course Phy202 I am going to submit this form and email it to you. Should I always do both? MYCʈ~assignment #001
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11:09:17 `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 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 `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 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 `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 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 `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 `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 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 `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 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 `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 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 `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 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 *&*& 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 `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 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 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 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 `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 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 `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 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 `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 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 `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 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 `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 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 `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 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 `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 ** 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 `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 ** 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 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