30 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.
......!!!!!!!!...................................
RESPONSE --> I understad the different notations self critique assessment: 2
.................................................
«U©€ÕÏ^€¶ì¾€ôLÉ߳ɀ{Þµqw¦ÆÁÑL—® assignment #001 001. typewriter notation qa initial problems 08-27-2008 Áû¶ä©ÞëˆÚ¹Œ¥ø‰‹°î£ª˜©å΀çü assignment #001 001. typewriter notation qa initial problems 08-27-2008
......!!!!!!!!...................................
21:42:40 `q001. Explain the difference between x - 2 / x + 4 and (x - 2) / (x + 4). The evaluate each expression for x = 2.
......!!!!!!!!...................................
RESPONSE --> The first is 5 the second is 0 confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:43:02 `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.
......!!!!!!!!...................................
RESPONSE --> I understand confidence assessment:
.................................................
......!!!!!!!!...................................
21:43:11 2 ^ 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.
......!!!!!!!!...................................
RESPONSE --> i understand self critique assessment:
.................................................
......!!!!!!!!...................................
21:44:15 `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?
......!!!!!!!!...................................
RESPONSE --> expression x - 3 / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x numerator 3 denominator [ (2x-5)^2 * 3x + 1 ] evatuate 95/7 confidence assessment:
.................................................
......!!!!!!!!...................................
21:44:33 `q004. Explain, step by step, how you evaluate the expression (x - 5) ^ 2x-1 + 3 / x-2 for x = 4.
......!!!!!!!!...................................
RESPONSE --> I understand confidence assessment:
.................................................
......!!!!!!!!...................................
21:46:21 *&*& Standard mathematics notation is easier to see. On the other hand 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 have to enter expressions through a typewriter, so it is well worth the trouble to get used to this notation. Indicate your understanding of the necessity to understand this notation.
......!!!!!!!!...................................
RESPONSE --> I understand self critique assessment:
.................................................
......!!!!!!!!...................................
21:47:05 `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.
......!!!!!!!!...................................
RESPONSE --> i understand the difference intypewriter notation and standard. confidence assessment:
.................................................
......!!!!!!!!...................................
21:47:17 You 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,
......!!!!!!!!...................................
RESPONSE --> i see the examples self critique assessment:
.................................................
......!!!!!!!!...................................
21:47:30 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.
......!!!!!!!!...................................
RESPONSE --> i understand self critique assessment:
.................................................
©“š¥¤˜ê®¡ÅŶ‰ÕÚvŒÖ§« assignment #001 001. Rates qa rates 08-27-2008
......!!!!!!!!...................................
21:50:42 `q001. You should copy and paste these instructions to a word processor for reference. However you can always view them, as well as everything else that has appeared in this box, by clicking the 'Display Everything' button. 1. For the next question or answer, you click on 'Next Question / Answer' button above the box at top left until a question has been posed. Once a question has been posed you are to answer before you click again on this button. 2. Before clicking for an answer, type your best answer to the current question into the box to the right, then clip on the 'Enter Answer' button. 3. After entering your answer you will click on 'Next Question / Answer' to view the answer to the question. Do not tamper with the information displayed in the left-hand box. 4. If your answer was incorrect, incomplete or would otherwise require revision, you will enter a self-critique. If you learned something from the answer, you need to restate it in your own words in order to reinforce your learning. If there is something you feel you should note for future reference, you should make a note in your own words. Go to the response box (the right-hand box) and type in a self-critique and/or notes, as appropriate. Do not copy and paste anything from the left-hand box, since that information will be saved in any case. 5. If you wish to save your response to your Notes file you may choose to click on the 'Save As Notes' button rather than the 'Enter Answer' button. Doing so will save your work for your future reference. Your work will be saved in a Notes file in the c:\vhmthphy folder. The title of the Notes file will also include the name you gave when you started the program. 6. After clicking either the 'Enter Response' or the 'Save as Notes' button, click on 'Next Question / Answer' and proceed in a similar manner. In the right-hand box briefly describe your understanding of these instructions, then click 'Enter Answer'.
......!!!!!!!!...................................
RESPONSE --> I understand the q/a thing along with display and enter response as save to notes. confidence assessment:
.................................................
......!!!!!!!!...................................
21:50:55 Your answer has been noted. Enter 'ok' in the Response Box and click on Enter Response, then click on Next Question/Answer for the first real question.
......!!!!!!!!...................................
RESPONSE --> ok confidence assessment:
.................................................
......!!!!!!!!...................................
21:51:36 `q002. 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. Continue as far as you can until you are completely lost. Students who are prepared for the highest-level math courses might not ever get lost. If you make $50 in 5 hr, then at what rate are you earning money?
......!!!!!!!!...................................
RESPONSE --> $10/hr confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:52:07 `q003.If you make $60,000 per year then how much do you make per month?
......!!!!!!!!...................................
RESPONSE --> ok confidence assessment:
.................................................
......!!!!!!!!...................................
21:53:08 `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?
......!!!!!!!!...................................
RESPONSE --> ok confidence assessment:
.................................................
......!!!!!!!!...................................
21:53:46 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.
......!!!!!!!!...................................
RESPONSE --> ok confidence assessment:
.................................................
......!!!!!!!!...................................
21:55:05 `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?
......!!!!!!!!...................................
RESPONSE --> 50 mi/hr it is the average becaus your instant anious velocity would be higher at some points. confidence assessment:
.................................................
......!!!!!!!!...................................
21:55:20 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.
......!!!!!!!!...................................
RESPONSE --> ok confidence assessment:
.................................................
......!!!!!!!!...................................
21:56:27 `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?
......!!!!!!!!...................................
RESPONSE --> 20 mi/gal confidence assessment:
.................................................
......!!!!!!!!...................................
21:57:52 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 those 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 require the use of more fuel on some miles than on others. 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 Notes.
......!!!!!!!!...................................
RESPONSE --> i converted it to the commonly used version as the instuctor would expected. confidence assessment:
.................................................
......!!!!!!!!...................................
21:59:32 `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?
......!!!!!!!!...................................
RESPONSE --> because there are highe and lower points in our rates data. inwhich we could find instantanious data but wed need more information. confidence assessment:
.................................................
......!!!!!!!!...................................
21:59:57 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.
......!!!!!!!!...................................
RESPONSE --> ok confidence assessment:
.................................................
......!!!!!!!!...................................
22:01:37 `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?
......!!!!!!!!...................................
RESPONSE --> 3/73 pushups per day confidence assessment:
.................................................
......!!!!!!!!...................................
22:02:35 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.
......!!!!!!!!...................................
RESPONSE --> i did the wrong rate though i see the answer is correct confidence assessment:
.................................................
......!!!!!!!!...................................
22:04:44 `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?
......!!!!!!!!...................................
RESPONSE --> 17/30 poundslifted /pounds added confidence assessment:
.................................................
......!!!!!!!!...................................
22:05:21 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.
......!!!!!!!!...................................
RESPONSE --> i was wrong in the puond lifted confidence assessment:
.................................................
......!!!!!!!!...................................
22:05:58 `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?
......!!!!!!!!...................................
RESPONSE --> 10m/s confidence assessment:
.................................................
......!!!!!!!!...................................
22:06:17 `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?
......!!!!!!!!...................................
RESPONSE --> ok confidence assessment:
.................................................
......!!!!!!!!...................................
22:06:43 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.
......!!!!!!!!...................................
RESPONSE --> ok confidence assessment:
.................................................
......!!!!!!!!...................................
22:07:25 `q012. We just averaged two quantities, adding them in dividing by 2, to find an average rate. We didn't do that before. Why we do it now?
......!!!!!!!!...................................
RESPONSE --> because it makes sense to do it like that. confidence assessment:
.................................................
......!!!!!!!!...................................
22:07:42 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.
......!!!!!!!!...................................
RESPONSE --> ok confidence assessment:
.................................................
course phy201 ’Çèný§”–¶ºíüÔ^îÀÓðë ÑŒÆR°‘assignment #001
......!!!!!!!!...................................
09:44:25 `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.
......!!!!!!!!...................................
RESPONSE --> 12m^2 confidence assessment:
.................................................
......!!!!!!!!...................................
09:44:30 A 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:44:47 `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?
......!!!!!!!!...................................
RESPONSE --> 6m^2 confidence assessment:
.................................................
......!!!!!!!!...................................
09:44:56 A 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.
......!!!!!!!!...................................
RESPONSE --> OK self critique assessment:
.................................................
......!!!!!!!!...................................
09:45:11 `q003. What is the area of a parallelogram whose base is 5.0 meters and whose altitude is 2.0 meters?
......!!!!!!!!...................................
RESPONSE --> 10M^2 confidence assessment:
.................................................
......!!!!!!!!...................................
09:45:18 A 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.
......!!!!!!!!...................................
RESPONSE --> OK self critique assessment:
.................................................
......!!!!!!!!...................................
09:45:30 `q004. What is the area of a triangle whose base is 5.0 cm and whose altitude is 2.0 cm?
......!!!!!!!!...................................
RESPONSE --> 5m^2 confidence assessment:
.................................................
......!!!!!!!!...................................
09:45:36 It 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:45:52 `q005. What is the area of a trapezoid with a width of 4.0 km and average altitude of 5.0 km?
......!!!!!!!!...................................
RESPONSE --> 20km^2 confidence assessment:
.................................................
......!!!!!!!!...................................
09:45:58 Any 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:47:29 `q006. What is the area of a trapezoid whose width is 4 cm in whose altitudes are 3.0 cm and 8.0 cm?
......!!!!!!!!...................................
RESPONSE --> 22cm^2 confidence assessment:
.................................................
......!!!!!!!!...................................
09:47:35 The 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:48:02 `q007. What is the area of a circle whose radius is 3.00 cm?
......!!!!!!!!...................................
RESPONSE --> 9 pi cm^2 confidence assessment:
.................................................
......!!!!!!!!...................................
09:48:06 The 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:48:25 `q008. What is the circumference of a circle whose radius is exactly 3 cm?
......!!!!!!!!...................................
RESPONSE --> 6 pi cm confidence assessment:
.................................................
......!!!!!!!!...................................
09:48:31 The 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:48:53 `q009. What is the area of a circle whose diameter is exactly 12 meters?
......!!!!!!!!...................................
RESPONSE --> 36 pi m^2 confidence assessment:
.................................................
......!!!!!!!!...................................
09:48:58 The 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:49:16 `q010. What is the area of a circle whose circumference is 14 `pi meters?
......!!!!!!!!...................................
RESPONSE --> 49 pi m^2 confidence assessment:
.................................................
......!!!!!!!!...................................
09:49:24 We 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:49:41 `q011. What is the radius of circle whose area is 78 square meters?
......!!!!!!!!...................................
RESPONSE --> r = 5.0 m confidence assessment:
.................................................
......!!!!!!!!...................................
09:49:47 Knowing 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:50:02 `q012. Summary Question 1: How do we visualize the area of a rectangle?
......!!!!!!!!...................................
RESPONSE --> Lx W the lenght times the width confidence assessment:
.................................................
......!!!!!!!!...................................
09:50:10 We 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:50:26 `q013. Summary Question 2: How do we visualize the area of a right triangle?
......!!!!!!!!...................................
RESPONSE --> b(h)(.5) it is half a square so half of lenght times width. confidence assessment:
.................................................
......!!!!!!!!...................................
09:50:30 We 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:50:45 `q014. Summary Question 3: How do we calculate the area of a parallelogram?
......!!!!!!!!...................................
RESPONSE --> b(h) bout the same as a rectangle be cause a rectange is a parrelagram lenghth times wodth. confidence assessment:
.................................................
......!!!!!!!!...................................
09:50:50 The area of a parallelogram is equal to the product of its base and its altitude. The altitude is measured perpendicular to the base.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:51:04 `q015. Summary Question 4: How do we calculate the area of a trapezoid?
......!!!!!!!!...................................
RESPONSE --> base times average eltitude witch you may have to divide to find the average. confidence assessment:
.................................................
......!!!!!!!!...................................
09:51:09 We think of the trapezoid being oriented so that its two parallel sides are vertical, and we multiply the average altitude by the width.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:51:22 `q016. Summary Question 5: How do we calculate the area of a circle?
......!!!!!!!!...................................
RESPONSE --> Pir^2 confidence assessment:
.................................................
......!!!!!!!!...................................
09:51:25 We use the formula A = pi r^2, where r is the radius of the circle.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:51:38 `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?
......!!!!!!!!...................................
RESPONSE --> 2Pi(r) confidence assessment:
.................................................
......!!!!!!!!...................................
09:51:44 We 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:51:55 `q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
......!!!!!!!!...................................
RESPONSE --> knowing the area and circumference formulas in a certain way help to work problems faster. confidence assessment:
.................................................
þr÷kë}ê¯Ô÷§Øç¹ù˜Ý|ß~éú| assignment #001 001. Areas qa areas volumes misc 08-28-2008
......!!!!!!!!...................................
09:17:48 `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.
......!!!!!!!!...................................
RESPONSE --> 12m^2 confidence assessment:
.................................................
......!!!!!!!!...................................
09:18:04 A 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:18:19 `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?
......!!!!!!!!...................................
RESPONSE --> 6m^2 confidence assessment:
.................................................
......!!!!!!!!...................................
09:18:29 A 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:18:48 `q003. What is the area of a parallelogram whose base is 5.0 meters and whose altitude is 2.0 meters?
......!!!!!!!!...................................
RESPONSE --> 10m^2 confidence assessment:
.................................................
......!!!!!!!!...................................
09:18:55 A 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:20:14 `q004. What is the area of a triangle whose base is 5.0 cm and whose altitude is 2.0 cm?
......!!!!!!!!...................................
RESPONSE --> 5m^2 confidence assessment: 0
.................................................
......!!!!!!!!...................................
09:20:41 It 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.
......!!!!!!!!...................................
RESPONSE --> wrong measurement self critique assessment:
.................................................
......!!!!!!!!...................................
09:22:10 `q005. What is the area of a trapezoid with a width of 4.0 km and average altitude of 5.0 km?
......!!!!!!!!...................................
RESPONSE --> 20km^2 confidence assessment:
.................................................
......!!!!!!!!...................................
09:22:18 Any 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:23:11 `q006. What is the area of a trapezoid whose width is 4 cm in whose altitudes are 3.0 cm and 8.0 cm?
......!!!!!!!!...................................
RESPONSE --> 22cm^2 confidence assessment:
.................................................
......!!!!!!!!...................................
09:23:18 The 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:25:18 `q007. What is the area of a circle whose radius is 3.00 cm?
......!!!!!!!!...................................
RESPONSE --> 9Pi 28.3 confidence assessment:
.................................................
......!!!!!!!!...................................
09:25:37 The 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.
......!!!!!!!!...................................
RESPONSE --> forgot mearsure ments self critique assessment:
.................................................
......!!!!!!!!...................................
09:28:41 `q008. What is the circumference of a circle whose radius is exactly 3 cm?
......!!!!!!!!...................................
RESPONSE --> 6Picm 18.8cm confidence assessment:
.................................................
......!!!!!!!!...................................
09:28:51 The 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:30:27 `q009. What is the area of a circle whose diameter is exactly 12 meters?
......!!!!!!!!...................................
RESPONSE --> 28.3m^2 confidence assessment:
.................................................
......!!!!!!!!...................................
09:31:02 The 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.
......!!!!!!!!...................................
RESPONSE --> I see my mistake self critique assessment:
.................................................
......!!!!!!!!...................................
09:32:47 `q010. What is the area of a circle whose circumference is 14 `pi meters?
......!!!!!!!!...................................
RESPONSE --> 49Pi m^2 confidence assessment:
.................................................
......!!!!!!!!...................................
09:32:58 We 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:36:09 `q011. What is the radius of circle whose area is 78 square meters?
......!!!!!!!!...................................
RESPONSE --> 4.98m confidence assessment:
.................................................
......!!!!!!!!...................................
09:36:32 Knowing 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.
......!!!!!!!!...................................
RESPONSE --> forgot sig figs self critique assessment:
.................................................
......!!!!!!!!...................................
09:37:06 `q012. Summary Question 1: How do we visualize the area of a rectangle?
......!!!!!!!!...................................
RESPONSE --> Lx W the lenght times the width confidence assessment:
.................................................
......!!!!!!!!...................................
09:37:18 We 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:38:14 `q013. Summary Question 2: How do we visualize the area of a right triangle?
......!!!!!!!!...................................
RESPONSE --> b(h)(.5) it is half a square so half of lenght times width. confidence assessment:
.................................................
......!!!!!!!!...................................
09:38:20 We 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:39:39 `q014. Summary Question 3: How do we calculate the area of a parallelogram?
......!!!!!!!!...................................
RESPONSE --> b(h) bout the same as a rectangle be cause a rectange is a parrelagram lenghth times wodth. confidence assessment:
.................................................
......!!!!!!!!...................................
09:39:47 The area of a parallelogram is equal to the product of its base and its altitude. The altitude is measured perpendicular to the base.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:40:45 `q015. Summary Question 4: How do we calculate the area of a trapezoid?
......!!!!!!!!...................................
RESPONSE --> base times average eltitude witch you may have to divide to find the average. confidence assessment:
.................................................
......!!!!!!!!...................................
09:40:54 We think of the trapezoid being oriented so that its two parallel sides are vertical, and we multiply the average altitude by the width.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:41:23 `q016. Summary Question 5: How do we calculate the area of a circle?
......!!!!!!!!...................................
RESPONSE --> Pir^2 confidence assessment:
.................................................
......!!!!!!!!...................................
09:41:32 We use the formula A = pi r^2, where r is the radius of the circle.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:42:00 `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?
......!!!!!!!!...................................
RESPONSE --> 2Pi(r) confidence assessment:
.................................................
......!!!!!!!!...................................
09:42:08 We 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:43:13 `q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
......!!!!!!!!...................................
RESPONSE --> knowing the area and circumference formulas in a certain way help to work problems faster. confidence assessment:
.................................................
½ˆ…Âðò«ñ¦ÀîãؾëÚ¥û¨ÆH¥w–Æ assignment #001 001. Areas qa areas volumes misc 08-28-2008 Án”„õÝÿÞ{ÞŽôƒÃü¿ˆœ|ã¾ assignment #001 001. Areas qa areas volumes misc 08-28-2008 è·¶g‡¾ú±…êÕĵüµ‚]Òö“¸Ù¢ assignment #001 001. Areas qa areas volumes misc 08-28-2008
......!!!!!!!!...................................
09:53:53 `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.
......!!!!!!!!...................................
RESPONSE --> 12m^2 confidence assessment:
.................................................
......!!!!!!!!...................................
09:54:00 A 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:55:17 `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?
......!!!!!!!!...................................
RESPONSE --> 6m^2 confidence assessment:
.................................................
......!!!!!!!!...................................
09:55:26 A 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:55:30 `q003. What is the area of a parallelogram whose base is 5.0 meters and whose altitude is 2.0 meters?
......!!!!!!!!...................................
RESPONSE --> 10M^2 confidence assessment:
.................................................
......!!!!!!!!...................................
09:55:35 A 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:55:45 `q004. What is the area of a triangle whose base is 5.0 cm and whose altitude is 2.0 cm?
......!!!!!!!!...................................
RESPONSE --> 5m^2 confidence assessment:
.................................................
......!!!!!!!!...................................
09:55:49 It 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:55:58 `q005. What is the area of a trapezoid with a width of 4.0 km and average altitude of 5.0 km?
......!!!!!!!!...................................
RESPONSE --> 20km^2 confidence assessment:
.................................................
......!!!!!!!!...................................
09:56:04 Any 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:56:12 `q006. What is the area of a trapezoid whose width is 4 cm in whose altitudes are 3.0 cm and 8.0 cm?
......!!!!!!!!...................................
RESPONSE --> 22cm^2 confidence assessment:
.................................................
......!!!!!!!!...................................
09:56:18 The 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:56:28 `q007. What is the area of a circle whose radius is 3.00 cm?
......!!!!!!!!...................................
RESPONSE --> 9 pi cm^2 confidence assessment:
.................................................
......!!!!!!!!...................................
09:56:33 The 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:56:42 `q008. What is the circumference of a circle whose radius is exactly 3 cm?
......!!!!!!!!...................................
RESPONSE --> 6 pi cm confidence assessment:
.................................................
......!!!!!!!!...................................
09:56:48 The 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:56:59 `q009. What is the area of a circle whose diameter is exactly 12 meters?
......!!!!!!!!...................................
RESPONSE --> 36 pi m^2 confidence assessment:
.................................................
......!!!!!!!!...................................
09:57:05 The 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:57:17 `q010. What is the area of a circle whose circumference is 14 `pi meters?
......!!!!!!!!...................................
RESPONSE --> 49 pi m^2 confidence assessment:
.................................................
......!!!!!!!!...................................
09:57:23 We 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:57:34 `q011. What is the radius of circle whose area is 78 square meters?
......!!!!!!!!...................................
RESPONSE --> r = 5.0 m confidence assessment:
.................................................
......!!!!!!!!...................................
09:57:38 Knowing 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:58:12 `q012. Summary Question 1: How do we visualize the area of a rectangle?
......!!!!!!!!...................................
RESPONSE --> Lx W the lenght times the width confidence assessment:
.................................................
......!!!!!!!!...................................
09:58:17 We 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:58:44 `q013. Summary Question 2: How do we visualize the area of a right triangle?
......!!!!!!!!...................................
RESPONSE --> b(h)(.5) it is half a square so half of lenght times width. confidence assessment:
.................................................
......!!!!!!!!...................................
09:58:48 We 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:59:01 `q014. Summary Question 3: How do we calculate the area of a parallelogram?
......!!!!!!!!...................................
RESPONSE --> b(h) bout the same as a rectangle be cause a rectange is a parrelagram lenghth times wodth. confidence assessment:
.................................................
......!!!!!!!!...................................
09:59:09 The area of a parallelogram is equal to the product of its base and its altitude. The altitude is measured perpendicular to the base.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:59:20 `q015. Summary Question 4: How do we calculate the area of a trapezoid?
......!!!!!!!!...................................
RESPONSE --> base times average eltitude witch you may have to divide to find the average. confidence assessment:
.................................................
......!!!!!!!!...................................
09:59:25 We think of the trapezoid being oriented so that its two parallel sides are vertical, and we multiply the average altitude by the width.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:59:35 `q016. Summary Question 5: How do we calculate the area of a circle?
......!!!!!!!!...................................
RESPONSE --> Pir^2 confidence assessment:
.................................................
......!!!!!!!!...................................
09:59:39 We use the formula A = pi r^2, where r is the radius of the circle.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
09:59:46 `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?
......!!!!!!!!...................................
RESPONSE --> 2Pi(r) confidence assessment:
.................................................
......!!!!!!!!...................................
09:59:51 We 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
10:00:02 `q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
......!!!!!!!!...................................
RESPONSE --> knowing the area and circumference formulas in a certain way help to work problems faster. confidence assessment:
.................................................