course Mth 173 This should be all the work requested on the assignments page, including all of assignment 0.Assignment zero said
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17:09:31 `q001. Explain the difference between x - 2 / x + 4 and (x - 2) / (x + 4).
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RESPONSE --> The first one has only 2 being divided by ""x"", according to the order of operations. The second one uses parenthesis to show that the difference of X and 2 should be taken, the sum of X and 4 should be taken, and the difference should be divided by the sum.
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17:10:17 The order of operations dictates that grouped expressions must be evaluated first, that exponentiation must be done before multiplication or division, which must be done before addition or subtraction. It makes a big difference whether you subtract the 2 from the 2 or divide the -2 by 4 first. If there are no parentheses you have to divide before you subtract: 2 - 2 / 2 + 4 = 2 - 1 + 4 (do multiplications and divisions before additions and subtractions) = 5 (add and subtract in indicated order) If there are parentheses you evaluate the grouped expressions first: (x - 2) / (x - 4) = (2 - 2) / ( 4 - 2) = 0 / 2 = 0.
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RESPONSE --> ok
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17:13:34 `q002. Explain the difference between 2 ^ x + 4 and 2 ^ (x + 4). Then evaluate each expression for x = 2. Note that a ^ b means to raise a to the b power. This process is called exponentiation, and the ^ symbol is used on most calculators, and in most computer algebra systems, to represent exponentiation.
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RESPONSE --> 2^x+4 you raise to the power of X first, then add four 2^(x+4) you take the sum first, then raise 2 to that quantity f(x) = 2^x+4 f(2) = 8 g(x) = 2^(x+4) g(2) = 64
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17:13:57 2 ^ x + 4 indicates that you are to raise 2 to the x power before adding the 4. 2 ^ (x + 4) indicates that you are to first evaluate x + 4, then raise 2 to this power. If x = 2, then 2 ^ x + 4 = 2 ^ 2 + 4 = 2 * 2 + 4 = 4 + 4 = 8. and 2 ^ (x + 4) = 2 ^ (2 + 4) = 2 ^ 6 = 2*2*2*2*2*2 = 64.
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RESPONSE --> I got it right
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17:17:48 `q003. What is the numerator of the fraction in the expression x - 3 / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x? What is the denominator? What do you get when you evaluate the expression for x = 2?
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RESPONSE --> The numerator is x-3 The denominator is [ (2x-5)^2 * 3x + 1 ] - 2 + 7x f(2) = -1/19
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17:19:24 The numerator is 3. x isn't part of the fraction. / indicates division, which must always precede subtraction. Only the 3 is divided by [ (2x-5)^2 * 3x + 1 ] and only [ (2x-5)^2 * 3x + 1 ] divides 3. If we mean (x - 3) / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x we have to write it that way. The preceding comments show that the denominator is [ (2x-5)^2 * 3x + 1 ] Evaluating the expression for x = 2: - 3 / [ (2 * 2 - 5)^2 * 3(2) + 1 ] - 2 + 7*2 = 2 - 3 / [ (4 - 5)^2 * 6 + 1 ] - 2 + 14 = evaluate in parenthese; do multiplications outside parentheses 2 - 3 / [ (-1)^2 * 6 + 1 ] -2 + 14 = add inside parentheses 2 - 3 / [ 1 * 6 + 1 ] - 2 + 14 = exponentiate in bracketed term; 2 - 3 / 7 - 2 + 14 = evaluate in brackets 13 4/7 or 95/7 or about 13.57 add and subtract in order. The details of the calculation 2 - 3 / 7 - 2 + 14: Since multiplication precedes addition or subtraction the 3/7 must be done first, making 3/7 a fraction. Changing the order of the terms we have 2 - 2 + 14 - 3 / 7 = 14 - 3/7 = 98/7 - 3/7 = 95/7.
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RESPONSE --> I got this problem wrong. I somehow mentally inserted parenthesis around what I thought were the numerator and the denominator. I know better than that and this problem should serve as a good wake-up call for me.
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17:19:31 The numerator is 3. x isn't part of the fraction. / indicates division, which must always precede subtraction. Only the 3 is divided by [ (2x-5)^2 * 3x + 1 ] and only [ (2x-5)^2 * 3x + 1 ] divides 3. If we mean (x - 3) / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x we have to write it that way. The preceding comments show that the denominator is [ (2x-5)^2 * 3x + 1 ] Evaluating the expression for x = 2: - 3 / [ (2 * 2 - 5)^2 * 3(2) + 1 ] - 2 + 7*2 = 2 - 3 / [ (4 - 5)^2 * 6 + 1 ] - 2 + 14 = evaluate in parenthese; do multiplications outside parentheses 2 - 3 / [ (-1)^2 * 6 + 1 ] -2 + 14 = add inside parentheses 2 - 3 / [ 1 * 6 + 1 ] - 2 + 14 = exponentiate in bracketed term; 2 - 3 / 7 - 2 + 14 = evaluate in brackets 13 4/7 or 95/7 or about 13.57 add and subtract in order. The details of the calculation 2 - 3 / 7 - 2 + 14: Since multiplication precedes addition or subtraction the 3/7 must be done first, making 3/7 a fraction. Changing the order of the terms we have 2 - 2 + 14 - 3 / 7 = 14 - 3/7 = 98/7 - 3/7 = 95/7.
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RESPONSE -->
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17:36:21 `q004. Explain, step by step, how you evaluate the expression (x - 5) ^ 2x-1 + 3 / x-2 for x = 4.
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RESPONSE --> f(x) = (x-5)^2x-1 + 3/x-2 f(4) = (4-5)^2(4)-1 + 3/(4)-2 I will solve the parenthesis first: (-1)^2(4)-1+3/(4)-2 Then, exponents should be solved: 1(4)-1+3/(4)-2 Then, solve multiplication and division parts: 4-1+.75-2 Then, add and subtract 1.75 or 7/4 f(4) = 1.75
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17:37:24 We get (4-5)^2 * 4 - 1 + 3 / 1 - 4 = (-1)^2 * 4 - 1 + 3 / 4 - 2 evaluating the term in parentheses = 1 * 4 - 1 + 3 / 4 - 2 exponentiating (2 is the exponent, which is applied to -1 rather than multiplying the 2 by 4 = 4 - 1 + 3/4 - 2 noting that 3/4 is a fraction and adding and subtracting in order we get = 1 3/4 = 7 /4 (Note that we could group the expression as 4 - 1 - 2 + 3/4 = 1 + 3/4 = 1 3/4 = 7/4). COMMON ERROR: (4 - 5) ^ 2*4 - 1 + 3 / 4 - 2 = -1 ^ 2*4 - 1 + 3 / 4-2 = -1 ^ 8 -1 + 3 / 4 - 2. INSTRUCTOR COMMENTS: There are two errors here. In the second step you can't multiply 2 * 4 because you have (-1)^2, which must be done first. Exponentiation precedes multiplication. Also it isn't quite correct to write -1^2*4 at the beginning of the second step. If you were supposed to multiply 2 * 4 the expression would be (-1)^(2 * 4). Note also that the -1 needs to be grouped because the entire expression (-1) is taken to the power. -1^8 would be -1 because you would raise 1 to the power 8 before applying the - sign, which is effectively a multiplication by -1.
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RESPONSE --> ok
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Hjȇ횔|G Student Name: assignment #002 002. Describing Graphs
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17:42:26 `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 --> y-intercept is -4 x-intercept is 4/3
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17:43:22 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 --> I just included the appropriate X and Y values for the intercept, and did not list each point as a coordinate (i.e. (0, -4)) as they are listed here.
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17:44:31 `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 --> The steepness (slope) of the graph does not change. No matter the value of X, the graph always has a slope of 3, rising three times for every one time it moves horizontally.
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17:44:36 The graph forms a straight line with no change in steepness.
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RESPONSE --> ok
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17:44:52 `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
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17:45:06 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
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17:46:04 `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 increasing. The steepness increases for greater values of X. The steepness rises exponentially. The graph is increasing at an increasing rate.
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17:46:16 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 --> ok
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17:47:20 `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. The graph becomes less steep as X values move from -3 to 0 The graph is decreasing at a decreasing rate.
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17:47:37 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 --> ok
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17:50:45 `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 graph becomes less steep for greater values of X The graph is increasing at a decreasing rate
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17:51:00 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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17:52:45 `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 graph becomes less steep for greater values of X The graph is decreasing at a decreasing rate
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17:53:02 ** 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 --> ok
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17:54:26 `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 --> y vs. t would be increasing the graph would be increasing at an increasing rate
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17:54:40 ** 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
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x~xcۥŝĆ Student Name: assignment #005 005. Calculus
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17:58:52 `q001. The graph of a certain function is a smooth curve passing through the points (3, 5), (7, 17) and (10, 29). Between which two points do you think the graph is steeper, on the average? Why do we say 'on the average'?
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RESPONSE --> The graph is steeper, on the average, between (7, 17) and (10, 29) It is on the average because the slope between these two points only factors in the x and y values at each of these two instances. Any values of x and y occurring between these two points are not known, so we are only taking an average steepness between points.
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17:59:35 Slope = rise / run. Between points (7, 17) and (10, 29) we get rise / run = (29 - 17) / (10 - 7) =12 / 3 = 4. The slope between points (3, 5) and (7, 17) is 3 / 1. (17 - 5) / (7 -3) = 12 / 4 = 3. The segment with slope 4 is the steeper. The graph being a smooth curve, slopes may vary from point to point. The slope obtained over the interval is a specific type of average of the slopes of all points between the endpoints.
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RESPONSE --> ok
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18:03:48 2. Answer without using a calculator: As x takes the values 2.1, 2.01, 2.001 and 2.0001, what values are taken by the expression 1 / (x - 2)? 1. As the process continues, with x getting closer and closer to 2, what happens to the values of 1 / (x-2)? 2. Will the value ever exceed a billion? Will it ever exceed one trillion billions? 3. Will it ever exceed the number of particles in the known universe? 4. Is there any number it will never exceed? 5. What does the graph of y = 1 / (x-2) look like in the vicinity of x = 2?
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RESPONSE --> 2. 10, 100, 1000, and 10000 1. They get infinitely larger 2. Yes it will 3. Yes 4. No 5. There is a vertical asymptote
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18:04:44 For x = 2.1, 2.01, 2.001, 2.0001 we see that x -2 = .1, .01, .001, .0001. Thus 1/(x -2) takes respective values 10, 100, 1000, 10,000. It is important to note that x is changing by smaller and smaller increments as it approaches 2, while the value of the function is changing by greater and greater amounts. As x gets closer in closer to 2, it will reach the values 2.00001, 2.0000001, etc.. Since we can put as many zeros as we want in .000...001 the reciprocal 100...000 can be as large as we desire. Given any number, we can exceed it. Note that the function is simply not defined for x = 2. We cannot divide 1 by 0 (try counting to 1 by 0's..You never get anywhere. It can't be done. You can count to 1 by .1's--.1, .2, .3, ..., .9, 1. You get 10. You can do similar thing for .01, .001, etc., but you just can't do it for 0). As x approaches 2 the graph approaches the vertical line x = 2; the graph itself is never vertical. That is, the graph will have a vertical asymptote at the line x = 2. As x approaches 2, therefore, 1 / (x-2) will exceed all bounds. Note that if x approaches 2 through the values 1.9, 1.99, ..., the function gives us -10, -100, etc.. So we can see that on one side of x = 2 the graph will approach +infinity, on the other it will be negative and approach -infinity.
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RESPONSE --> ok
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18:08:13 `q003. One straight line segment connects the points (3,5) and (7,9) while another connects the points (10,2) and (50,4). From each of the four points a line segment is drawn directly down to the x axis, forming two trapezoids. Which trapezoid has the greater area? Try to justify your answer with something more precise than, for example, 'from a sketch I can see that this one is much bigger so it must have the greater area'.
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RESPONSE --> The area of a trapezoid is the (width)*(average height) The first trapezoid has a width of 4 (7-3) and heights of 5 and 9, or an average height of 7 so its area is 4*7 or 28 The second trapezoid has a width of 40 (50-40) and heights of 2 and 4, or an average height of 3 so its area is 40*3 or 120 The second trapezoid has a larger area
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18:08:53 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
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18:13:07 `q004. If f(x) = x^2 (meaning 'x raised to the power 2') then which is steeper, the line segment connecting the x = 2 and x = 5 points on the graph of f(x), or the line segment connecting the x = -1 and x = 7 points on the same graph? Explain the basisof your reasoning.
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RESPONSE --> f(x) = x^2 f(2) = 4 f(5) = 25 f(-1) = 1 f(7) = 49 So we have segments from (2, 4) to (5, 25) and from (-1, 1) to (7, 49) The slope of the first segment is (25-4)/(5-2) or 7 The slope of the second is (49-1)/(7+1) or 6 The line segment connecting x=2 and x=5 is steeper
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18:13:19 The line segment connecting x = 2 and the x = 5 points is steeper: Since f(x) = x^2, x = 2 gives y = 4 and x = 5 gives y = 25. The slope between the points is rise / run = (25 - 4) / (5 - 2) = 21 / 3 = 7. The line segment connecting the x = -1 point (-1,1) and the x = 7 point (7,49) has a slope of (49 - 1) / (7 - -1) = 48 / 8 = 6. The slope of the first segment is greater.
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RESPONSE --> ok
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18:15:25 `q005. Suppose that every week of the current millenium you go to the jewler and obtain a certain number of grams of pure gold, which you then place in an old sock and bury in your backyard. Assume that buried gold lasts a long, long time ( this is so), that the the gold remains undisturbed (maybe, maybe not so), that no other source adds gold to your backyard (probably so), and that there was no gold in your yard before.. 1. If you construct a graph of y = the number of grams of gold in your backyard vs. t = the number of weeks since Jan. 1, 2000, with the y axis pointing up and the t axis pointing to the right, will the points on your graph lie on a level straight line, a rising straight line, a falling straight line, a line which rises faster and faster, a line which rises but more and more slowly, a line which falls faster and faster, or a line which falls but more and more slowly? 2. Answer the same question assuming that every week you bury 1 more gram than you did the previous week. {}3. Answer the same question assuming that every week you bury half the amount you did the previous week.
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RESPONSE --> 1. A rising straight line 2. A line which rises faster and faster 3. A line which rises slower and slower
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18:15:48 1. If it's the same amount each week it would be a straight line. 2. Buying gold every week, the amount of gold will always increase. Since you buy more each week the rate of increase will keep increasing. So the graph will increase, and at an increasing rate. 3. Buying gold every week, the amount of gold won't ever decrease. Since you buy less each week the rate of increase will just keep falling. So the graph will increase, but at a decreasing rate. This graph will in fact approach a horizontal asymptote, since we have a geometric progression which implies an exponential function.
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RESPONSE --> ok
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18:17:16 `q006. Suppose that every week you go to the jewler and obtain a certain number of grams of pure gold, which you then place in an old sock and bury in your backyard. Assume that buried gold lasts a long, long time, that the the gold remains undisturbed, and that no other source adds gold to your backyard. 1. If you graph the rate at which gold is accumulating from week to week vs. tne number of weeks since Jan 1, 2000, will the points on your graph lie on a level straight line, a rising straight line, a falling straight line, a line which rises faster and faster, a line which rises but more and more slowly, a line which falls faster and faster, or a line which falls but more and more slowly? 2. Answer the same question assuming that every week you bury 1 more gram than you did the previous week. 3. Answer the same question assuming that every week you bury half the amount you did the previous week.
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RESPONSE --> 1. A level straight line 2. A rising straight line 3. A falling straight line
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18:18:11 This set of questions is different from the preceding set. This question now asks about a graph of rate vs. time, whereas the last was about the graph of quantity vs. time. Question 1: This question concerns the graph of the rate at which gold accumulates, which in this case, since you buy the same amount eact week, is constant. The graph would be a horizontal straight line. Question 2: Each week you buy one more gram than the week before, so the rate goes up each week by 1 gram per week. You thus get a risingstraight line because the increase in the rate is the same from one week to the next. Question 3. Since half the previous amount will be half of a declining amount, the rate will decrease while remaining positive, so the graph remains positive as it decreases more and more slowly. The rate approaches but never reaches zero.
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RESPONSE --> I went too fast on number three. I understand what I did wrong, though.
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18:22:45 7. If the depth of water in a container is given, in centimeters, by 100 - 2 t + .01 t^2, where t is clock time in seconds, then what are the depths at clock times t = 30, t = 40 and t = 60? On the average is depth changing more rapidly during the first time interval or the second?
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RESPONSE --> f(x) = 100-2t+.01t^2 f(30) = 49 f(40) = 36 f(60) = 16 First interval the average change is 1.3 Second interval the avg change is 1 The average depth changes more rapidly in the first interval.
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18:22:54 At t = 30 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 30 + .01 * 30^2 = 49. At t = 40 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 40 + .01 * 40^2 = 36. At t = 60 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 60 + .01 * 60^2 = 16. 49 cm - 36 cm = 13 cm change in 10 sec or 1.3 cm/s on the average. 36 cm - 16 cm = 20 cm change in 20 sec or 1.0 cm/s on the average.
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RESPONSE --> ok
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18:24:39 8. If the rate at which water descends in a container is given, in cm/s, by 10 - .1 t, where t is clock time in seconds, then at what rate is water descending when t = 10, and at what rate is it descending when t = 20? How much would you therefore expect the water level to change during this 10-second interval?
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RESPONSE --> f(t) = 10-.1t f(10) = 9 f(20) = 8 In this 10-second interval the rate of change goes from 9cm/s to 8cm/s, for an average rate of change of 8.5cm/s. The interval is 10 seconds so I would expect the water level to lower 85cm.
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18:25:19 At t = 10 sec the rate function gives us 10 - .1 * 10 = 10 - 1 = 9, meaning a rate of 9 cm / sec. At t = 20 sec the rate function gives us 10 - .1 * 20 = 10 - 2 = 8, meaning a rate of 8 cm / sec. The rate never goes below 8 cm/s, so in 10 sec the change wouldn't be less than 80 cm. The rate never goes above 9 cm/s, so in 10 sec the change wouldn't be greater than 90 cm. Any answer that isn't between 80 cm and 90 cm doesn't fit the given conditions.. The rate change is a linear function of t. Therefore the average rate is the average of the two rates, or 9.5 cm/s. The average of the rates is 8.5 cm/sec. In 10 sec that would imply a change of 85 cm.
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RESPONSE --> ok
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_ P Student Name: assignment #001
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18:30:26 `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 -->
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18:31:40 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 --> ok
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18:33:25 In your own words briefly summarize the instructions and the intent of this exercise.
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RESPONSE --> For the given problem, create a visualization of the problem and describe your visualization. This exercise serves as an introduction into important visualization and reasoning techniques necessary for the course.
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18:36:41 `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 --> There are 50 dollars laid out one after another on a large table, and the table is divided into five even portions each representing one hour of work. Each of these portions has 50/5 or 10 dollars in it. This gives an answer of 10 dollars per hour.
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18:40:21 If you travel 300 miles in 6 hours, at what average rate are you traveling, in miles per hour?
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RESPONSE --> There is a distance representing 300 miles long drawn out. Along this distance there are 6 markers evenly spaced from beginning to end, making 6 even segments denoting each hour traveled. Each of these segments contains a representation of 300/6 or 50mi/hr.
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18:41:56 `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 --> There is a distance representing 40 centimeters drawn out. Along this distance there are 5 markers evenly spaced creating five even segments each representing one second of travel. Each segment contains 40/5 or 8cm/s
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18:43:16 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 --> A rate is a relationship. It is useful in simplifying information such as a large number of dollars earned in a large number of hours down to a rate of a certain number of dollars earned in just one hour.
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18:44:36 `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 --> There are 15 dollars in a group on a table. There are six of these groups, one for each hour worked. The total dollars on the table is 90 15*6
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18:46:10 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 --> There is a measured piece of paper representing 60 miles. Nine of these pieces of paper, one for each hour of travel, are laid end-to-end. The resulting length of these pieces of paper is representative of 540 miles 60*9
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18:47:39 `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 --> A string is measured to 13cm. Three of these strings are stretched out lengthwise on a table, one for each second of travel. The resulting amount of string measures 39cm 13*3
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18:49:42 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 --> A rate is a representation of a measure divided by a time, so r = m/t given any two parts of this relationship, the third can be found in the first exercises the M and T were given and the R was calculated in the second exercises a value of R and a measure of T was given, and the M could be found by algebraically switching the equation to r*t = m
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18:51:59 `q005. How long does it take to earn 100 dollars at an average rate of 4 dollars per hour?
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RESPONSE --> Four dollars are grouped together. These groups are laid on a table one-by-one until there are 100 total dollars on the table. After exactly 25 groups are put down there are exactly 100 dollars. This represents 25 hours of work.
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18:53:30 How long does it take to travel 500 miles at an average rate of 25 miles per hour?
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RESPONSE --> A piece of string is cut to represent 25 miles. This distance is laid out on a table, and more strings are laid end-to-end until a representation of 500 miles of string has been laid on the table. At that point there are 20 strings on the table, representing a time of 20 hours.
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18:54:17 `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 --> A 16cm string is cut, and more of these strings are laid end-to-end until there are 80 cm of string total laid down. At this point there are 5 strings on the table, representing 5 seconds of travel.
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18:57:59 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 --> Again, a rate is a value of a measure divided by a time, so r = m/t Just like if the rate and the time are given, measure can be calculated...if a rate and a measure are given, the time can be calculated by algebraically manipulating the formula to read t = m/r So the time is equal to the measure divided by the rate.
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pb;ߣہݑ Student Name: assignment #001 001. Rates LUC|[\蟿 Student Name: assignment #001 001. Rates
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19:23:56 `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 --> I understand
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19:24:04 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
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19:24:49 `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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19:25:20 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 --> I got it
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19:25:41 `q003.If you make $60,000 per year then how much do you make per month?
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RESPONSE --> $5,000
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19:25:52 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 --> got it
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19:26:11 `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 --> An average of
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19:26:23 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 --> got it
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19:27:23 `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 --> 50mi/hr This is an average rate because no matter how you are traveling, this rate is likely to include periods of faster and slower travel, or even periods of complete stops. It is the average over the time period.
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19:27:44 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 --> got it
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19:28:22 `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 --> 0.05 gallons/mile
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19:29:34 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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RESPONSE --> got it
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19:32:19 `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 adding that is references is being done to determine a total of something. If there are two things, you want to divide by the total of two things. In an average in the sense of these rates problems, the total has already been calculated and is given. A problem could say that 25 miles were traveled in 5 hours, or it could say that 4 miles were traveled in 1 hour, 6 miles in the next, etc. etc. breaking it down that way.
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19:32:51 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 --> got it
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19:40:49 `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 --> We want to find average rate of lifting strength increase per daily pushup, so our answer will be in terms of (lifting strength increase)/(daily pushup) or pounds/pushup The lifting strength increased from 147 to 162 pounds, or 15 pounds The pushups increased from 10 daily pushups to 50 daily pushups, or 40 pushups Imagine this information on a graph of strength vs. pushups, giving the coordinates (10, 147) and (50, 162) The slope between these points, or rate of change in pounds/pushup, is (162-147)/(50-10) or 15/40 Lifting strength increased at an average of 0.375 pounds per daily pushup.
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19:40:59 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 --> got it
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19:45:25 `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 --> We want to find the avg rate that lifting strength increased with respect to added shoulder weight. The difference in strength for the two groups was 17 pounds. The difference in shoulder weight for the two groups was 20 pounds. The average rate that lifting strength increased (17) with respect to the added shoulder weight (20) is 17/20 pounds Average of 0.85 pounds increase of strength per pound of shoulder weight increase
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19:45:34 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 --> got it
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19:46:38 `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 --> Average rate = distance divided by time He ran 100 meters (200-100) in 10 seconds (22-12) His average rate was 10m/s
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19:46:51 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 --> got it
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19:51:41 `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 --> If at one point the runner is going 10m/s and at another they are going 9m/s their average rate should be 9.5m/s At an average rate of 9.5m/s it should take 100/9.5 seconds to travel the 100m approximately 10.5263 seconds
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19:52:16 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 --> got it
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19:55:32 `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 --> This problem is different. Before we were given total average values. Now, we are given multiple values and a calculation is necessary to determine the average value, and that calculation requires adding and dividing by two. Previous problems had effectively done the calculation for us.
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19:56:04 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 --> I was a little off on the specifics of the question, but I understand the concept.
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ܯF̡z^ Student Name: assignment #001 001. Depth vs. Clock Time and Rate of Depth Change
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20:29:24 Note that there are four questions in this assignment. `q001. If your stocks are worth $5000 in mid-March, $5300 in mid-July and $5500 in mid-December, during which period, March-July or July-December was your money growing faster?
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RESPONSE --> March-July
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20:29:42 The first period (4 months) was shorter than the second (5 months), and the value changed by more during the shorter first period (increase of $300) than during the longer second perios (increase of $200). A greater increase in a shorter period implies a greater rate of change. So the rate was greater during the first period.
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RESPONSE --> got it right
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ηLzMSibP}֭䣸 Student Name: assignment #001 001. Depth vs. Clock Time and Rate of Depth Change
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20:32:26 `q002. What were the precise average rates of change during these two periods?
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RESPONSE --> The first period was a $300 increase over 4 months, or an average rate of change of $75/month The second period was a $200 increase over 5 months, or an average rate of change of $40/month
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20:32:38 From mid-March thru mid-July is 4 months. A change of $300 in four months gives an average rate of change of $300 / (4 months) = $75/month. From mid-July through mid-December is 5 months, during which the value changes by $200, giving an average rate of $200 / (5 months) = $40 / month. Thus the rate was greater during the first period than during the second.
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RESPONSE --> ok
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20:37:01 `q003. If the water in a uniform cylinder is leaking from a hole in the bottom, and if the water depths at clock times t = 10, 40 and 90 seconds are respectively 80 cm, 40 cm and 20 cm, is the depth of the water changing more quickly or less quickly, on the average, between t = 10 sec and t = 40 sec, or between t = 40 sec and t = 90 sec?
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RESPONSE --> We have values of 10, 80 40, 40 90, 20 The rate of change between 10s and 40s is (40-80)/(40-10) = -4/3 The rate of change between 40s and 90s is (20-40)/(90-40) = -2/5 So, the depth of water is changing more quickly, on the average, between 10s and 40s
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20:37:11 Between clock times t = 10 sec and t = 40 sec the depth changes by -40 cm, from 80 cm to 40 cm, in 30 seconds. The average rate is therefore -40 cm / (30 sec) = -1.33 cm/s. Between t = 40 sec and t = 90 sec the change is -20 cm and the time interval is 50 sec so the average rate of change is -20 cm/ 50 sec = -.4 cm/s, approx. The depth is changing more quickly during the first time interval.
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RESPONSE --> ok
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20:42:20 `q004. How are the two preceding questions actually different versions of the same question? How it is the mathematical reasoning the same in both cases?
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RESPONSE --> In order to solve either problem an average rate of change must be found. No matter if it is cm per second or dollars per month, it's the same concept of a rate of change. Also, comparing the rates of change between the two periods is used in both problems. Each one is essentially different numbers and/or units, but the same mathematical concepts applied.
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20:42:45 In each case we were given a quantity that changed with time. We were given the quantities at three different clock times and we were asked to compare the rates of change over the corresponding intervals. We did this by determining the changes in the quantities and the changes in the clock times, and for each interval dividing change in quantity by change in clock time. We could symbolize this process by representing the change in clock time by `dt and the change in the quantity by `dQ. The rate is thus rate = `dQ /`dt.
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RESPONSE --> ok
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