course Phy 201 This includes more than one assignment. Thank you. ܲ烤doypу虡yɒStudent Name:
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22:03:35 `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. 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 --> The x intercept is (4/3, 0) and the y intercept is (0,-4).
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22:04:11 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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RESPONSE --> Yes, my graph confirmed this. When x=0, y=-4 and when y=0, x=4/3
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22:04:44 `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 --> This is a linear expression, and so the slope is constant. The slope remains at 3 for the duration of the line.
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22:04:49 The graph forms a straight line with no change in steepness.
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RESPONSE --> ok
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22:05:08 `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 is 3 for the duration of the line.
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22:05:49 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 --> y=mx+b, m=slope... so it is 3. correct.
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22:07:51 `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 --> This graph is increasing. The steepness of the graph changes exponentially as the slope approaches infinity. The graph is increasing at an increasing rate.
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22:07:59 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.
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RESPONSE -->
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22:09:18 `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 is decreasing in this case. the steepness of the graph is decreasing at a decreasing rate until it reaches 0.
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22:09:24 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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RESPONSE -->
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22:11:33 `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 appears to be increasing, but the steepness is decreasing, so the graph is said to be increasing at a decreasing rate.
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22:11:47 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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RESPONSE --> ok
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22:15:48 `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 increasing at a decreasing rate in this example, but it is concave up rather than concave down like the last example.
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22:17:04 ** 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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RESPONSE --> I looked at the graph too quickly and mistook the graph for increasing when it is infact decreasing. This is evident by the concavity, which i did identify as concave up. The rate of change decreases, but the graph itself also decreases.
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22:19:27 `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 --> Since the car's speed is constantly increasing as the distance increases, then the graph of y vs. t would likewise be increasing. The speed of the car would have to determine whether the rate of change is increasing or remaining constant or decreasing. If the car constantly gets faster in set increments, then it is a constant change, represented by a linear graph. If the rate of change is not constant and is increasing faster and faster, then it is an increasing rate. The inverse is true if it is a decreasing rate.
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22:19:59 ** 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 i may have looked too deep into this problem, but it makes sense.
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lXoقY̡S Student Name: assignment #001
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22:20:57 `q001. It will be very important in this course for your instructor to see and understand the process of visualization and reasoning you use when you solve problems. This exercise is designed to give you a first experience with these ideas, and your instructor a first look at your work. Answer the following questions and explain in commonsense terms why your answer makes sense.
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RESPONSE --> ok
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22:22:59 For each question draw a picture to make sense out of the situation, and include a description of the picture. Samples Sample question and response Question: If a bundle of shingles covers 30 square feet, how many bundles are required to cover a 600 square foot roof? Response: We might draw a picture of a rectangle representing the area, dividing the rectangle into a number of smaller rectangles each representing the area covered by a single bundle. This makes it clear that we are dividing the roof area into 1-bundle areas, and makes it clear why we are going to have to divide. Reasoning this problem out in words, we can say that a single bundle would cover 30 square feet. Two bundles would cover 60 square feet. Three bundles would cover 90 square feet. We could continue in this manner until we reach 600 square feet. However, this would be cumbersome. It is more efficient to use the ideas of multiplication and division. We imagine grouping the 600 square feet into 30 square foot patches. There will be 600 / 30 patches and each will require exactly one bundle. We therefore require 600 / 30 bundles = 20 bundles. {}Your responses might not be as clear as the above, though they might be even more clear. I won't be looking for perfection, though I wouldn't object to it, but for a first effort at visualizing a situation and communicating a reasoning process. This is not something you are used to doing and it might take a few attempts before you can achieve good results, but you will get better every time you try. {}You might be unsure of what to do on a specific question. In such a case specific questions and expressions of confusion are also acceptable responses. Such a response must include your attempts to come up with a picture and reason out an explanation. For example your response might be Sample expression of confusion: I've drawn a picture of a pile of bundles and a roof but I'm not sure how to connect the two. I tried multiplying the number of bundles by the square feet of the roof but I got 18,000, and I know it won't take 18,000 bundles to cover the roof. How do you put the area covered by a bundle together with the roof area to get the number of bundles required? A poor response would be something like 'I don't know how to do #17'. This response reveals nothing of your attempt to understand the question and the situation. Nor does it ask a specific question. Incidentally, you might be tempted to quote rules or formulas about rates and velocities in answering these questions. Don't. This exercise isn't about being able to memorize rules and quote them. It is about expanding your ability to visualize, reason and communicate.
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RESPONSE --> This is an exercise to work on my ability to explain a thought process.
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22:24:39 In your own words briefly summarize the instructions and the intent of this exercise.
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RESPONSE --> This exercise is meant to familiarize the student with the process necessary to communicate a thoughtprocess and rationale, while also showing the teacher what is the style of that particular student. The student needs to show a clear, concise representation of the process that took them to the answer which they are giving. This is a description, not an equation.
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22:26:00 `q001. If you earn 50 dollars in 5 hours, at what average rate are you earning money, in dollars per hour?
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RESPONSE --> The total amount is divided by the time needed to find the average: $50/ 5 hours= $10/hour. This gives our answer in $/hour, which is appropriate. So an average of $10/hour is earned.
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22:27:27 If you travel 300 miles in 6 hours, at what average rate are you traveling, in miles per hour?
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RESPONSE --> The total distance is divided by the time taken to achieve the average speed, or rate, used to achieve the 300 miles in 6 hours: 300miles/6 hours= 50miles/hour or 50mph. This has the appropriate units and shows an average rate of 50 miles/hour.
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22:28:47 `q002. If a ball rolling down a grooved track travels 40 centimeters in 5 seconds, at what average rate is the ball moving, in centimeters per second?
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RESPONSE --> The ball travels a total of 40 cm in 5 s, which, when divided will show an average of the rate the ball travels: 40cm/5s= 8 cm/s which is an average given in the appropriate units.
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22:30:10 The preceding three questions illustrate the concept of a rate. In each case, to find the rate we divided the change in some quantity (the number of dollars or the distance, in these examples) by the time required for the change (the number of hours or seconds, in these examples). Explain in your own words what is meant by the idea of a rate.
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RESPONSE --> The rate is the ratio of two values. The rate is the change in a value in relation to the time required for that change to occur.
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22:31:33 `q003. If you are earning money at the average rate of 15 dollars per hour, how much do you earn in 6 hours?
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RESPONSE --> Since the denominator of the rate is the time, if you multiply by the time, you achieve the value for which the rate takes place (in this case dollars): (15 $/hr)(6 hr)=$90. The hours cancel leaving only $ as the unit.
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22:32:20 If you are traveling at an average rate of 60 miles per hour, how far do you travel in 9 hours?
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RESPONSE --> THe same method proves true here. When multiplied together, the hours cancel, leaving only the miles traveled: (60 miles/hour)(9 hours)= 540 miles.
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22:33:28 `q004. If a ball travels at and average rate of 13 centimeters per second, how far does it travel in 3 seconds?
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RESPONSE --> Multiply the rate (13 cm/s) by the time (3 seconds) to find the distance (39 cm) since the time units cancel, leaving only distance units.
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22:35:49 In the preceding three exercises you turned the concept of a rate around. You were given the rate and the change in the clock time, and you calculated the change in the quantity. Explain in your own words how this increases your understanding of the concept of a rate.
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RESPONSE --> This fortifies that the rate is a ratio of a time required to represent a value. The ratio can be used to find the relationship between a known value and an unknown value. The unknown value has the relationship of the RATE to the known value.
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22:37:23 `q005. How long does it take to earn 100 dollars at an average rate of 4 dollars per hour?
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RESPONSE --> If the value ($100) is divided by the rate ($4/hour) the time is left. If a fraction is in the denominator of a fraction, then this gives way to the inverse. $100/ ($4/hr)= 25 hours.
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22:38:13 How long does it take to travel 500 miles at an average rate of 25 miles per hour?
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RESPONSE --> 500 miles/ (25 miles/hour) gives way to an answer in terms of hours since the rate is a ratio between the miles and the hours. This equals 20 hours.
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22:39:23 `q006. How long does it take a rolling ball to travel 80 centimeters at an average rate of 16 centimeters per second?
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RESPONSE --> The average rate represents how far the ball can go in a second. This ratio is used to determine how long it takes the ball to go a certain distance. 80 cm / (16 cm/s)= 5 seconds.
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22:39:48 In the preceding three exercises you again expanded your concept of the idea of a rate. Explain how these problems illustrate the concept of a rate.
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RESPONSE --> This also shows that the rate is simply a ratio that can be manipulated to show the relationship between two values.
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㽗˸yƁӗ Student Name: assignment #001 001. Rates
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22:40:45 `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 -->
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22:42:41 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 --> The instructions for a problem appear to the left and I enter the answer to the right and then click the ""enter response"" button. At this point it will be saved to my SEND file for the professor.
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22:43:08 `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 --> $10/hour
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22:43:37 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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RESPONSE --> ok
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22:44:12 `q003.If you make $60,000 per year then how much do you make per month?
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RESPONSE --> $60,000 / 12 months = 5000 dollars/month
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22:44:21 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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RESPONSE --> ok
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22:45:06 `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 --> It would be more appropriate to call it an average since some months may make more and other months less. The $5000 is not guaranteed, merely an average.
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22:45:20 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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RESPONSE --> ok
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22:46:15 `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 --> 300 miles/ 6 hours = 50 miles/hour. This is an average because sometimes you will be traveling at more than 50 mph and sometimes less than 50 mph, but the 50 mph is a good representation of the AVERAGE speed.
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22:46:39 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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RESPONSE --> ok
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22:48:18 `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 --> Since this question is asking for the rate of the gasoline used, this is the numerator: 60 gallons/ 1200 miles = 0.05 gallons/mile. If the question was phrased asking for the number of miles achieved per gallon, then the roles would have been reversed: 1200 miles/ 60 gallons = 20 miles/gallon.
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22:49:22 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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RESPONSE --> Again, a rate is a ratio between two values. It represents the relationship between those two units of conversion at a specific moment in time.
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22:51:27 `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 --> The average rates we are calculating are with the assumption that the addition is being calculated. When you multiply two numbers, you are simply adding those numbers a lot of times. These calculations are under the assumption that if real data was available, we would go through each measurement taken, find the rate, then add those rates up and divide by the numer of rates taken. We are simply learning the theory, not the literal practice.
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22:51:46 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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RESPONSE --> ok
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22:58:55 `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 --> For this one, the question is asking for LIFTING STRENGTH in relation to the daily singular pushup. First the number of daily pushups was calculated as 3650 for group 1 and 18259 for group two. This is done because each pushup is important, not just the daily exercise. Then, the final lifting strength is divided by the pushups completed to find the rate of increase in lifting strength. (this is under the assumption that all the lifting strength is new from the past year and that the initial lifting strength was 0). When 147 lbs is divided by the 3650 pushups done, it give a rate of 0.04 lbs/pushup. When 162 lbs is divided by the 18250 pushups done, it gives a rate of 0.009 lbs/pushup.
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23:00:27 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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RESPONSE --> I am not sure where I went wrong with this one. I showed the rates of lbs addition per pushup for both groups. When in fact, it was asking for the rate of the difference of lifting strength. I did a year long rate as opposed to a daily rate.
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23:01:51 `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 --> The 17 lb difference in lifting strength is divided by the addition 10 lb of weight used. this gives a rate of 0.85 lb lifting strength/lb weight.
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23:01:57 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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RESPONSE --> ok
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23:03:17 `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 --> The question is only asking for the rate of the runner between the two markers, so it is 10 seconds to cover 100 meters, or 10 meters/second.
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23:03:22 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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RESPONSE -->
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23:05:24 `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 --> It takes the runner 10 seconds to reach the first marker and an approximately 11.1 seconds to cover the second 100 meters. This adds up to an approximate 21.1 seconds to cover 200 meters, however the rates were taken only at the time when the runner crossed the marker, not the average rates.
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23:06:52 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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RESPONSE --> ok. I did not average the two times, but instead took the times recorded as a possible average of the past 100 meters run. I did not have the values to assume an average, so I didn't.
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23:07:47 `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 --> The previous example, we were given the average rates, and so we had more exact data. In this case we had to work with what we had and try to give the runner the middle time and not an extreme time in either direction.
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23:08:17 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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RESPONSE --> ok. we had two different rates in one problem and so had to take them both into account.
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