course Mth 164 J??????????? ?assignment #001
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01:58:47 `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'.
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RESPONSE --> Always enter a response unless saving notes. After entering a response, click next question/answer. confidence assessment: 3
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01:59:01 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.
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RESPONSE --> ok confidence assessment: 3
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02:04:06 `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?
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RESPONSE --> To get the rate at which money is earned, you would find out how much you are earning in an hour. To do this, you would set it up as $50 = 5hr. After doing that, you would then attempt to get 1 hour on one side of the equation. After that has been done, the rate at which you earn money is the result. $50 = 5hr $50/(5) = 5hr/(5) $10 = hr The rate at which money is earned is $10 per hour. confidence assessment: 3
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02:04:46 01-15-2009 02:04:46 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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NOTES -------> OK
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02:07:23 `q003.If you make $60,000 per year then how much do you make per month?
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RESPONSE --> In order to get the answer, you need to understand that one year is 12 months. To set up the equation, you would write $60,000 = 12mo. Now it's neceessary to get one month at one side of the equation by dividing by 12. After that is done, the rate at which money is earned is the result. $60,000 = 12mo $60,000/(12) = 12mo/(12) $5000 = 1mo. You would be earning $5000 per month. confidence assessment: 3
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02:07:43 01-15-2009 02:07:43 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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NOTES -------> OK
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02:08:50 `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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RESPONSE --> Seeing as how there is no way to know if exactly $5000 was made each month, it would be wise to assume the business makes an average of $5000 per month. confidence assessment: 3
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02:09:05 01-15-2009 02:09:05 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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NOTES -------> OK
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02:11:46 `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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RESPONSE --> In order to get the answer, it's required to isolate hours. You would divide both sides by 6 in order to get a single hour isolated. After that, you will have your answer in miles per hour. 300mi = 6hr 300mi/(6) = 6hr/(6) 50mi = 1hr It's unlikely that the rate would be a constant 50mph the entire 300 mile distance so it is necessary to say that is the average rate. confidence assessment: 3
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02:12:04 01-15-2009 02:12:04 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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NOTES -------> ok
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02:14:43 `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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RESPONSE --> In order to find this equation, you must set up the equation as 1200m = 60g where m = miles and g = gallons. After that, you would need to divide both sides by 60 in order to get how many miles traveled to a single gallon of gasoline. 1200m = 60g 20m = 1g Therefore, the average rate of gasoline used per mile is 1 gallon per 20 miles. confidence assessment: 3
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02:15:34 01-15-2009 02:15:34 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.
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NOTES -------> always read the question and make sure you understand it
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02:18:22 `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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RESPONSE --> We are trying to figure out the rate at which something happens, not the mean of two numbers. confidence assessment: 1
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02:19:04 01-15-2009 02:19:04 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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NOTES -------> ok
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02:19:44 `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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RESPONSE --> I have no idea. confidence assessment: 0
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02:20:35 01-15-2009 02:20:35 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.
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NOTES -------> I forgot that's what I needed to do to solve it.
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02:25:01 `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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RESPONSE --> I have no idea. confidence assessment: 0
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02:26:03 01-15-2009 02:26:03 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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NOTES -------> i needed to find out the difference in total weight, then the total weight dfference in the shoulder weights. then i needed to divide them to find out the average rate.
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02:30:15 `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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RESPONSE --> In this problem, you would have to find out the difference between the time in seconds and the distance in meters. The difference between 100 and 200 meters is 100 meters. The difference between 12 seconds and 22 seconds is 10. After that, you would divide 100 by 10 to find out he was running 10 meters per second. confidence assessment: 2
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02:30:24 01-15-2009 02:30:24 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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NOTES -------> ok
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02:31:25 `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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RESPONSE --> You would need to find the average of 10 and 9 meters/second. You would add 10 and 9, then divide by 2. My estimate would be he was running 9.5 meters/second confidence assessment: 3
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02:31:50 01-15-2009 02:31:50 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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NOTES -------> ok
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02:33:00 `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?
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RESPONSE --> Two values of the same unit were given. In order to find out the average rate between two other average rates, we needed to average them. confidence assessment: 2
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02:33:16 01-15-2009 02:33:16 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.
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NOTES -------> ok
.................................................??S???K???????F???assignment #002
002. Describing Graphs qa initial problems 01-15-2009
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02:40:53 `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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RESPONSE --> After making the table and graphing the function, the x-intercept was (2,0) and the y-intercept was (0, -4) confidence assessment: 3
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02:41:24 01-15-2009 02:41:24 The 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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NOTES -------> ok
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02:41:52 `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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RESPONSE --> It doesn't change. It slope is 3 the entire time. confidence assessment: 3
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02:42:04 The graph forms a straight line with no change in steepness.
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RESPONSE --> ok self critique assessment: 3
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02:42:45 `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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RESPONSE --> The slope of y=3x-4 is 3. confidence assessment: 3
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02:42:56 Between 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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RESPONSE --> ok self critique assessment: 3
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02:44:17 `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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RESPONSE --> The graph is definitely growing steeper. It's increasing at an increasing rate. confidence assessment: 3
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02:45:40 Graph 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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RESPONSE --> ok self critique assessment: 2
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02:46:43 `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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RESPONSE --> The graph grows steeper as the line grows. It is decreasing at an increasing rate. confidence assessment: 2
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02:47:38 01-15-2009 02:47:38 From 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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NOTES -------> oops, didn't pay attention
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02:48:52 `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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RESPONSE --> The graph is increasing. The steepness changes but very slowly. It is increasing at an increasing rate. confidence assessment: 2
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02:53:44 01-15-2009 02:53:44 If 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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NOTES -------> oops, increasing at an increasing rate means the rate at which it actually increases is bigger, not that it continues to go up.
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03:20:12 `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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RESPONSE --> The graph is decreasing. The steepness of the slope is growing. The graph is decreasing at an increasing rate. confidence assessment: 2
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03:20:32 01-15-2009 03:20:32 ** 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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NOTES -------> ok
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03:21:09 `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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RESPONSE --> The graph is increasing at an increasing rate. confidence assessment: 2
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03:21:22 ** 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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RESPONSE --> ok self critique assessment: 2
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???????`?m??U???? assignment #003 003. PC1 questions qa initial problems 01-15-2009
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03:25:01 `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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RESPONSE --> The line connecting (7, 17) and (10, 29). The diference between 7 and 10 is 3. The difference in first line[(3,5) and (7, 17)] is 4. Therefore, it would make sense that the lower number would be steeper as it is less space for it to rise. confidence assessment: 2
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03:25:29 The 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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RESPONSE --> ok self critique assessment: 2
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03:27:39 `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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RESPONSE --> You would plug both values of x into the expression. If either parts in the parenthesis comes out to zero, it makes the entire expression zero. confidence assessment: 2
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03:28:01 If 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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RESPONSE --> ok self critique assessment: 2
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03:29:19 `q003. For what x values will the expression (3x - 6) * (x + 4) * (x^2 - 4) be zero?
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RESPONSE --> The values that will make the expression (3x - 6) * (x + 4) * (x^2 - 4) be zero are x = 2 and x = -4. The reasoning behind this is if anything in the parenthesis turns out to be a zero, it makes the entire expression zero. confidence assessment: 2
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03:29:35 01-15-2009 03:29:35 In 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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NOTES -------> forgot about -2
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03:34:10 `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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RESPONSE --> The trapezoid including the points (10, 2) and (50, 4). The distance between the two x values in it are larger than the distance between the x values of the trapezoid containing the points (3,5) and (7, 9) confidence assessment: 2
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03:34:24 Your 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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RESPONSE --> ok self critique assessment: 2
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03:43:55 * `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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RESPONSE --> For y = x^2, as the graph moves left to right the graph increases as its slope increases. For y = 1/x, as the graph moves left to right the graph decreases as the slope decreases. For y = sqrt(x), as the graph moves left to right, the graph increases as its slope decreases. self critique assessment: 2
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03:44:54 01-15-2009 03:44:54 For 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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NOTES -------> ok
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04:03:26 `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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RESPONSE --> At the end of 3 months, you would have 26.62 frogs in the frog pond. To find that out, I took 10% of 20 and got 2. I added 2 to 20, and then did 10% of every number after that until I got 3 months. As far as getting the number of frogs after 300 months, I have no clue. confidence assessment: 1
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04:12:11 01-15-2009 04:12:11 At 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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NOTES -------> ok
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04:21:40 `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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RESPONSE --> 1/1 = 1 1/.1 = 10 1/.01 = 100 1/.001 = 1000 The pattern is simply another zero added to the number. These x values are approaching zero because they are getting closer to zero. If we were to continue with the pattern, we could use .0001 and .00001. The values of 1/x continue to grow larger. confidence assessment: 2
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04:22:19 01-15-2009 04:22:19 If 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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NOTES -------> ok
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04:41:37 * `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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RESPONSE --> First, you would solve for v. v= 3(5)+9 v=15+9 v=24 Then you would plug 24 in for v in E= 800 v^2. E = 800(24^2) E = 800 ^ 576 The amount of energy is 460800. self critique assessment: 2
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04:41:47 01-15-2009 04:41:47 ?
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NOTES -------> ok
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04:41:54 For t=5, v = 3 t + 9 = (3*5) + 9 = 24. Therefore E = 800 * 24^2 = 460800.
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RESPONSE --> ok self critique assessment: 2
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04:42:01 ?
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RESPONSE --> ok self critique assessment: 2
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04:45:05 * `q009. Continuing the preceding problem, can you give an expression for E in terms of t?
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RESPONSE --> I don't know how. self critique assessment: 1
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04:45:28 01-15-2009 04:45:28 Since 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).
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NOTES -------> ok
.................................................????????|????assignment #001
001. Areas qa areas volumes misc 01-15-2009
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05:11:47 `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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RESPONSE --> To get the area, you would do 4m^3m and get 12m^2 confidence assessment: 3
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05:12:03 01-15-2009 05:12:03 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.
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NOTES -------> ok
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05:13:48 `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?
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RESPONSE --> To find the area of the triangle, you would do a = 1/2(4^3) and come up with 6m^2 confidence assessment: 3
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05:14:21 01-15-2009 05:14:21 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.
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NOTES -------> ok
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05:15:01 `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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RESPONSE --> You would do a=bh where b = 5 and h = 2. The area is 10m^2 confidence assessment: 3
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05:15:08 01-15-2009 05:15:08 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.
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NOTES -------> ok
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05:16:02 `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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RESPONSE --> You would use a = 1/2bh where b is 5cm and h is 2cm. You would end up with 5cm^2 confidence assessment: 3
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05:16:08 01-15-2009 05:16:08 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.
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NOTES -------> ok
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05:19:16 `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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RESPONSE --> You would use a = h(b1+b2)/2. From the problem, the base is already given as 4. From there, you would just plug the numbers in. a = (5km(4km))/2 a = 20km^2/2 a = 10km^2 The area is 10km^2 confidence assessment: 3
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05:19:28 01-15-2009 05:19:28 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.
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NOTES -------> ok
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23:29:47 `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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RESPONSE --> The formula for the area of a trapezoid is A=1/2a(b1+b2). We just need to plug in numbers for the values. A=1/2(4)(3+8). A=2(3+8) A=2(11) A=22 The area of the trapezoid is 22cm^2 confidence assessment: 2
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23:29:56 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.
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RESPONSE --> ok self critique assessment: 2
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23:33:45 `q007. What is the area of a circle whose radius is 3.00 cm?
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RESPONSE --> To find the area of the circle, we need to use the formula A = pi * r * r and substitute in values. A = 3.14 * 3cm * 3cm A=28.26cm^2 The area of the circle is 28.26cm^2 confidence assessment: 2
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23:34:06 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.
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RESPONSE --> ok self critique assessment: 2
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23:39:59 `q008. What is the circumference of a circle whose radius is exactly 3 cm?
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RESPONSE --> To find the answer, we must find the diameter first. As the radius is half of the diameter, we need to double the radius to get 6cm. After that, you would just need to use the circumference formula. C=pi * d C = 3.14 * 6cm C= 18.84cm The circumference of the circle is 18.84 confidence assessment: 2
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23:42:08 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.
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RESPONSE --> ok self critique assessment: 2
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23:43:35 `q009. What is the area of a circle whose diameter is exactly 12 meters?
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RESPONSE --> We would need to find the radius. To get it, we would divide 12m by 2 to get 6m then just use the area formula. A = pi * r * r A = 3.14 * 6 * 6 A= 113.04 The area of the circle is 113.04cm^2 confidence assessment: 2
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23:43:42 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.
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RESPONSE --> ok self critique assessment: 2
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23:57:26 `q010. What is the area of a circle whose circumference is 14 `pi meters?
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RESPONSE --> confidence assessment: 0
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23:57:40 01-15-2009 23:57:40 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.
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NOTES -------> ok
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23:59:55 `q011. What is the radius of circle whose area is 78 square meters?
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RESPONSE --> We need to set 78 equal to the area formula. 78m^2 = pi * r *r 78m^2/3.14 = r*r 24.84m^2 = r*r sqrt(24.84m) = r confidence assessment: 2
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00:00:34 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.
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RESPONSE --> ok self critique assessment: 2
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00:01:07 `q012. Summary Question 1: How do we visualize the area of a rectangle?
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RESPONSE --> Visualizing the area of the rectangle is all the space inside of it. confidence assessment: 2
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00:01:16 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.
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RESPONSE --> ok self critique assessment: 2
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00:05:45 `q013. Summary Question 2: How do we visualize the area of a right triangle?
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RESPONSE --> A rectangle is a right triangle with a diagonal line from one corner to another. The area would be half of the area of the rectangle. confidence assessment: 2
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00:05:50 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.
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RESPONSE --> ok self critique assessment: 2
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00:06:44 `q014. Summary Question 3: How do we calculate the area of a parallelogram?
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RESPONSE --> You would use the formula A= bh confidence assessment: 2
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00:06:52 The 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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RESPONSE --> ok self critique assessment: 2
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00:09:16 `q015. Summary Question 4: How do we calculate the area of a trapezoid?
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RESPONSE --> To calculate the area of a trapezoid, you would find the average of the two altitudes then multiply it by the base. confidence assessment: 2
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00:09:22 We 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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RESPONSE --> ok self critique assessment: 2
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00:09:58 `q016. Summary Question 5: How do we calculate the area of a circle?
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RESPONSE --> To find the rea of the circle, you would find the radius then use it in the formula A = pi * r * r confidence assessment: 2
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00:10:04 We use the formula A = pi r^2, where r is the radius of the circle.
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RESPONSE --> ok self critique assessment: 2
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00:11:17 `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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RESPONSE --> You would use the formula C = 2pi*r. If you get an answer in square roots, it's wrong. confidence assessment: 2
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00:11:20 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.
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RESPONSE --> ok self critique assessment: 2
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00:12:06 `q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> I memorized all the formulas and then used them. confidence assessment: 2
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