#$&* course PHY201 1-14-2014 7:45pm Here are the remaining ten questions:*********************************************
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Given Solution: Many students simply know, at the level of common sense, that if we divide $72 by $8 / hour we get 9 hours, so 9 hours are required. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q004. Calculate (8 + 3) * 5 and 8 + 3 * 5, indicating the order of your steps. Explain, as best you can, the reasons for the difference in your results. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (type in your solution starting in the next line) I still remember one of my teachers using the “Please Excuse My Dear Aunt Sally” mnemonic to help us with the order of operations. (8 + 3) * 5 = 55, parenthesis first than multiply. 8 + 3 * 5 = 23, Multiply first than add. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: (8 + 3) * 5 and 8 + 3 * 5 To evaluate (8 + 3) * 5, you will first do the calculation in parentheses. 8 + 3 = 11, so (8 + 3) * 5 = 11 * 5 = 55. To evaluate 8 + 3 * 5 you have to decide which operation to do first, 8 + 3 or 3 * 5. You should be familiar with the order of operations, which tells you that multiplication precedes addition. The first calculation to do is therefore 3 * 5, which is equal to 15. Thus 8 + 3 * 5 = 8 + 15 = 23 The results are different because the grouping in the first expression dictates that the addition be done first. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q005. Calculate (2^4) * 3 and 2^(4 * 3), indicating the order of your steps. Explain, as best you can, the reasons for the difference in your results. Note that the symbol '^' indicates raising to a power. For example, 4^3 means 4 raised to the third power, which is the same as 4 * 4 * 4 = 64. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ( 2 ^ 4 ) * 3 = 48. Parenthesis than multiply. ( 2 ^ 4 ) = 16. 16 * 3 = 48. 2 ^ ( 4 * 3 ) = 4096. Parenthesis than exponents . ( 4 * 3 ) = 12. 2 ^ 12 = 4096. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: To evaluate (2^4) * 3 we first evaluate the grouped expression 2^4, which is the fourth power of 2, equal to 2 * 2 * 2 * 2 = 16. So we have (2^4) * 3 = 16 * 3 = 48. To evaluate 2^(4 * 3) we first do the operation inside the parentheses, obtaining 4 * 3 = 12. We therefore get 2^(4 * 3) = 2^12 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 = 4096. It is easy to multiply by 2, and the powers of 2 are important, so it's appropriate to have asked you to do this problem without using a calculator. Had the exponent been much higher, or had the calculation been, say, 3^12, the calculation would have become tedious and error-prone, and the calculator would have been recommended. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ********************************************* Question: `q006. Calculate 3 * 5 - 4 * 3 ^ 2 and 3 * 5 - (4 * 3)^2 according to the standard order of operations, indicating the order of your steps. Explain, as best you can, the reasons for the difference in your results. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 3 * 5 - 4 * 3 ^ 2 = -21. Exponent, multiply, then subtract. 3 ^ 2 = 9. 3 * 5 = 15 and 4 * 9 = 36. 15 - 36 = -21. 3 * 5 - (4 * 3)^2 = 129. Parenthesis, Exponents, multiply, subtract. ( 4 * 3 ) = 12. 12 ^ 2 = 144. 3 * 5 = 15. 15 - 144 = -129. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: To calculate 3 * 5 - 4 * 3 ^ 2, the first operation is the exponentiation operation ^. • The two numbers involved in the exponentiation are 3 and 2; the 4 is 'attached' to the 3 by multiplication, and this multiplication can't be done until the exponentiation has been performed. • The exponentiation operation is therefore 3^2 = 9, and the expression becomes 3 * 5 - 4 * 9. Evaluating this expression, the multiplications 3 * 5 and 4 * 9 must be performed before the subtraction. 3 * 5 = 15 and 4 * 9 = 36 so we now have 3 * 5 - 4 * 3 ^ 2 = 3 * 5 - 4 * 9 = 15 - 36 = -21. To calculate 3 * 5 - (4 * 3)^2 we first do the operation in parentheses, obtaining 4 * 3 = 12. Then we apply the exponentiation to get 12 ^2 = 144. Finally we multiply 3 * 5 to get 15. Putting this all together we get 3 * 5 - (4 * 3)^2 = 3 * 5 - 12^2 = 3 * 5 - 144 = 15 - 144 = -129. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 In the next three problems, the graphs will be of one of the basic shapes listed below. You will be asked to construct graphs for three simple functions, and determine which of the depicted graphs each of your graphs most closely resembles. At this point you won't be expected to know these terms or these graph shapes; if at some point in your course you are expected to know these things, they will be presented at that point. Linear:  Quadratic or parabolic:  Exponential:  Odd power:  Fractional positive power:  Even negative power:  partial graph of polynomial of degree 3  more extensive graph of polynomial of degree 3  ********************************************* Question: `q007. Let y = 2 x + 3. (Note: Liberal Arts Mathematics students are encouraged to do this problem, but are not required to do it). • Evaluate y for x = -2. What is your result? In your solution explain the steps you took to get this result. • Evaluate y for x values -1, 0, 1 and 2. Write out a copy of the table below. In your solution give the y values you obtained in your table. x y -2 -1 0 1 2 • Sketch a graph of y vs. x on a set of coordinate axes resembling the one shown below. You may of course adjust the scale of the x or the y axis to best depict the shape of your graph.  • In your solution, describe your graph in words, and indicate which of the graphs depicted previously your graph most resembles. Explain why you chose the graph you did. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Replacing the variable X with the number then solving for Y. For X = -2, the equation would become Y = 2 (-2) + 3 = -1. (X, Y) -2, -1 -1, 1 0, 3 1, 5 2, 7 The Graph is linear with the intersection of the X-axis at -1.5 and the Y-axis at 3. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Two slightly different explanations are give below, one by a student and one by the instructor. Neither format is inherently better than the other. GOOD SOLUTION BY STUDENT: First we need to complete the table. I have added a column to the right of the table to show the calculation of “y” when we us the “x” values as given. x y Calculation: If y = 2x + 3 -2 -1 If x = -2, then y = 2(-2)+3 = -4+3 = -1 -1 1 If x= -1, then y = 2(-1)+3 = -2+3 = 1 0 3 If x= 0, then y = 2(0)+3 = 0+3 = 3 1 5 If x= 1, then y = 2(1)+3 = 2+3 = 5 2 7 If x= 2, then y = 2(2)+3 = 4+3 = 7 Once an answer has been determined, the “y” value can be filled in. Now we have both the “x” and “y” values and we can begin our graph. The charted values continue on a straight line representing a linear function as shown above. INSTRUCTOR'S SOLUTION: We easily evaluate the expression: • When x = -2, we get y = 2 x + 3 = 2 * (-2) + 3 = -4 + 3 = -1. • When x = -1, we get y = 2 x + 3 = 2 * (-1) + 3 = -2 + 3 = 1. • When x = 0, we get y = 2 x + 3 = 2 * (0) + 3 = 0 + 3 = 3. • When x = 1, we get y = 2 x + 3 = 2 * (1) + 3 = 2 + 3 = 5. • When x = 2, we get y = 2 x + 3 = 2 * (2) + 3 = 4 + 3 = 7. Filling in the table we have x y -2 -1 -1 1 0 3 1 5 2 7 When we graph these points we find that they lie along a straight line. Only one of the depicted graphs consists of a straight line, and we conclude that the appropriate graph is the one labeled 'linear'. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q008. Let y = x^2 + 3. (Note: Liberal Arts Mathematics students are encouraged to do this problem, but are not required to do it). • Evaluate y for x = -2. What is your result? In your solution explain the steps you took to get this result. • Evaluate y for x values -1, 0, 1 and 2. Write out a copy of the table below. In your solution give the y values you obtained in your table. x y -2 -1 0 1 2 • Sketch a graph of y vs. x on a set of coordinate axes resembling the one shown below. You may of course adjust the scale of the x or the y axis to best depict the shape of your graph.  • In your solution, describe your graph in words, and indicate which of the graphs depicted previously your graph most resembles. Explain why you chose the graph you did. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Same as previous equation, replace X variable with number, solve. X = -2 becomes Y = (-2) ^ 2 + 3 = 7. (X, Y) -2, 7 -1, 4 0, 3 1, 4 2, 7 The graph is parabolic and symmetrical along the Y axis. The line bottoms out at 3. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Evaluating y = x^2 + 3 at the five points: • If x = -2 then we obtain y = x^2 + 3 = (-2)^2 + 3 = 4 + 3 = 7. • If x = -1 then we obtain y = x^2 + 3 = (-1)^2 + 3 = ` + 3 = 4. • If x = 0 then we obtain y = x^2 + 3 = (0)^2 + 3 = 0 + 3 = 3. • If x = 1 then we obtain y = x^2 + 3 = (1)^2 + 3 = 1 + 3 = 4. • If x = 2 then we obtain y = x^2 + 3 = (2)^2 + 3 = 4 + 3 = 7. The table becomes x y -2 7 -1 4 0 3 1 4 2 7 We note that there is a symmetry to the y values. The lowest y value is 3, and whether we move up or down the y column from the value 3, we find the same numbers (i.e., if we move 1 space up from the value 3 the y value is 4, and if we move one space down we again encounter 4; if we move two spaces in either direction from the value 3, we find the value 7). A graph of y vs. x has its lowest point at (0, 3). If we move from this point, 1 unit to the right our graph rises 1 unit, to (1, 4), and if we move 1 unit to the left of our 'low point' the graph rises 1 unit, to (-1, 4). If we move 2 units to the right or the left from our 'low point', the graph rises 4 units, to (2, 7) on the right, and to (-2, 7) on the left. Thus as we move from our 'low point' the graph rises up, becoming increasingly steep, and the behavior is the same whether we move to the left or right of our 'low point'. This reflects the symmetry we observed in the table. So our graph will have a right-left symmetry. Two of the depicted graphs curve upward away from the 'low point'. One is the graph labeled 'quadratic or parabolic'. The other is the graph labeled 'partial graph of degree 3 polynomial'. If we look closely at these graphs, we find that only the first has the right-left symmetry, so the appropriate graph is the 'quadratic or parabolic' graph. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK. ####I understood the graph to be parabolic and didn’t even notice the partial graph of degree 3 polynomial until I had read the Given solution. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q009. Let y = 2 ^ x + 3. (Note: Liberal Arts Mathematics students are encouraged to do this problem, but are not required to do it). • Evaluate y for x = 1. What is your result? In your solution explain the steps you took to get this result. • Evaluate y for x values 2, 3 and 4. Write out a copy of the table below. In your solution give the y values you obtained in your table. x y 1 2 3 4 • Sketch a graph of y vs. x on a set of coordinate axes resembling the one shown below. You may of course adjust the scale of the x or the y axis to best depict the shape of your graph.  • In your solution, describe your graph in words, and indicate which of the graphs depicted previously your graph most resembles. Explain why you chose the graph you did. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Same as previous equation, replace X variable with number, solve. X = 1 becomes Y = 2 ^ 1 + 3 = 5. (X, Y) 1, 5 2, 7 3, 11 4, 19 The graph is exponential with the graphed points not intersecting the X or Y axis. The line increases in slope as it moves to the right. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Recall that the exponentiation in the expression 2^x + 1 must be done before, not after the addition. When x = 1 we obtain y = 2^1 + 3 = 2 + 3 = 5. When x = 2 we obtain y = 2^2 + 3 = 4 + 3 = 7. When x = 3 we obtain y = 2^3 + 3 = 8 + 3 = 11. When x = 4 we obtain y = 2^4 + 3 = 16 + 3 = 19. x y 1 5 2 7 3 11 4 19 Looking at the numbers in the y column we see that they increase as we go down the column, and that the increases get progressively larger. In fact if we look carefully we see that each increase is double the one before it, with increases of 2, then 4, then 8. When we graph these points we find that the graph rises as we go from left to right, and that it rises faster and faster. From our observations on the table we know that the graph in fact that the rise of the graph doubles with each step we take to the right. The only graph that increases from left to right, getting steeper and steeper with each step, is the graph labeled 'exponential'. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q010. If you divide a certain positive number by 1, is the result greater than the original number, less than the original number or equal to the original number, or does the answer to this question depend on the original number? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Any positive number divided by one is itself. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If you divide any number by 1, the result is the same as the original number. Doesn't matter what the original number is, if you divide it by 1, you don't change it. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q011. If you divide a certain positive number by a number greater than 1, is the result greater than the original number, less than the original number or equal to the original number, or does the answer to this question depend on the original number? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Any positive number divided by a number greater than 1 will decrease. If the dividing number is less than 1, the number would increase. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If you split something up into equal parts, the more parts you have, the less will be in each one. Dividing a positive number by another number is similar. The bigger the number you divide by, the less you get. Now if you divide a positive number by 1, the result is the same as your original number. So if you divide the positive number by a number greater than 1, what you get has to be smaller than the original number. Again it doesn't matter what the original number is, as long as it's positive. Students will often reason from examples. For instance, the following reasoning might be offered: OK, let's say the original number is 36. Let's divide 36 be a few numbers and see what happens: 36/2 = 18. Now 3 is bigger than 2, and 36 / 3 = 12. The quotient got smaller. Now 4 is bigger than 3, and 36 / 4 = 9. The quotient got smaller again. Let's skip 5 because it doesn't divide evenly into 36. 36 / 6 = 4. Again we divided by a larger number and the quotient was smaller. I'm convinced. That is a pretty convincing argument, mainly because it is so consistent with our previous experience. In that sense it's a good argument. It's also useful, giving us a concrete example of how dividing by bigger and bigger numbers gives us smaller and smaller results. However specific examples, however convincing and however useful, don't actually prove anything. The argument given at the beginning of this solution is general, and applies to all positive numbers, not just the specific positive number chosen here. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q012. If you divide a certain positive number by a positive number less than 1, is the result greater than the original number, less than the original number or equal to the original number, or does the answer to this question depend on the original number? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I actually answered this along with the previous answer. Any positive number divided by a positive number less than 1 will increase. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If you split something up into equal parts, the more parts you have, the less will be in each one. Dividing a positive number by some other number is similar. The bigger the number you divide by, the less you get. The smaller the number you divide by, the more you get. Now if you divide a positive number by 1, the result is the same as your original number. So if you divide the positive number by a positive number less than 1, what you get has to be larger than the original number. Again it doesn't matter what the original number is, as long as it's positive. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q013. Students often get the basic answers to nearly all, or even all these questions, correct. Your instructor has however never seen anyone who addressed all the subtleties in the given solutions in their self-critiques, and it is very common for a student to have given no self-critiques. It is very likely that there is something in the given solutions that is not expressed in your solution. This doesn't mean that you did a bad job. If you got most of the 'answers' right, you did fine. However, in order to better understand the process, you are asked here to go back and find something in one of the given solutions that you did not address in your solution, and insert a self-critique. You should choose something that isn't trivial to you--something you're not 100% sure you understand. If you can't find anything, you can indicate this below, and the instructor will point out something and request a response (the instructor will select something reasonable, but will then expect a very good and complete response). However it will probably be less work for you if you find something yourself. Your response should be inserted at the appropriate place in this document, and should be indicated by preceding it with ####. As an answer to this question, include a copy of whatever you inserted above, or an indication that you can't find anything. your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv ####I understood the graph to be parabolic and didn’t even notice the partial graph of degree 3 polynomial until I had read the Given solution. This was the only response I gave in the Self-Critique section. Feel free to point something out as I believe I have understood everything, but could be wrong. Thank You. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q013. Students often get the basic answers to nearly all, or even all these questions, correct. Your instructor has however never seen anyone who addressed all the subtleties in the given solutions in their self-critiques, and it is very common for a student to have given no self-critiques. It is very likely that there is something in the given solutions that is not expressed in your solution. This doesn't mean that you did a bad job. If you got most of the 'answers' right, you did fine. However, in order to better understand the process, you are asked here to go back and find something in one of the given solutions that you did not address in your solution, and insert a self-critique. You should choose something that isn't trivial to you--something you're not 100% sure you understand. If you can't find anything, you can indicate this below, and the instructor will point out something and request a response (the instructor will select something reasonable, but will then expect a very good and complete response). However it will probably be less work for you if you find something yourself. Your response should be inserted at the appropriate place in this document, and should be indicated by preceding it with ####. As an answer to this question, include a copy of whatever you inserted above, or an indication that you can't find anything. your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv ####I understood the graph to be parabolic and didn’t even notice the partial graph of degree 3 polynomial until I had read the Given solution. This was the only response I gave in the Self-Critique section. Feel free to point something out as I believe I have understood everything, but could be wrong. Thank You. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!