course I did not think I had submitted this lesson yet. I apologize if this has already been sent. I am still in the learning stage. MYCʈ~assignment #001
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11:09:17 `questionNumber 10000 `q001. Explain the difference between x - 2 / x + 4 and (x - 2) / (x + 4). The evaluate each expression for x = 2.
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RESPONSE --> The first expression will divide 2 by x first. The second expression will divide (x-2) by (x+4). The result will be different because the order of operations are different. confidence assessment: 3
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11:10:54 `questionNumber 10000 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 x or divide the -2 by 4 first. If there are no parentheses you have to divide before you subtract. Substituting 2 for x we get 2 - 2 / 2 + 4 = 2 - 1 + 4 (do multiplications and divisions before additions and subtractions) = 5 (add and subtract in indicated order) If there are parentheses you evaluate the grouped expressions first: (x - 2) / (x - 4) = (2 - 2) / ( 4 - 2) = 0 / 2 = 0.
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RESPONSE --> self critique assessment: 2
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11:17:10 `questionNumber 10000 `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 --> In the first expression 2 is raised to the x power and then you add four. Exponents must be done before addition. There are no parentheses around the x+4 so x is the exponent. In the second expression, 2 is raised to the (x+4) power. The parentheses around the (x+4). The result for the first expression 2 ^ x + 4 for x=2 is: 2^2 + 4 = 4 + 4 which is equal to 8. The result for the second expression for x=2 is: 2^(2+4) = 2 ^ 6 which is equal to 64. confidence assessment: 3
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11:18:03 `questionNumber 10000 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 understood the order of operations and showed my way of thinking in my answer. self critique assessment: 2
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11:37:13 `questionNumber 10000 `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 of the fraction is 3. Note that the x-3 is not in parentheses so 3 is the only thing being divided. The denominator is ((2x-5)^2 *3x +1. Note that there are no parentheses around the 3x+1 so we multiply by 3x before adding 1 in the denominator. The result is: 2-3/((2*2-5)^2*3*2+1)-2+7*2 2-3/((4-5)^2*3*6+1)-2+7*2 2 - 3/((-1)^2*6 + 1)-2 +14 2-3/((1*6 + 1) -2 + 14 2-3/((6+1))-2+14 2-3/(7))-2+14 1 + 4/7 - 2 + 14 -3/7 + 14 13 4/7 By order of operations, we must do what is in parentheses first, then exponents, then multiplication and division from left to right (whichever comes first), then addition and subtraction from left to right (whichever comes first). confidence assessment: 3
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11:38:26 `questionNumber 10000 The numerator is 3. x isn't part of the fraction. / indicates division, which must always precede subtraction. Only the 3 is divided by [ (2x-5)^2 * 3x + 1 ] and only [ (2x-5)^2 * 3x + 1 ] divides 3. If we mean (x - 3) / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x we have to write it that way. The preceding comments show that the denominator is [ (2x-5)^2 * 3x + 1 ] Evaluating the expression for x = 2: - 3 / [ (2 * 2 - 5)^2 * 3(2) + 1 ] - 2 + 7*2 = 2 - 3 / [ (4 - 5)^2 * 6 + 1 ] - 2 + 14 = evaluate in parenthese; do multiplications outside parentheses 2 - 3 / [ (-1)^2 * 6 + 1 ] -2 + 14 = add inside parentheses 2 - 3 / [ 1 * 6 + 1 ] - 2 + 14 = exponentiate in bracketed term; 2 - 3 / 7 - 2 + 14 = evaluate in brackets 13 4/7 or 95/7 or about 13.57 add and subtract in order. The details of the calculation 2 - 3 / 7 - 2 + 14: Since multiplication precedes addition or subtraction the 3/7 must be done first, making 3/7 a fraction. Changing the order of the terms we have 2 - 2 + 14 - 3 / 7 = 14 - 3/7 = 98/7 - 3/7 = 95/7. COMMON STUDENT QUESTION: ok, I dont understand why x isnt part of the fraction? And I dont understand why only the brackets are divided by 3..why not the rest of the equation? INSTRUCTOR RESPONSE: Different situations give us different algebraic expressions; the situation dictates the form of the expression. If the above expression was was written otherwise it would be a completely different expression and most likely give you a different result when you substitute. If we intended the numerator to be x - 3 then the expression would be written (x - 3) / [(2x-5)^2 * 3x + 1 ] - 2 + 7x, with the x - 3 grouped. If we intended the numerator to be the entire expression after the / the expression would be written x - 3 / [(2x-5)^2 * 3x + 1 - 2 + 7x ].
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RESPONSE --> self critique assessment: 2
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11:39:42 `questionNumber 10000 `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 --> confidence assessment:
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}ᢝْyܬ[ assignment #001 001. typewriter notation qa initial problems 08-17-2008 ϗe i assignment #001 001. typewriter notation qa initial problems 08-17-2008 Ӵ|ԋvǍ assignment #001 001. typewriter notation qa initial problems 08-17-2008 хCƾ̿Lƥ assignment #001 001. typewriter notation qa initial problems 08-17-2008
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11:46:29 `questionNumber 10000 `q001. Explain the difference between x - 2 / x + 4 and (x - 2) / (x + 4). The evaluate each expression for x = 2.
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RESPONSE --> confidence assessment:
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11:46:34 `questionNumber 10000 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 x or divide the -2 by 4 first. If there are no parentheses you have to divide before you subtract. Substituting 2 for x we get 2 - 2 / 2 + 4 = 2 - 1 + 4 (do multiplications and divisions before additions and subtractions) = 5 (add and subtract in indicated order) If there are parentheses you evaluate the grouped expressions first: (x - 2) / (x - 4) = (2 - 2) / ( 4 - 2) = 0 / 2 = 0.
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RESPONSE --> self critique assessment:
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11:46:39 `questionNumber 10000 `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 --> confidence assessment:
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11:46:48 `questionNumber 10000 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 --> self critique assessment:
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11:46:54 `questionNumber 10000 `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 --> confidence assessment:
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11:46:59 `questionNumber 10000 The numerator is 3. x isn't part of the fraction. / indicates division, which must always precede subtraction. Only the 3 is divided by [ (2x-5)^2 * 3x + 1 ] and only [ (2x-5)^2 * 3x + 1 ] divides 3. If we mean (x - 3) / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x we have to write it that way. The preceding comments show that the denominator is [ (2x-5)^2 * 3x + 1 ] Evaluating the expression for x = 2: - 3 / [ (2 * 2 - 5)^2 * 3(2) + 1 ] - 2 + 7*2 = 2 - 3 / [ (4 - 5)^2 * 6 + 1 ] - 2 + 14 = evaluate in parenthese; do multiplications outside parentheses 2 - 3 / [ (-1)^2 * 6 + 1 ] -2 + 14 = add inside parentheses 2 - 3 / [ 1 * 6 + 1 ] - 2 + 14 = exponentiate in bracketed term; 2 - 3 / 7 - 2 + 14 = evaluate in brackets 13 4/7 or 95/7 or about 13.57 add and subtract in order. The details of the calculation 2 - 3 / 7 - 2 + 14: Since multiplication precedes addition or subtraction the 3/7 must be done first, making 3/7 a fraction. Changing the order of the terms we have 2 - 2 + 14 - 3 / 7 = 14 - 3/7 = 98/7 - 3/7 = 95/7. COMMON STUDENT QUESTION: ok, I dont understand why x isnt part of the fraction? And I dont understand why only the brackets are divided by 3..why not the rest of the equation? INSTRUCTOR RESPONSE: Different situations give us different algebraic expressions; the situation dictates the form of the expression. If the above expression was was written otherwise it would be a completely different expression and most likely give you a different result when you substitute. If we intended the numerator to be x - 3 then the expression would be written (x - 3) / [(2x-5)^2 * 3x + 1 ] - 2 + 7x, with the x - 3 grouped. If we intended the numerator to be the entire expression after the / the expression would be written x - 3 / [(2x-5)^2 * 3x + 1 - 2 + 7x ].
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RESPONSE --> self critique assessment:
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11:52:30 `questionNumber 10000 `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 --> I first write the problem as: (x-5)^2x -1 + 3/x -2 for x=4 then substitue 4 for x (4-5)^(2*4) -1 + (3/4) - 2 do inside par. first (-1)^8 -1 + (3/4) -2 calc. exponents 1-1 + (3/4) -2 we then do add. and sub. 0 + (3/4) -2 (3/4) - 2 -1 1/4 or -5/4 confidence assessment: 3
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11:55:31 `questionNumber 10000 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 --> I misunderstood the exponent. I took 2x as the exponent instead of just 2. I will also learn to write down the original problem so I can compare it to how I wrote it out in standard form. I do see where I made the mistake. self critique assessment: 1
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11:56:48 `questionNumber 10000 *&*& Standard mathematics notation is easier to see. On the other hand it's very important to understand order of operations, and students do get used to this way of doing it. You should of course write everything out in standard notation when you work it on paper. It is likely that you will at some point use a computer algebra system, and when you do you will have to enter expressions through a typewriter, so it is well worth the trouble to get used to this notation. Indicate your understanding of the necessity to understand this notation.
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RESPONSE --> I wrote how I made my mistake in the previous question. self critique assessment: 1
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12:05:54 `questionNumber 10000 `q005. At the link http://www.vhcc.edu/dsmith/genInfo/introductory problems/typewriter_notation_examples_with_links.htm (copy this path into the Address box of your Internet browser; alternatively use the path http://vhmthphy.vhcc.edu/ > General Information > Startup and Orientation (either scroll to bottom of page or click on Links to Supplemental Sites) > typewriter notation examples and you will find a page containing a number of additional exercises and/or examples of typewriter notation.Locate this site, click on a few of the links, and describe what you see there.
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RESPONSE --> These links show you some examples of how to translate the problems. Also, there is a link for you to practice and then check your answer. confidence assessment: 3
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12:07:10 `questionNumber 10000 You should see a brief set of instructions and over 30 numbered examples. If you click on the word Picture you will see the standard-notation format of the expression. The link entitled Examples and Pictures, located in the initial instructions, shows all the examples and pictures without requiring you to click on the links. There is also a file which includes explanations. The instructions include a note indicating that Liberal Arts Mathematics students don't need a deep understanding of the notation, Mth 173-4 and University Physics students need a very good understanding,
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RESPONSE --> I think I answered it correctly in the previous question. self critique assessment: 2
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12:07:32 `questionNumber 10000 while students in other courses should understand the notation and should understand the more basic simplifications. There is also a link to a page with pictures only, to provide the opportunity to translated standard notation into typewriter notation.
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RESPONSE --> self critique assessment: 2
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12:07:55 `questionNumber 10000 end program
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RESPONSE --> self critique assessment: 2
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ΧxSĻpy|xzѓ assignment #001 001. Areas qa areas volumes misc 08-17-2008 cڢ̟⬥ assignment #002 002. Describing Graphs qa initial problems 08-17-2008
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20:51:45 `questionNumber 20000 `q001. You will frequently need to describe the graphs you have constructed in this course. This exercise is designed to get you used to some of the terminology we use to describe graphs. Please complete this exercise and email your work to the instructor. Note that you should do these graphs on paper without using a calculator. None of the arithmetic involved here should require a calculator, and you should not require the graphing capabilities of your calculator to answer these questions. Problem 1. We make a table for y = 2x + 7 as follows: We construct two columns, and label the first column 'x' and the second 'y'. Put the numbers -3, -2, -1, -, 1, 2, 3 in the 'x' column. We substitute -3 into the expression and get y = 2(-3) + 7 = 1. We substitute -2 and get y = 2(-2) + 7 = 3. Substituting the remaining numbers we get y values 5, 7, 9, 11 and 13. These numbers go into the second column, each next to the x value from which it was obtained. We then graph these points on a set of x-y coordinate axes. Noting that these points lie on a straight line, we then construct the line through the points. Now make a table for and graph the function y = 3x - 4. Identify the intercepts of the graph, i.e., the points where the graph goes through the x and the y axes.
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RESPONSE --> We make a table for y=3x-4 .We make a table with 2 columns. One for the x values (input values) and the other for the y values (output values). The first column is labeled x and the second column is labeled y. Put the numbers -3, -2, -1, 0, 1, 2, 3 in the x column. We substitute -3 into the expression and get y=3(-3) - 4 = -13. We substitute -2 and get y = 3(-2) - 4 = -10. Substituting the remaining numbers we get y values -7, -4, -1, 2, 5. These numbers go into the second column, each next to the value from which it was obtained. We then graph these points on a set of x-y coordinate axes. Noting that this is a linear equation and the points lie on a straight line, we then construct the line through the points. The x-intercept is a point where the y value is zero. The y-intercept is a point where the x-value is zero. We can find the x-intercept by substituting 0 for y. y=3x-4 0=3x-4 add 4 to both sides 4=3x divide both sides by 3 4/3 = x Thus, the (4/3, 0) is the x-intercept We can find the y-intercept by substituting 0 for x. y=3x-4 y=3(0) - 4 Sub. 0 for x multiply 3(0) y=0-4 subtract y = -4 Thus, the y-intercept is (0, -4) Note: The y-intercept can be found just by looking at the linear equation when the equation is in slope intercept form y=mx + b where b represents the y-intercept. confidence assessment: 3
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20:53:07 `questionNumber 20000 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 feel that I correctly answered and explained my answer sufficiently. I did not know whether to try to graph this on a word program. self critique assessment: 2
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20:58:03 `questionNumber 20000 `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 --> As the value of x increases, so does the value of y increase. The slope (steepness) of the line is 3/1. This means for every 3 units we move up on the y axis, we only move one to the right on the x axis. This graph has a positive slope. As we go to the right, the value of y increases. confidence assessment: 3
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20:59:14 `questionNumber 20000 The graph forms a straight line with no change in steepness.
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RESPONSE --> I should have paid more attention to the word ""change"" in the question. I think I might have explained more than was necessary. self critique assessment: 2
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21:00:57 `questionNumber 20000 `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 --> Since the equation is in slope intercept form, y=mx + b where m represents the slope it is obvious that the slope is 3/1 confidence assessment: 3
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21:02:40 `questionNumber 20000 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 --> I used the fact that is was in slope intercept form. I did not choose two points and work out the equation m =( y2-y1)/(x2-x1) self critique assessment: 2
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21:12:24 `questionNumber 20000 `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 --> We first make an x, y table for the equation y=x^2 for x integers between x=0 and x=3. We substitute the 0 for x in the equation y=x^2 y=(0)^2 y=0 Thus a point is (0,0) (The origin) We substitute the value of 1 in for x y=x^2 y=(1)^2 y=1 Thus a point is (1,1) We substitute the numbers 2 and 3 in for x and we get 4 and 9, respectively. The graph is increasing. The steepness of the graph does change but it is not at a constant rate. The steepness of the graph is increasing at an increasing rate. It increases exponentially. confidence assessment: 3
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21:16:25 `questionNumber 20000 Graph points include (0,0), (1,1), (2,4) and (3,9). The y values are 0, 1, 4 and 9, which increase as we move from left to right. The increases between these points are 1, 3 and 5, so the graph not only increases, it increases at an increasing rate STUDENT QUESTION: I understand increasing...im just not sure at what rate...how do you determine increasing at an increasing rate or a constant rate? INSTRUCTOR RESPONSE: Does the y value increase by the same amount, by a greater amount or by a lesser amount every time x increases by 1? In this case the increases get greater and greater. So the graph increases, and at an increasing rate. *&*&.
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RESPONSE --> For a constant rate, the y values would increase by the same number for each sequential integer substituted for x. The graph of the equation would be a straight line for a constant rate of change. The curve gets steeper as we go to the right of the graph, thus it is increasing at an increasing rate. self critique assessment: 2
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21:26:11 `questionNumber 20000 `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 --> We first make an x-y chart by substituting the values -3, -2, -1, 0 into the equation, y=x^2, for x. We substitute -3 for x, y=(-3)^2 y=9 Thus, (-3, 9) is a point on the graph. We substitute -2 for x y=(-2)^2 y=4 Thus, (-2,4) is a point on the graph We substitute -1 and get 1. Thus (-1,1) is a point on the graph. We substitute 0 for x and get 0. Thus (0,0) is a point on the graph. Notice in the ordered pairs, as the value of x increases the value of y decreases. Thus, the graph is decreasing. The values of y get closer together as we move to the right. So the graph decreases at a decreasing rate. confidence assessment: 3
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21:26:46 `questionNumber 20000 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 --> I feel my answer was sufficient. self critique assessment:
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21:35:46 `questionNumber 20000 `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 --> We first make an x-y chart for y='sqrt(x) We substitute the values of 0, 1, 2, 3 for x and find the y-values. We get the y values of 0, 1, and approximately 1.414, 1.732. The graph is increasing. As the values of x increase, the y values also increase. The graph is steeper in the beginning and then starts to level out. Thus the graph is increasing at a decreasing rate. confidence assessment: 3
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21:37:26 `questionNumber 20000 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 --> I felt that I answered the question sufficiently. I am still in doubt whether I should learn to graph it on the word processer and insert it in the answers. self critique assessment: 2
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22:15:23 `questionNumber 20000 `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 --> I first rewrite the equation as y= 5 (2)^(-x). I then make an x-y chart. Substitute 0 for x y=5(1) y=5 Thus, a point on the graph is (0,5) Sub. 1 for x y=5(2)^(-1) y=5 (1/2) y=5/2 Thus, a point on the graph is (1, 5/2) Substituting 2 for x, I get 5/4 for y Thus, (2,5/4) is a point on the graph. Substituting 3 for x, I get 5/8 for y. thus (3, 5/8) is a point on the graph. The graph is decreasing. The steepness of the graph does change. The steepness decreases. The graph is decreasing at a decreasing rate. confidence assessment: 3
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22:16:18 `questionNumber 20000 ** 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 am going to try to see if it is sufficient to give #3 for this reply. self critique assessment: 3
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22:20:32 `questionNumber 20000 `q008. Suppose you stand still in front of a driveway. A car starts out next to you and moves away from you, traveling faster and faster. If y represents the distance from you to the car and t represents the time in seconds since the car started out, would a graph of y vs. t be increasing or decreasing? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> The graph would be increasing. As time increases the distance also increases. Since the car gains speed, traveling faster and faster, I would translate that to mean that the car is gaining momentum and it is getting farther away at an increasing rate. Thus, I would say it is increasing at an increasing rate. confidence assessment: 2
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22:20:54 `questionNumber 20000 ** 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 --> self critique assessment: 2
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22:21:09 `questionNumber 20000 end program
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RESPONSE --> self critique assessment:
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WĉyJӥ assignment #002 002. Volumes qa areas volumes misc 08-17-2008 UUʷeOy assignment #006 006. Physics qa initial problems 08-18-2008
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17:26:48 `questionNumber 60000 `q001. There are two parts to this problem. Reason them out using common sense. If the speed of an automobile changes by 2 mph every second, then how long will it take the speedometer to move from the 20 mph mark to the 30 mph mark? Given the same rate of change of speed, if the speedometer initially reads 10 mph, what will it read 7 seconds later?
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RESPONSE --> If the initial speed is 20mph and it changes 2mph every second, then at 1 sec = 22 mph 2 sec = 24 mph 3 sec = 26 mph 4 sec = 28 mph 5 sec = 30 mph So it will take the speedometer 5 seconds to move from the 20 mph to the 30 mph mark. There is an easier way to get the answer: The interval the speedometer should move is 10 mph to get from 20 mph to 30 mph. Since it takes 1 second to move 2 mph, then you could use ratios and proportions to solve. (1 sec/2 mph) = (x /10 mph). Use cross multiplication and solve. The answer is 5 seconds Since the speedometer moves 2mph every second, then it should move 14 mph in 7 seconds. Since the speedometer initially reads 10 mph and you add 14 mph to that, you get 24 mph. confidence assessment: 3
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17:27:10 `questionNumber 60000 It will take 5 seconds to complete the change. 30 mph - 20 mph = 10 mph change at 2 mph per second (i.e., 2 mph every second) implies 5 seconds to go from 20 mph to 30 mph Change in speed is 2 mph/second * 7 seconds = 14 mph Add this to the initial 10 mph and the speedometer now reads 24 mph.
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RESPONSE --> self critique assessment: 3
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17:39:07 `questionNumber 60000 `q002. An automobile traveling down a hill passes a certain milepost traveling at a speed of 10 mph, and proceeds to coast to a certain lamppost further down the hill, with its speed increasing by 2 mph every second. The time required to reach the lamppost is 10 seconds. It then repeats the process, this time passing the milepost at a speed of 20 mph. Will the vehicle require more or less than 10 seconds to reach the lamppost? Since its initial speed was 10 mph greater than before, does it follow that its speed at the lamppost will be 10 mph greater than before?
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RESPONSE --> It will require less than 10 seconds because it is traveling faster. Yes, it would follow that its speed at the lamppost will be 10 mph greater than before because the speed is increasing at a constant rate of 2mph. confidence assessment: 1
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17:40:42 `questionNumber 60000 If it starts coasting down the same section of road at 20 mph, and if velocity changes by the same amount every second, the automobile should always be traveling faster than if it started at 10 mph, and would therefore take less than 10 seconds. The conditions here specify equal distances, which implies less time on the second run. The key is that, as observed above, the automobile has less than 10 seconds to increase its speed. Since its speed is changing at the same rate as before and it has less time to change it will therefore change by less.
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RESPONSE --> I was not thinking in terms of same distance. The explanation was necessary. self critique assessment: 1
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17:52:17 `questionNumber 60000 `q003. The following example shows how we can measure the rate at which an automobile speeds up: If an automobile speeds up from 30 mph to 50 mph as the second hand of a watch moves from the 12-second position to the 16-second position, and its speed changes by 20 mph in 4 seconds. This gives us an average rate of velocity change equal to 20 mph / 4 seconds = 5 mph / second. We wish to compare the rates at which two different automobiles increase their speed: Which automobile speeds up at the greater rate, one which speeds up from 20 mph to 30 mph in five seconds or one which speeds up from 40 mph to 90 mph in 20 seconds?
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RESPONSE --> An automobile which speeds up from 20 mph to 30 mph in 5 seconds has an average rate of velocity change equal to 10 mph / 5 sec = 2 mph / sec An automoblie which speeds up from 40 mph to 90 mph in twenty seconds has an averge rate of velocity change equal to 50 mph / 20 sec = 2.5 mph / sec So the automobile which speeds up from 40 mph to 90 mph in twenty seconds speeds up at the greater rate. confidence assessment: 3
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17:52:45 `questionNumber 60000 The first automobile's speed changes from 20 mph to 30mph, a 10 mph difference, which occurs in 5 seconds. So the rate of chage in 10 mph / (5 sec) = 2 mph / sec. = rate of change of 2 mph per second. }{The second automobile's speed changes from 40 mph to 90 mph, a 50 mph difference in 20 seconds so the rate of change is 50 mph / (20 sec) = 2.5 mph per second. Therefore, the second auto is increasing its velocity ar a rate which is .5 mph / second greater than that of the first.
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RESPONSE --> self critique assessment: 3
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18:05:06 `questionNumber 60000 4. If an automobile of mass 1200 kg is pulled by a net force of 1800 Newtons, then the number of Newtons per kg is 1800 / 1200 = 1.5. The rate at which an automobile speeds up is determined by the net number of Newtons per kg. Two teams pulling on ropes are competing to see which can most quickly accelerate their initially stationary automobile to 5 mph. One team exerts a net force of 3000 Newtons on a 1500 kg automobile while another exerts a net force of 5000 Newtons on a 2000 kg automobile. Which team will win and why? If someone pulled with a force of 500 Newtons in the opposite direction on the automobile predicted to win, would the other team then win?
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RESPONSE --> The team that exerted a net force of 3,000 Newtons on a 1,500 kg automobile: 3,000 Newtons/ 1,500 kg = 2 Newtons/kg The other team exerted a net force of 5,000 Newtons/2,000 kg = 2.5 Newtons / kg The team that exerted a net force of 5,000 Newtons on a 2,000 kg automobile. If a team pulled with a force of 500 Newtons in the opposite direction, then the net force would be 5,000 Newtons - 500 Newtons = 4,500 Newtons 4,500 Newtons / 2,000 kg = 2.25 Newtons/kg No, the other team still would not win. self critique assessment: 2
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18:05:49 `questionNumber 60000 The first team's rate is 3000 Newtons divided by 1500 kg or 2 Newtons per kg, while the second team's rate is 5000 Newtons divided by 2000 kg or 2.5 Newtons per kg. The second team therefore increases velocity more quickly. Since both start at the same velocity, zero, the second team will immediately go ahead and will stay ahead. The second team would still win even if the first team was hampered by the 500 Newton resistance, because 5000 Newtons - 500 Newtons = 4500 Newtons of force divided by 2000 kg of car gives 2.25 Newtons per kg, still more than the 2 Newtons / kg of the first team
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RESPONSE --> self critique assessment: 3
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18:12:09 `questionNumber 60000 `q005. Both the mass and velocity of an object contribute to its effectiveness in a collision. If a 250-lb football player moving at 10 feet per second collides head-on with a 200-lb player moving at 20 feet per second in the opposite direction, which player do you precidt will be moving backward immediately after the collision, and why?
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RESPONSE --> The 250 lb player movting at 10 ft per sec has a force of 250 lb/10 sec = 25 lb/ft. The other play is 200 lb at 20 feet per second has a force of 200 lb/20 sec = 10 lb/ft. Thus, the 250 lb player has more force. Consequently, the 200 lb player will be moving backward immediately after the collision. confidence assessment: 2
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18:13:23 `questionNumber 60000 Greater speed and greater mass both provide advantages. In this case the player with the greater mass has less speed, so we have to use some combination of speed and mass to arrive at a conclusion. It turns out that if we multiply speed by mass we get the determining quantity, which is called momentum. 250 lb * 10 ft/sec = 2500 lb ft / sec and 200 lb * 20 ft/sec = 4000 lb ft / sec, so the second player will dominate the collision. In this course we won't use pounds as units, and in a sense that will become apparent later on pounds aren't even valid units to use here. However that's a distinction we'll worry about when we come to it.
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RESPONSE --> The explanation was necessary. I got the right answer but was thinking incorrectly. self critique assessment: 1
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18:19:26 `questionNumber 60000 `q006. Two climbers eat Cheerios for breakfast and then climb up a steep mountain as far as they can until they use up all their energy from the meal. All other things being equal, who should be able to climb further up the mountain, the 200-lb climber who has eaten 12 ounces of Cheerios or the 150-lb climber who has eaten 10 ounces of Cheerios?
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RESPONSE --> 200 lb/ 12 oz = 16 2/3 lb/oz 150 lb / 10 oz = 15 lb/oz The 150 lb climber is only having to carry 15 lb/oz and the 200 lb climber is having to carry 16 2/3 lb/oz So the 150 lb climber should be able to climb further up the mountain. confidence assessment: 0
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18:20:32 `questionNumber 60000 The comparison we make here is the number of ounces of Cheerios per pound of body weight. We see that the first climber has 12 oz / (200 lb) = .06 oz / lb of weight, while the second has 10 0z / (150 lb) = .067 oz / lb. The second climber therefore has more energy per pound of body weight. It's the ounces of Cheerios that supply energy to lift the pounds of climber. The climber with the fewer pounds to lift for each ounce of energy-producing Cheerios will climb further.
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RESPONSE --> The correct answer for the wrong reason. self critique assessment: 1
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18:23:49 `questionNumber 60000 `q007. Two automobiles are traveling up a long hill with an steepness that doesn't change until the top, which is very far away, is reached. One automobile is moving twice as fast as the other. At the instant the faster automobile overtakes the slower their drivers both take them out of gear and they coast until they stop. Which automobile will take longer to come to a stop? Will that automobile require about twice as long to stop, more than twice as long or less than twice as long? Which automobile will have the greater average coasting velocity? Will its average coasting velocity by twice as great as the other, more than twice as great or less than twice as great? Will the distance traveled by the faster automobile be equal to that of the slower, twice that of the slower or more than twice that of the slower?
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RESPONSE --> The automobile moving at the faster rate will take longer to stop because it has more momentum when it overtakes the slower one. It will take about twice as long to stop. confidence assessment: 1
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18:24:27 `questionNumber 60000 It turns out that, neglecting air resistance, since the slope is the same for both, both automobiles will change velocity at the same rate. So in this case the second would require exactly twice as long. If you include air resistance the faster car experiences more so it actually takes a bit less than twice as long as the slower. For the same reasons as before, and because velocity would change at a constant rate (neglecting air resistance) it would be exactly twice as great if air resistance is neglected. Interestingly if it takes twice as much time and the average velocity is twice as great the faster car travels four times as far. If there is air resistance then it slows the faster car down more at the beginning than at the end and the average velocity will be a bit less than twice as great and the coasting distance less than four times as far.
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RESPONSE --> self critique assessment: 2
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18:35:56 `questionNumber 60000 `q008. When a 100 lb person hangs from a certain bungee cord, the cord stretches by 5 feet beyond its initial unstretched length. When a person weighing 150 lbs hangs from the same cord, the cord is stretched by 9 feet beyond its initial unstretched length. When a person weighing 200 lbs hangs from the same cord, the cord is stretched by 12 feet beyond its initial unstretched length. Based on these figures, would you expect that a person of weight 125 lbs would stretch the cord more or less than 7 feet beyond its initial unstretched length?
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RESPONSE --> I think that a person of weight 125 lbs would stretch the cord more than 7 feet beyond its initial unstretched length. The rate that the cord stretches is not a constant rate. I think there is less resistance with the lighter person and the cord stretches more. confidence assessment: 0
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18:37:01 `questionNumber 60000 From 100 lbs to 150 lbs the stretch increased by 4 feet, from 150 lbs to 200 lbs the increase was only 3 feet. Thus it appears that at least in the 100 lb - 200 lb rands each additional pound results in less increase in length than the last and that there would be more increase between 100 lb and 125 lb than between 125 lb and 150 lb. This leads to the conclusion that the stretch for 125 lb would be more than halfway from 5 ft to 9 ft, or more than 7 ft. A graph of stretch vs. weight would visually reveal the nature of the nonlinearity of this graph and would also show that the stretch at 125 lb must be more than 7 feet (the graph would be concave downward, or increasing at a decreasing rate, so the midway stretch would be higher than expected by a linear approximation).
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RESPONSE --> The right answer but the wrong explanation. self critique assessment: 1
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18:44:36 `questionNumber 60000 `q009. When given a push of 10 pounds, with the push maintained through a distance of 4 feet, a certain ice skater can coast without further effort across level ice for a distance of 30 feet. When given a push of 20 pounds (double the previous push) through the same distance, the skater will be able to coast twice as far, a distance of 60 feet. When given a push of 10 pounds for a distance of 8 feet (twice the previous distance) the skater will again coast a distance of 60 feet. The same skater is now accelerated by a sort of a slingshot consisting of a bungee-type cord slung between two posts in the ice. The cord, as one might expect, exerts greater and greater force as it is pulled back further and further. Assume that the force increases in direct proportion to pullback (ie.g., twice the pullback implies twice the force). When the skater is pulled back 4 feet and released, she travels 20 feet. When she is pulled back 8 feet and released, will she be expected to travel twice as far, more than twice as far or less than twice as far as when she was pulled back 4 feet?
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RESPONSE --> Since twice the pull back inplies twice the force, the skater will be expected to travel twice as far. confidence assessment: 0
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18:45:24 `questionNumber 60000 The distance through which the force acts will be twice as great, which alone would double the distance; because of the doubled pullback and the linear proportionality relationship for the force the average force is also twice as great, which alone would double the distance. So we have to double the doubling; she will go 4 times as far
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RESPONSE --> I did not get this one right. self critique assessment: 1
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18:47:30 `questionNumber 60000 `q010. Two identical light bulbs are placed at the centers of large and identically frosted glass spheres, one of diameter 1 foot and the other of diameter 2 feet. To a moth seeking light from half a mile away, unable to distinguish the difference in size between the spheres, will the larger sphere appear brighter, dimmer or of the same brightness as the first? To a small moth walking on the surface of the spheres, able to detect from there only the light coming from 1 square inch of the sphere, will the second sphere appear to have the same brightness as the first, twice the brightness of the first, half the brightness of the first, more than twice the brightness of the first, or less than half the brightness of the first?
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RESPONSE --> The larger sphere will appear brighter. It will appear to have the same. confidence assessment: 0
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18:48:11 `questionNumber 60000 Both bulbs send out the same energy per second. The surface of the second bulb will indeed be dimmer than the first, as we will see below. However the same total energy per second reaches the eye (identically frosted bulbs will dissipate the same percent of the bulb energy) and from a great distance you can't tell the difference in size, so both will appear the same. The second sphere, while not as bright at its surface because it has proportionally more area, does have the extra area, and that exactly compensates for the difference in brightness. Specifically the brightness at the surface will be 1/4 as great (twice the radius implies 4 times the area which results in 1/4 the illumination at the surface) but there will be 4 times the surface area. Just as a 2' x 2' square has four times the area of a 1' x 1' square, a sphere with twice the diameter will have four times the surface area and will appear 1 / 4 as bright at its surface. Putting it another way, the second sphere distributes the intensity over four times the area, so the light on 1 square inch has only 1 / 4 the illumination.
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RESPONSE --> self critique assessment: 0
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18:51:24 `questionNumber 60000 `q011. The water in a small container is frozen in a freezer until its temperature reaches -20 Celsius. The container is then placed in a microwave oven, which proceeds to deliver energy at a constant rate of 600 Joules per second. After 10 seconds the ice is still solid and its temperature is -1 Celsius. After another 10 seconds a little bit of the cube is melted and the temperature is 0 Celsius. After another minute most of the ice is melted but there is still a good bit of ice left, and the ice and water combination is still at 0 Celsius. After another minute all the ice is melted and the temperature of the water has risen to 40 degrees Celsius. Place the following in order, from the one requiring the least energy to the one requiring the most: Increasing the temperature of the ice by 20 degrees to reach its melting point. Melting the ice at its melting point. Increasing the temperature of the water by 20 degrees after all the ice melted. At what temperature does it appear ice melts, and what is the evidence for your conclusion?
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RESPONSE --> Melting the ice at its melting point Increasing the temp. of the water by 20 degrees after all the ice melted. Increasing the temp of the ice by 20 degrees to reach its melting point. confidence assessment: 0
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18:52:30 `questionNumber 60000 Since the temperature is the same when a little of the ice is melted as when most of it is melted, melting takes place at this temperature, which is 0 Celsius. The time required to melt the ice is greater than any of the other times so melting at 0 C takes the most energy. Since we don't know how much ice remains unmelted before the final minute, it is impossible to distinguish between the other two quantities, but it turns out that it takes less energy to increase the temperature of ice than of liquid water.
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RESPONSE --> ??????? self critique assessment: 0
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18:54:42 `questionNumber 60000 `q012. Suppose you are in the center of a long, narrow swimming pool (e.g., a lap pool). Two friends with kickboards are using them to push waves in your direction. Their pushes are synchronized, and the crests of the waves are six feet apart as they travel toward you, with a 'valley' between each pair of crests. Since your friends are at equal distances from you the crests from both directions always reach you at the same instant, so every time the crests reach you the waves combine to create a larger crest. Similarly when the valleys meet you experience a larger valley, and as a result you bob up and down further than you would if just one person was pushing waves at you. Now if you move a bit closer to one end of the pool the peak from that end will reach you a bit earlier, and the peak from the other end will reach you a little later. So the peaks won't quite be reaching you simultaneously, nor will the valleys, and you won't bob up and down as much. If you move far enough, in fact, the peak from one end will reach you at the same time as the valley from the other end and the peak will 'fll in' the valley, with the result that you won't bob up and down very much. If the peaks of the approaching waves are each 6 inches high, how far would you expect to bob up and down when you are at the center point? How far would you have to move toward one end or the other in order for peaks to meet valleys, placing you in relatively calm water?
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RESPONSE --> None. ???? confidence assessment: 0
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18:55:35 `questionNumber 60000 If the two 6-inch peaks meet and reinforce one another completely, the height of the 'combined' peak will be 6 in + 6 in = 12 in. If for example you move 3 ft closer to one end you move 3 ft further from the other and peaks, which are 6 ft apart, will still be meeting peaks. [ Think of it this way: If you move 3 ft closer to one end you move 3 ft further from the other. This shifts your relative position to the two waves by 6 feet (3 feet closer to the one you're moving toward, 3 feet further from the other). So if you were meeting peaks at the original position, someone at your new position would at the same time be meeting valleys, with two peaks closing in from opposite directions. A short time later the two peaks would meet at that point. ] However if you move 1.5 ft the net 'shift' will be 3 ft and peaks will be meeting valleys so you will be in the calmest water.
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RESPONSE --> ??????? self critique assessment: 0
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nyzSإzʋ assignment #001 001. Rates qa rates 08-18-2008
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19:08:12 `questionNumber 10000 `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 --> Note: I do not see a ""Display Everything"" button. The third button looks messed up...I cannot tell what it says it is. I should copy and paste these instructions to a word processor. Click on next question to get a new question to pop up. I should answer that question in the right box before looking at the answer. I click on enter response to record my answer. Do not tamper with the information in the left box. If my answer is incorrect, incomplete or requires revision, I need to enter a self-critique. If I learned something from the answer, I should note it. If I want my response in my Notes file, I click on the Save As Notes button. I then click on the next question button and proceed in the same manner. confidence assessment: 3
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19:08:44 `questionNumber 10000 Your answer has been noted. Enter 'ok' in the Response Box and click on Enter Response, then click on Next Question/Answer for the first real question.
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RESPONSE --> ok confidence assessment: 3
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19:09:43 `questionNumber 10000 `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 --> 50 $ / 5 hr = 10 $/ hr confidence assessment: 3
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19:10:19 `questionNumber 10000 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 --> Understood. confidence assessment: 3
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19:11:17 `questionNumber 10000 `q003.If you make $60,000 per year then how much do you make per month?
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RESPONSE --> 60,000 $ / 12 mth = 5,000 $ / mth confidence assessment: 3
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19:11:45 `questionNumber 10000 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 --> Understood. confidence assessment: 3
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19:13:00 `questionNumber 10000 `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 --> The business makes an average of 5,000 $ / mth We only know what it made the entire year. So we are only giving an average. confidence assessment: 3
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19:13:24 `questionNumber 10000 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 --> Understood. confidence assessment: 3
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19:15:50 `questionNumber 10000 `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 --> While traveling, the speeds are at different rates because of traffic conditions, interstate vs. state routes, and etc. 300 mi/ 6 hr = 50 mi/hr We are looking at the total distance, so we can only say what the average is. confidence assessment: 3
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19:16:20 `questionNumber 10000 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 --> Understood. confidence assessment: 3
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19:18:09 `questionNumber 10000 `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 --> 1200 mi/ 60 gal = 20 mi/gal confidence assessment: 3
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19:19:32 08-18-2008 19:19:32 The rate of change of one quantity with respect to another is the change in the first quantity, divided by the change in the second. As in previous examples, we found the rate at which money was made with respect to time by dividing the amount of money made by the time required to make it. By analogy, the rate at which we use fuel with respect to miles traveled is the change in the amount of fuel divided by the number of miles traveled. In this case we use 60 gallons of fuel in 1200 miles, so the average rate it 60 gal / (1200 miles) = .05 gallons / mile. Note that this question didn't ask for miles per gallon. Miles per gallon is an appropriate and common calculation, but it measures the rate at which miles are covered with respect to the amount of fuel used. Be sure you see the difference. Note that in this problem we again have here an example of a rate, but unlike previous instances this rate is not calculated with respect to time. This rate is calculated with respect to the amount of fuel used. We divide the accumulated quantity, in this case miles, by the amount of fuel required to cover those miles. Note that again we call the result of this problem an average rate because there are always at least subtle differences in driving conditions that require the use of more fuel on some miles than on others. It's very important to understand the phrase 'with respect to'. Whether the calculation makes sense or not, it is defined by the order of the terms. In this case gallons / mile tells you how many gallons you are burning, on the average, per mile. This concept is not as familiar as miles / gallon, but except for familiarity it's technically no more difficult. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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NOTES -------> I inverted the miles and the gallons.
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19:20:12 `questionNumber 10000 `q007. The word 'average' generally connotes something like adding two quantities and dividing by 2, or adding several quantities and dividing by the number of quantities we added. Why is it that we are calculating average rates but we aren't adding anything?
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RESPONSE --> We are already given the totals, so we aren't having to add anything. confidence assessment: 3
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19:20:51 `questionNumber 10000 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 --> Understood. confidence assessment: 3
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19:24:44 `questionNumber 10000 `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 --> 2.23 lbs/day confidence assessment: 0
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19:26:28 08-18-2008 19:26:28 The second group had 15 pounds more lifting strength as a result of doing 40 more daily pushups than the first. The desired rate is therefore 15 pounds / 40 pushups = .375 pounds / pushup. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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NOTES -------> I should have looked at the difference of the two groups for the number of pounds and the number of pushups. Then dividing 15 lb/40 PU would give me .375 lbs/PU
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21:05:16 `questionNumber 10000 `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 --> 30 PU - 10 PU = 20 PU 188 lb - 171 lb = 17 lb 20 PU/17 lb = 1.17 PU/lb confidence assessment: 0
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21:06:32 `questionNumber 10000 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 --> with respect to the added weight....I should have put the added weight in the denominator. confidence assessment:
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21:09:10 `questionNumber 10000 `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 --> 200 m - 100 m = 100 m 22 sec - 12 sec = 10 sec 100m/10sec = 10 m / sec confidence assessment: 2
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21:09:36 `questionNumber 10000 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 --> Understood confidence assessment: 3
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21:16:40 `questionNumber 10000 `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 10 seconds to cover the first 100 yards. It takes 11.1 seconds to cover the next 100 yards. (10 sec + 11.1 sec)/2 = 10.5 sec confidence assessment: 1
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21:18:34 `questionNumber 10000 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 --> confidence assessment:
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21:19:18 `questionNumber 10000 `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 --> We already had the rates so all we had to do was average the rates. confidence assessment: 1
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21:19:44 `questionNumber 10000 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 --> Understood confidence assessment: 2
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