course MTH 163 Here are all of the assignments due up until 02/04/06. It should be exercises 0-5 and other assignments. Am I doing the right assignments? I have separated each assignment by a line (_______).
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15:17:52 `q001. Part 1 includes six activities. If you have completed an activity, just enter the answer 'completed'. This question is appearing in the Question box. The box to the right is the Answer box, where you will type in your answers to the questions posed here. To use this program you read a question, then enter your answer in the Answer box and click on Enter Answer. In your answers give what is requested, but don't go into excruciating detail. Try to give just enough that the instructor can tell that you understand an item. After entering an answer click on Next Question/Answer above the Question box. Do you understand these instructions?
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RESPONSE --> Yes, I understand
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15:27:34 This program has created the folder c:\vhmthphy on your hard drive. Browse to that folder and locate the file whose name begins with SEND. The name of this file will also include your name, as you gave it to the program, and the file will show as a Text file. Never tamper with a SEND file in any way. It contains internal codes as if these codes are tampered with you won't get credit for the assignment. However you are welcome to copy this file to another location and view it, make changes, etc. Just be sure that when requested to do so you send the instructor the original, tamper-free file. State in the Answer box whether or not you have been able to locate the SEND file. Don't send the SEND file yet. Note that more questions/instructions remain in the q_a_prelim.
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RESPONSE --> Yes, I am able to locate the SEND file.
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15:30:21 `q002. Note that every time you click on Enter Answer the program writes your response to your SEND file. Even if the program disappears all the information you have entered with the Enter Answer button will remain in that file. This program never 'unwrites' anything. Even if this program crashes your information will still be there in the SEND file. Explain this in your own words.
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RESPONSE --> Everytime I click on the Enter Answer button, my response will be written in the SEND file. The program won't 'unwrite' anything, therefore I won't lose any of the information in the SEND file, even if this program was lost.
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15:35:28 Any time you do not receive a reply from the instructor by the end of the following day, you should resubmit your work using the Resubmit Form at http://www.vhcc.edu/dsmith/genInfo/. You have already seen that page, but take another look at that page and be sure you see the Submit Work form, the Resubmit Form and a number of other forms that will be explained later. Enter a sentence or two describing the related links you see at that location.
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RESPONSE --> I see the 'Submit Work Form,' the 'Resubmit Form,' the 'Request Access Code,' the 'Class Work Submission Form,' and a link to 'Acces your Portfolio.'
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15:38:31 `q003. If you are working on a VHCC computer, it is probably set up in such a way as to return to its original configuration when it is rebooted. To avoid losing information it is suggested that you back up your work frequently, either by emailing yourself a copy or by using a key drive or other device. This is a good idea on any computer. Please indicate your understanding of this suggestion.
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RESPONSE --> I understand what is being said; to avoid losing information, I should frequently back up my work by emailing myself a copy, or by using a key drive or other device.
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________________________________________________________________________________ ćծֹŷ`F| Student Name: assignment #003 003. PC1 questions ý[hc{b̲ Student Name: assignment #001 001. Rates Fmy풤Qy Student Name: assignment #002 001. Rates
zqfP哓 Student Name: assignment #001 001. Areas ѱڧJذŘo Student Name: assignment #005 005. Calculus ____________________________________________________________________________________ 䚨ФϠVvװݗ Student Name: assignment #001 001. typewriter notation......!!!!!!!!...................................
14:41:51 `q001. Explain the difference between x - 2 / x + 4 and (x - 2) / (x + 4).
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RESPONSE -->
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NDfywG̚eﱟ Student Name: assignment #001 001. typewriter notation
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14:45:32 `q001. Explain the difference between x - 2 / x + 4 and (x - 2) / (x + 4).
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RESPONSE --> The difference between x - 2 / x + 4 and (x - 2) / (x + 4) is that, in the first problem, 2 / x is being subtracted from x and then 4 is added. In the second problem, (x - 2) is being divided by (x + 4).
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14:50:15 The order of operations dictates that grouped expressions must be evaluated first, that exponentiation must be done before multiplication or division, which must be done before addition or subtraction. It makes a big difference whether you subtract the 2 from the 2 or divide the -2 by 4 first. If there are no parentheses you have to divide before you subtract: 2 - 2 / 2 + 4 = 2 - 1 + 4 (do multiplications and divisions before additions and subtractions) = 5 (add and subtract in indicated order) If there are parentheses you evaluate the grouped expressions first: (x - 2) / (x - 4) = (2 - 2) / ( 4 - 2) = 0 / 2 = 0.
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. The order of operations tells me what is to be done first; parentheses, exponents, multiplication/division, and addition/subtraction. To receive the correct answer, I must use this ""rule of thumb.""
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15:01:12 `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 problem, 4 is being added to 2 ^ x and since there are no parentheses, the exponent (2 ^ x) should be done first. For ex.: If x = 2, then 2^2 + 4 = 4 + 4 = 8 In the second problem, since there are parentheses, they should be done first, followed by exponents and multiplication. For ex.: If x = 2, then 2 ^ (2 + 4) = 2 ^ (6) = 64
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15:02:12 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 understand what is being said.I got the same answer.
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15:16:32 `q003. What is the numerator of the fraction in the expression x - 3 / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x? What is the denominator? What do you get when you evaluate the expression for x = 2?
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RESPONSE --> The numerator is everything that precedes the division ( / ) sign (or what is on top of the fraction), which in this case would be x - 3. For example: If x = 2, then 2 - 3 = -1 The denominator is everything that follows the division ( / ) sign (or what is on the bottom of the fraction), which in this case would be [ (2x-5)^2 * 3x + 1 ] - 2 + 7x. For example: If x = 2, then = [(2(2)-5)^2 * 3(2) + 1] - 2 + 7(2) = [(4-5)^2 * 6 + 1] - 2 + 14 = [(-1)^2 * 7] - 2 + 14 = [1 * 7] - 2 + 14 = 7 - 2 + 14 = 19
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15:22:14 The numerator is 3. x isn't part of the fraction. / indicates division, which must always precede subtraction. Only the 3 is divided by [ (2x-5)^2 * 3x + 1 ] and only [ (2x-5)^2 * 3x + 1 ] divides 3. If we mean (x - 3) / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x we have to write it that way. The preceding comments show that the denominator is [ (2x-5)^2 * 3x + 1 ] Evaluating the expression for x = 2: - 3 / [ (2 * 2 - 5)^2 * 3(2) + 1 ] - 2 + 7*2 = 2 - 3 / [ (4 - 5)^2 * 6 + 1 ] - 2 + 14 = evaluate in parenthese; do multiplications outside parentheses 2 - 3 / [ (-1)^2 * 6 + 1 ] -2 + 14 = add inside parentheses 2 - 3 / [ 1 * 6 + 1 ] - 2 + 14 = exponentiate in bracketed term; 2 - 3 / 7 - 2 + 14 = evaluate in brackets 13 4/7 or 95/7 or about 13.57 add and subtract in order. The details of the calculation 2 - 3 / 7 - 2 + 14: Since multiplication precedes addition or subtraction the 3/7 must be done first, making 3/7 a fraction. Changing the order of the terms we have 2 - 2 + 14 - 3 / 7 = 14 - 3/7 = 98/7 - 3/7 = 95/7.
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RESPONSE --> I didn't realize that x wasn't part of the fraction. I stated that the numerator was x - 3 and the denominator was [ (2x-5)^2 * 3x + 1 ] - 2 + 7x. In doing so, I got the wrong answer of course. I understand what is coreectly being done, but why isn't x counted as part of the numerator?
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15:36:38 `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 --> (x - 5)^2x-1 + 3 / x-2 = (4 - 5) ^2(4)-1 + 3 / 4-2 plug in 4 for x = (-1) ^8-1 + 3 / 4-2 Do parentheses first = (-1)^7 + 3 / 4-2 next do exponents = -1 + 3/2 Next do division; (3 / 2) = -1 + 1.5 next do addition = 0.5 or 1/2 turn decimal back into fraction form
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15:42:50 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'm not 100% sure what is done here. I'm not sure if I got the right answer. One thing that I did wrong was that I applied the exponent 2 to 4, rather than the -1.
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ůexہ Student Name: assignment #002 002. Describing Graphs
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15:54:45 `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. 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 --> How do I make the graph and send it through e-mail? I worked the problem out on paper and sketched my graph, but now I don't know how to get it drawn on the computer and then sent though e-mail. The X and Y values for y = 3x - 4 are: (X): -3, -2, -1, 0, 1, 2, 3 (Y): -13, -10, -7, -4, -1, 2, 5
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16:00:41 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 --> My graph starts with (-3, -13) and continues rising towards the right at an angle with a slope of 3/1. Is this correct? If not, then what am I doing wrong? I think I keep misinterpreting the questions to mean something else. Oh! OK, now I see that the x intercept (4/3, 0) and the y intercept (0, -4) are the same x and y intercept that I got in my graph.
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16:00:53 01-30-2006 16:00:53 `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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NOTES ------->
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16:01:38 `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 --> yes, the steepness changes. the lines rises towards the right with a slope of 3/1.
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16:03:15 The graph forms a straight line with no change in steepness.
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RESPONSE --> Apparently my mind isn't working. I can see that the graph forms a straight line, so if the line is straight, then then the steepness will always be the same and will never change? Is this correct?
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16:03:56 `q003. What is the slope of the graph of the preceding two exercises (the function ia y = 3x - 4;slope is rise / run between two points of the graph)?
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RESPONSE --> The slope of the graph is 3 / 1. It rises 3 and runs 1.
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16:05:31 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 --> Yes, I was correct. But instead of writing it down as 3/1, i can just write it as 3, because 3 divided by 1 is 3.
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16:11:10 `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 --> For y = x^2 and between x = 0 and x = 3, then: (X): 0, 1, 2, 3 (Y): 0, 1, 4, 9 The graph is increasing and the steepness does change, because it isn't a straight line. The steepness first increases by 1 then 3 then 5. The graph is increasing at an increasing rate.
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16:11:32 Graph points include (0,0), (1,1), (2,4) and (3,9). The y values are 0, 1, 4 and 9, which increase as we move from left to right. The increases between these points are 1, 3 and 5, so the graph not only increases, it increases at an increasing rate.
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RESPONSE --> YAY! I got it right!
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16:18:04 `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 --> For y = x^2 (X): -3, -2, -1, 0 (Y): 9, 4, 1, 0 The graph is decreasing. The line is straight, therefore the steepness doesn't change. Since the line decreases by 4, then we can say that the graph decreases at a constant rate.
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16:19:50 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 missed the steepness of the graph. I understand now . Since I missed that the graph had steepness, I also know why I missed that the graph is decreasing at a decreasing rate.
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16:26:55 `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 --> For y = 'sqrt(x) (X): 0, 1, 2, 3 (Y): 0, 1, 1.41, 1.73 The graph is slowly increasing. The steepness of the graph is changing, but not steadily. First it changed by .41 then it changed by .32; therefore, the steepness is decreasing. The graph is increasing at a decreasing rate.
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16:27:28 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 --> Yay! I got it right!
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17:21:41 `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 --> For y = 5 * 2^(-x) (X): 0, 1, 2, 3 (Y): 0, 1, 1/4, 1/8 The graph is slowly decreasing. Yes, the steepness of the graph is changing by a 1/4. The graph would then be decreasing at a decreasing rate.
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17:22:57 ** 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 got it right, but I believe my steepness numbers were a little off. I understand how my instructor reached the answer.
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17:25:24 `q008. Suppose you stand still in front of a driveway. A car starts out next to you and moves away from you, traveling faster and faster. If y represents the distance from you to the car and t represents the time in seconds since the car started out, would a graph of y vs. t be increasing or decreasing? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> Since the car is moving away from me, the distance will increase, as will the seconds passing. Therefore both the distance and time will increase as the car drives away from me. Therefore the graph is increasing at an increasing rate.
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17:25:54 ** 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 --> YAY! I got it!
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eĂÕ\}YxbۂۇR Student Name: assignment #003 003. PC1 questions
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17:33:56 `q001 A straight line connects the points (3, 5) and (7, 17), while another straight line continues on from (7, 17) to the point (10, 29). Which line is steeper and on what basis to you claim your result?
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RESPONSE --> For this problem, since (X): 3, 7, 10 (Y): 5, 17, 29 The difference between the Y values (5 to 17 and 17 to 29) are both 12. Therefore we can say that the steepness for each line is the same.
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17:35:23 The point (3,5) has x coordinate 3 and y coordinate 5. The point (7, 17) has x coordinate 7 and y coordinate 17. To move from (3,5) to (7, 17) we must therefore move 4 units in the x direction and 12 units in the y direction. Thus between (3,5) and (7,17) the rise is 12 and the run is 4, so the rise/run ratio is 12/4 = 3. Between (7,10) and (10,29) the rise is also 12 but the run is only 3--same rise for less run, therefore more slope. The rise/run ratio here is 12/3 = 4.
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RESPONSE --> I got it half right. I didn't calculate the differences between the x values. For some reason, I thought that steepness was measured by the y-values. I understand how to reach the correct answer now.
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17:41:42 `q002. The expression (x-2) * (2x+5) is zero when x = 2 and when x = -2.5. Without using a calculator verify this, and explain why these two values of x, and only these two values of x, can make the expression zero.
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RESPONSE --> When x = 2, The first set of parentheses (x -2) becomes zero because 2 -2 = 0. Then when it is multiplied by (2x+5) the answer is still zero, because anything multiplied by zero is zero. When x = -2.5 The first set of parentheses (x - 2) becomes -4.5 because (-2.5 - 2) = -4.5. In the second set of parentheses 2*-2.5 = -5 and -5 plus 5 is zero. Therefore when (-4.5) is multiplied to zero, the answer is still zero.
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17:43:01 If x = 2 then x-2 = 2 - 2 = 0, which makes the product (x -2) * (2x + 5) zero. If x = -2.5 then 2x + 5 = 2 (-2.5) + 5 = -5 + 5 = 0.which makes the product (x -2) * (2x + 5) zero. The only way to product (x-2)(2x+5) can be zero is if either (x -2) or (2x + 5) is zero. Note that (x-2)(2x+5) can be expanded using the Distributive Law to get x(2x+5) - 2(2x+5). Then again using the distributive law we get 2x^2 + 5x - 4x - 10 which simplifies to 2x^2 + x - 10. However this doesn't help us find the x values which make the expression zero. We are better off to look at the factored form.
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RESPONSE --> I got the same answer, but I didn't use the distributive property. I did the math in my head as it is laid out. I understand how the Distributive property can be used to get the same answer.
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17:51:34 `q003. For what x values will the expression (3x - 6) * (x + 4) * (x^2 - 4) be zero?
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RESPONSE --> If the x values are -4 or 0 (zero) then the answer for the problem would be zero.
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17:54:26 In order for the expression to be zero we must have 3x-6 = 0 or x+4=0 or x^2-4=0. 3x-6 = 0 is rearranged to 3x = 6 then to x = 6 / 3 = 2. So when x=2, 3x-6 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero. x+4 = 0 gives us x = -4. So when x=-4, x+4 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero. x^2-4 = 0 is rearranged to x^2 = 4 which has solutions x = + - `sqrt(4) or + - 2. So when x=2 or when x = -2, x^2 - 4 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero. We therefore see that (3x - 6) * (x + 4) * (x^2 - 4) = 0 when x = 2, or -4, or -2. These are the only values of x which can yield zero.**
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RESPONSE --> I was right about -4 and of course zero is a given, but I didn't think to break -4 down in to 2 and -2. I put all three problems together and tried figuring it out that way, instead of making each one equal zero and solving the equations.
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18:02:59 `q004. One straight line segment connects the points (3,5) and (7,9) while another connects the points (10,2) and (50,4). From each of the four points a line segment is drawn directly down to the x axis, forming two trapezoids. Which trapezoid has the greater area? Try to justify your answer with something more precise than, for example, 'from a sketch I can see that this one is much bigger so it must have the greater area'.
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RESPONSE --> The trapezoid which connects points (10, 2) and (50, 4) has the greater area because it overs a greater distance than the one with points (3, 5) and (7, 9). The straight line that connects (3,5) and (7,9) and the trapezoid that it makes is taller rather than wider, like the trapezoid formed by points (10,2) and (50,4). Covering a wider distance rather than by height, causes the second set of points to cover a greater area.
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18:04:23 Your sketch should show that while the first trapezoid averages a little more than double the altitude of the second, the second is clearly much more than twice as wide and hence has the greater area. To justify this a little more precisely, the first trapezoid, which runs from x = 3 to x = 7, is 4 units wide while the second runs from x = 10 and to x = 50 and hence has a width of 40 units. The altitudes of the first trapezoid are 5 and 9,so the average altitude of the first is 7. The average altitude of the second is the average of the altitudes 2 and 4, or 3. So the first trapezoid is over twice as high, on the average, as the first. However the second is 10 times as wide, so the second trapezoid must have the greater area. This is all the reasoning we need to answer the question. We could of course multiply average altitude by width for each trapezoid, obtaining area 7 * 4 = 28 for the first and 3 * 40 = 120 for the second. However if all we need to know is which trapezoid has a greater area, we need not bother with this step.
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RESPONSE --> I got it right. I answered the question based on the height and width of both trapezoids.
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18:16:43 `q005. Sketch graphs of y = x^2, y = 1/x and y = `sqrt(x) [note: `sqrt(x) means 'the square root of x'] for x > 0. We say that a graph increases if it gets higher as we move toward the right, and if a graph is increasing it has a positive slope. Explain which of the following descriptions is correct for each graph: As we move from left to right the graph increases as its slope increases. As we move from left to right the graph decreases as its slope increases. As we move from left to right the graph increases as its slope decreases. As we move from left to right the graph decreases as its slope decreases.
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RESPONSE --> For the first graph, the correct description would be that as we move from left to right, the graph increases as its slope increases. (X): 0, 1, 2, 3 (Y): 0, 1, 4, 9 For the second graph, the correct description would be that as we move from left to right, the graph decreases as its slope decreases. (X): 0,1, 2, 3 (Y): 0, 1, 1/2, 1/3
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18:19:07 For x = 1, 2, 3, 4: The function y = x^2 takes values 1, 4, 9 and 16, increasing more and more for each unit increase in x. This graph therefore increases, as you say, but at an increasing rate. The function y = 1/x takes values 1, 1/2, 1/3 and 1/4, with decimal equivalents 1, .5, .33..., and .25. These values are decreasing, but less and less each time. The decreasing values ensure that the slopes are negative. However, the more gradual the decrease the closer the slope is to zero. The slopes are therefore negative numbers which approach zero. Negative numbers which approach zero are increasing. So the slopes are increasing, and we say that the graph decreases as the slope increases. We could also say that the graph decreases but by less and less each time. So the graph is decreasing at a decreasing rate. For y = `sqrt(x) we get approximate values 1, 1.414, 1.732 and 2. This graph increases but at a decreasing rate.
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RESPONSE --> I used the x values of 0, 1, 2, and 3 but I didn't get the exact answers that you did. I understand what the instructor is saying as to how he got his answers, but I can't understand why I didn't get the same descriptions. ???
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18:27:57 `q006. If the population of the frogs in your frog pond increased by 10% each month, starting with an initial population of 20 frogs, then how many frogs would you have at the end of each of the first three months (you can count fractional frogs, even if it doesn't appear to you to make sense)? Can you think of a strategy that would allow you to calculate the number of frogs after 300 months (according to this model, which probably wouldn't be valid for that long) without having to do at least 300 calculations?
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RESPONSE --> 20 / 10% = 200 + 20 (number of frogs to begin with) = 220 frogs by the end of the first month. 220 / 10% = 2200 frogs by the end of the second month. 2200 / 10% = 22000 frogs by the end of the third month. 200 * 10 = 2200 * 10 = 22000 frogs. I'm not sure of the strategy to use to calculate how many frogs I would have after 300 months. ???
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18:35:47 At the end of the first month, the number of frogs in the pond would be (20 * .1) + 20 = 22 frogs. At the end of the second month there would be (22 * .1) + 22 = 24.2 frogs while at the end of the third month there would be (24.2 * .1) + 24.2 = 26.62 frogs. The key to extending the strategy is to notice that multiplying a number by .1 and adding it to the number is really the same as simply multiplying the number by 1.1. 10 * 1.1 = 22; 22 * 1.1 = 24.2; etc.. So after 300 months you will have multiplied by 1.1 a total of 300 times. This would give you 20 * 1.1^300, whatever that equals (a calculator will easily do the arithmetic). A common error is to say that 300 months at 10% per month gives 3,000 percent, so there would be 30 * 20 = 600 frogs after 30 months. That doesn't work because the 10% increase is applied to a greater number of frogs each time. 3000% would just be applied to the initial number, so it doesn't give a big enough answer.
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RESPONSE --> When I calculated 10%, instead of using .1, I used .10 which would cause my calculations to be majorly off. Since I used to wrong fraction, I understand now how to find the answer. I had the strategy right, just not the percentage.
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18:40:50 `q007. Calculate 1/x for x = 1, .1, .01 and .001. Describe the pattern you obtain. Why do we say that the values of x are approaching zero? What numbers might we use for x to continue approaching zero? What happens to the values of 1/x as we continue to approach zero? What do you think the graph of y = 1/x vs. x looks for x values between 0 and 1?
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RESPONSE --> The pattern is that as the percentage gets smaller, the answer gets larger. The answers: 1, 10, 100, and 1000 increase by a multiplication of 10. For ex.: 1 * 10 = 10 * 10 = 100 * 10 = 1000 As 1/x continues to approach zero, the values increase.
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18:42:33 If x = .1, for example, 1 / x = 1 / .1 = 10 (note that .1 goes into 1 ten times, since we can count to 1 by .1, getting.1, .2, .3, .4, ... .9, 10. This makes it clear that it takes ten .1's to make 1. So if x = .01, 1/x = 100 Ithink again of counting to 1, this time by .01). If x = .001 then 1/x = 1000, etc.. Note also that we cannot find a number which is equal to 1 / 0. Deceive why this is true, try counting to 1 by 0's. You can count as long as you want and you'll ever get anywhere. The values of 1/x don't just increase, they increase without bound. If we think of x approaching 0 through the values .1, .01, .001, .0001, ..., there is no limit to how big the reciprocals 10, 100, 1000, 10000 etc. can become. The graph becomes steeper and steeper as it approaches the y axis, continuing to do so without bound but never touching the y axis. This is what it means to say that the y axis is a vertical asymptote for the graph .
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RESPONSE --> I understand how to reach the conclusion of the problem and why the graph becomes steeper and steeper as it approaches the y axis.
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18:47:09 * `q008. At clock time t the velocity of a certain automobile is v = 3 t + 9. At velocity v its energy of motion is E = 800 v^2. What is the energy of the automobile at clock time t = 5?
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RESPONSE --> At clock time t = 5 the velocity is v = 3t + 9 v = 3(5) + 9 v = 15 + 9 v = 24 At clock time t = 5, the energy is E = 800 v^2 E = 800 (24^2) E = 800 (576) E = 460800 So at clock time t = 5, the energy of the automobile is 460800
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18:49:16
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RESPONSE --> I was on problem number 8 and I must have clicked on something, but the next question wouldn't come up. it jsut kept telling me to enter a response. What should I do about this? Do I need to do the section all over again? I'm sorry, but I don't know what happened.
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18:49:28 For t=5, v = 3 t + 9 = (3*5) + 9 = 24. Therefore E = 800 * 24^2 = 460800.
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RESPONSE -->
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18:50:19
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RESPONSE --> Ok, the next question/answer come back, but then it left again. I got problem number 8 correct.
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18:51:44 * `q009. Continuing the preceding problem, can you give an expression for E in terms of t?
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RESPONSE --> As time (t) increases, the velocity (v) also increases. and since the velocity increases, the automobile must exhert more energy, therefore it also increases.
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18:52:58 Since v = 3 t + 9 the expression would be E = 800 v^2 = 800 ( 3t + 9) ^2. This is the only answer really required here. For further reference, though, note that this expression could also be expanded by applying the Distributive Law:. Since (3t + 9 ) ^ 2 = (3 t + 9 ) * ( 3 t + 9 ) = 3t ( 3t + 9 ) + 9 * (3 t + 9) = 9 t^2 + 27 t + 27 t + 81 = 9 t^2 + 54 t + 81, we get E = 800 ( 9 t^2 + 54 t + 81) = 7200 t^2 + 43320 t + 64800 (check my multiplication because I did that in my head, which isn't always reliable).
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RESPONSE --> I'm not sure if I understand completely how you could use the Distributive property to obtain the answer. ???
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__________________________________________________________________________________ ^ޖ}ʼnὫ{ٕ路II Student Name: assignment #001
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15:14:43 `q001. Note that this assignment has 10 questions Solve the following system of simultaneous linear equations: 3a + 3b = 9 6a + 5b = 16.
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RESPONSE --> To solve the following system of simultaneous linear equations, 3a + 3b = 9 6a + 5b = 16 We first must make one pair of coefficients negatives of one another, therefore we must decide which of the unknowns to eliminate, a or b. Whichever one, we have to find the least common multiple and make that the new coefficients. If I eliminate a, then the new coefficients will be (6, -6) and if I eliminate b, the new coefficients will be (15, -15). I'll choose to eliminate b: 3a + 3b = 9 (* 5) = 15a + 15b = 45 6a + 6b = 16 (* -3) = -18a + -15b = -48 -3a = -3 a = 1 Equation 1) has been multiplied by 5. Equation 2) has been multiplied by -3 -- because we want to make those coefficients 15 and -15, so that on adding, they will cancel. To solve for b, we will substitute a = 1 in the original equation 1): 3(1) + 3b = 9 3 + 3b = 9 3b = 9 - 3 3b = 6 b = 2 I then plugged in the a-values for a and the b-values for b to check my work and make sure that (1, 2) made the equation true.
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15:17:02 The system 3a + 3b = 9 6a + 5b = 16 can be solved by adding an appropriate multiple of one equation in order to eliminate one of the variables. Since the coefficient of a in the second equation (the coefficient of a in the second equation is 6)) is double that in the first (the coefficient of a in the first equation is 3), we can multiply the first equation by -2 in order to make the coefficients of a equal and opposite: -2 * [ 3a + 3b ] = -2 [ 9 ] 6a + 5b = 16 gives us -6a - 6 b = -18 6a + 5b = 16 . Adding the two equations together we obtain -b = -2, or just b = 2. Substituting b = 2 into the first equation we obtain 3 a + 3(2) = 9, or 3 a + 6 = 9 so that 3 a = 3 and a = 1. Our solution is therefore a = 1, b = 2. This solution is verified by substituting these values into the second equation, where we get 6 * 1 + 5 * 2 = 6 + 10 = 16.
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RESPONSE --> I got the same answer, but I didn't solve the equations entirely the same way as my instructor. I kind of understand though, how my instructor solved it.
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15:33:45 `q002. Solve the following system of simultaneous linear equations using the method of elimination: 4a + 5b = 18 6a + 9b = 30.
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RESPONSE --> To solve the equations using the method of elimination, we must make one pair of coefficients negatives of one another. In this problem, we must decide which of the unknowns to eliminate, a or b. Whichever i decide to eliminate, i will make the new coefficients, the least common multiple of the original, but with opposite signs. For the equations: 4a + 5b = 18 6a + 9b = 30 The least common multiple for a = (12, -12) and for b = (45, -45). I'll choose to eliminate a. 4a + 5b = 18 (*6)= 24a + 30b = 108 6a + 9b = 30 (*-4)= -24a + -36b = -120 -6b = -12 b = 2 Equation 1) has been multiplied by 6. Equation 2) has been multiplied by -4 -- because we want to make those coefficients 24 and -24, so that on adding, they will cancel. To solve for a, we will substitute b = 2 in the original equation 1: 4a + 5(2) = 18 4a + 10 = 18 4a = 18 - 10 4a = 8 a = 2 I then substituted the a-value for a (2) and the b-value for b (2) in both of the equations to make sure that they made the equation true.
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15:38:02 In the system 4a + 5b = 18 6a + 9b = 30 we see that the coefficients of b are relatively prime and so have a least common multiple equal to 5 * 9, whereas the coefficients 4 and 6 of a have a least common multiple of 12. We could therefore 'match' the coefficients of a and b by multiplying the first equation by 9 in the second by -5 in order to eliminate b, or by multiplying the first equation by 3 and the second by -2 in order to eliminate a. Choosing the latter in order to keep the number smaller, we obtain 3 * [4a + 5b ] = 3 * 18 -2 * [ 6a + 9b ] = -2 * 30, or 12 a + 15 b = 54 -12 a - 18 b = -60. Adding the two we get -3 b = -6, so b = 2. Substituting this value into the first equation we obtain 4 a + 5 * 2 = 18, or 4 a + 10 = 18, which we easily solve to obtain a = 2. Substituting this value of a into the second equation we obtain 6 * 2 + 9 * 2 = 30, which verifies our solution.
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RESPONSE --> I got the same answers as my instructor, but I still solved it a little different. Instead of multiplying the first equation by 9 and the second by -5, i multiplied the first by 6 and the second by -4, to eliminate a. Why did you choose to use the values 3 and -2 to eliminate a? Where did these values come from?
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15:45:10 `q003. If y = 5x + 8, then for what value of x will we have y = 13?
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RESPONSE --> If I substitute y in the equation: y = 5x + 8 Then the equation will read 13 = 5x + 8 To solve for x, I got x (or 5x) alone on one side by subtracting 8 from both sides. y = 5x + 8 13 = 5x + 8 13 - 8 = 5x + 8 - 8 5 = 5x 1 = x or x = 1 I then checked to make sure that x = 1 made the equation true. 13 = 5(1) + 8 13 = 5 + 8 13 = 13
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15:45:32 We first substitute y = 13 into the equation y = 5 x + 8 to obtain 13 = 5 x + 8. Subtracting 8 from both equations and reversing the equality we obtain 5 x = 5, which we easily solve to obtain x = 1.
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RESPONSE --> yes, I did it correctly.
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16:06:22 `q004. Sketch a set of coordinate axes representing y vs. x, with y on the vertical axis and x on the horizontal axis, and plot the points (1, -2), (3, 5) and (7, 8). Sketch a smooth curve passing through these three points. On your curve, what are the y coordinates corresponding to x coordinates 1, 3, 5 and 7? Estimate these coordinates as accurately as you can from your graph. Retain your sketch for use in future assignments.
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RESPONSE --> I'm not sure if I understand this problem correctly. I think that the y coordinates corresponding to the x coordinates 1, 3, 5, and 7 should be -2, 5, 6 and 8. ???
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16:08:34 The x coordinates 1, 3 and 7 match the x coordinates of the three given points, the y coordinates will be the y coordinates -2, 3 and 8, respectively, of those points. At x = 5 the precise value of x, for a perfect parabola, would be 8 1/3, or about 8.333. Drawn with complete accuracy a parabola through these points will peak between x = 3 in and x = 7, though unless you have a very fine sense of the shape of a parabola your sketch might well peak somewhere to the right of x = 7. The peak of the actual parabola will occur close to x = 6, and the value at x = 7 will be just a bit greater than 8, perhaps 8.5 or so. If your peak was to the right of x = 7, your x = 5 value will be lass than 7.
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RESPONSE --> I don't understand this problem or how my intructor got 8 1/3 or 8.333. I also did't get a parabola when I sketched my graph, but I'm certain I plotted the values right. ???
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16:14:37 `q005. Using your sketch from the preceding exercise, estimate the x coordinates corresponding to y coordinates 1, 3, 5 and 7. Also estimate the x values at which y is 0.
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RESPONSE --> i don't think my graph matches my instructors sketch, but i will try and solve the problem. I estimate that the x coordinates corresponding to y coordinates 1, 3, 5 and 7 will be 1.5, 2, 3 and 5. My estimate for the x value when y is zero would be 0.25. I'm not sure if this is correct. ???
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16:15:18 The easiest way to estimate your points would be to make horizontal lines on your graph at y = 1, 3, 5 and 7. You would easily locate the points were these lines intersect your graph, then estimate the x coordinates of these points. For the actual parabola passing through the given points, y will be 1 when x = 1.7 (and also, if your graph extended that far, near x = 10). y = 3 near x = 2.3 (and near x = 9.3). y = 5 at the given point (3, 5), where x = 3. y = 7 near x = 4 (and also near x = 7.7).
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RESPONSE -->
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16:16:20 The easiest way to estimate your points would be to make horizontal lines on your graph at y = 1, 3, 5 and 7. You would easily locate the points were these lines intersect your graph, then estimate the x coordinates of these points. For the actual parabola passing through the given points, y will be 1 when x = 1.7 (and also, if your graph extended that far, near x = 10). y = 3 near x = 2.3 (and near x = 9.3). y = 5 at the given point (3, 5), where x = 3. y = 7 near x = 4 (and also near x = 7.7).
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RESPONSE --> To find the x values when y is 1, 3, 5 and 7, I drew a horizontal line until it reached my orginal parabola line and i got about the same answers as my instructor. I must've gotten it right.
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16:21:13 `q006. Suppose the graph you used in the preceding two exercises represents the profit y on an item, with profit given in cents, when the selling price is x, with selling price in dollars. According to your graph what would be the profit if the item is sold for 4 dollars? What selling price would result in a profit of 7 cents? Why is this graph not a realistic model of profit vs. selling price?
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RESPONSE --> According to my graph, I estimate that the profit of an item that is sold for 4 dollars, will be about 6 cents. A selling price of about $5 would result in a profit of 7 cents, I estimate. The graph isn't a realistic model of profit vs. selling price because if my estimates are right, then I would be making less than one cent per every dollar that I sold. I would spend more than I would make.
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16:22:13 To find the profit for a selling price of x = 4 dollars, we would look at the x = 4 point on the graph. This point is easily located by sketching a vertical line through x = 4. Projecting over to the y-axis from this point, you should have obtained an x value somewhere around 7. The profit is the y value, so to obtain the selling price x corresponding to a profit of y = 7 we sketch the horizontal line at y = 7, which as in a preceding exercise will give us x values of about 4 (or x = 7.7, approx.).
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RESPONSE --> I believe my estimate was about the same as my instructors. (about 7)
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16:25:11 `q007. On another set of coordinate axes, plot the points (-3, 4) and (5, -2). Sketch a straight line through these points. We will obtain an approximate equation for this line: First substitute the x and y coordinates of the first point into the form y = m x + b. What equation do you obtain?
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RESPONSE --> If I substitute the x and y coordinates of the first point into the form y = m x + b, I get the equation: 4 = m (-3) + b
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16:26:33 Substituting x = -3 and y = 4 into the form y = m x + b, we obtain the equation 4 = -3 m + b. Reversing the sides we have -3 m + b = 4.
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RESPONSE --> I was almost correct. I wrote it out as 4 = m(-3) + b instead of 4 = -3m + b. Common sense tells me to do this, I just forgot in the previous question.
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16:28:07 `q008. Substitute the coordinates of the point (5, -2) into the form y = m x + b. What equation do you get?
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RESPONSE --> If I substitute the coordinates of the point (5, -2) into the form y = m x + b, I get the equation: -2 = 5m + b
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16:28:26 Substituting x = 5 and y = -2 into the form y = m x + b, we obtain the equation -2 = 5 m + b. Reversing the sides we have 5 m + b = -2
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RESPONSE --> I was correct.
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16:29:17 `q009. You have obtained the equations -3 m + b = 4 and 5 m + b = -2
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RESPONSE --> correct, i have obtained the equations -3m + b = 4 and 5m + b = -2
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17:05:38 . Use the method of elimination to solve these simultaneous equations for m and b.
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RESPONSE --> To solve these equations for m and b by using the elimination method, since the equation is -3m + b = 4 5m + b = -2 and the b coefficient is by itself, then I just multiplied the top equation by 1 and the bottom equation by -1 so that the b coefficients would cancel out. -3m + b = 4 (*1)= -3m + b = 4 5m + b = -2 (*-1)= -5m - b = -2 -8m = 2 m = -4 Then I substitute m = -4 into the first equation to solve for b: -3*(-4) + b = 4 12 + b = 4 b = 4 -12 b = -8 When I checked the equation to see if the values m = -4 and b = -8 made it true, I found that it made the first equation true, but not the second. 5*(-4) + (-8) = -2 -20 - 8 = -28 I don't know why it made the first equation true and not the second.
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17:07:16 Starting with the system -3 m + b = 4 5 m + b = -2 we can easily eliminate b by subtracting the equations. If we subtract the first equation from the second we obtain -8 m = 6, with solution m = -3/4. Substituting this value into the first equation we obtain (-3/4) * -3 + b = 4, which we easily solve to obtain b = 7/4. To check our solution we substitute m = -3/4 and b = 7/4 into the second equation, obtaining 5 ( -3/4) + 7/4 = -2, which gives us -8/4 = -2 or -2 = -2.
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RESPONSE --> I didn't subtract the first equation from the second. Therefore I didn't get the same answer as my instructor. But even when I did it wrong, how come it made the first equation true?
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17:09:14 `q010. Substitute your solutions b = 7/4 and m = -3/4 into the original form y = m x + b. What equation do you obtain? What is the significance of this equation?
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RESPONSE --> When I substitute b = 7/4 and m = -3/4 into the original equation form y = mx + b, I get the equation: y = -3/4x + 7/4
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17:10:27 Substituting b = 7/4 and m = -3/4 into the form y = m x + b, we obtain the equation y = -3/4 x + 7/4. This is the equation of the straight line through the given points (-3, 4) and (5, -2).
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RESPONSE --> I got the equation right, but I didn't know it gave the points (5, -2). Where did the 5 and -2 come from? Why isn't it (7, 4)??
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___________________________________________________________________________________ V~ȯoΖ՞ Student Name: assignment #002
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13:11:02 `q001. Note that this assignment has 8 questions Begin to solve the following system of simultaneous linear equations by first eliminating the variable which is easiest to eliminate. Eliminate the variable from the first and second equations, then from the first and third equations to obtain two equations in the remaining two variables: 2a + 3b + c = 128 60a + 5b + c = 90 200a + 10 b + c = 0. NOTE SOLN IS -1, 10, 100.
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RESPONSE --> After eliminating variable c from the first and second equations, I obtained the equation: -58a - 2b = 218 And after eliminating variable c from the first and third equations, I obtained the equation: -198a - 7b = 128
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13:14:31 The variable c is most easily eliminated. We accomplish this if we subtract the first equation from the second, and the first equation from the third, replacing the second and third equations with respective results. Subtracting the first equation from the second, are left-hand side will be the difference of the left-hand sides, which is (60a + 5b + c )- (2a + 3b + c ) = 58 a + 2 b. The right-hand side will be the difference 90 - 128 = -38, so the second equation will become 'new' 2d equation: 58 a + 2 b = -38. The 'new' third equation by a similar calculation will be 'new' third equation: 198 a + 7 b = -128. You might have obtain this system, or one equivalent to it, using a slightly different sequence of calculations.
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RESPONSE --> for the 'new' 2d equation, i got -58a + 2b = -38 after checking over and recalculating my previous answer. I realize that in my previous answer, I forgot to multiply 90 by -1 to make it -90. after doing so, I came up with the above answer. This is almost the same as my instructors answers, except I used a different sequence of calculations.
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13:22:43 `q002. Solve the two equations 58 a + 2 b = -38 198 a + 7 b = -128 , which can be obtained from the system in the preceding problem, by eliminating the easiest variable.
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RESPONSE --> To solve the two equations: 58a + 2b = -38 198a + 7b = -128 I first eliminated the variable b by multiplying the top equation by -7 and the bottom one by 2, to obtain two new equations: -406a - 14b = 266 396a + 14b = -256 Since the variable b cancels out, i solved for a -10a = 10 a = -1
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13:23:11 Neither variable is as easy to eliminate as in the last problem, but the coefficients of b a significantly smaller than those of a. So we will eliminate b. To eliminate b we will multiply the first equation by -7 and the second by 2, which will make the coefficients of b equal and opposite. The first step is to indicate the program multiplications: -7 * ( 58 a + 2 b) = -7 * -38 2 * ( 198 a + 7 b ) = 2 * (-128). Doing the arithmetic we obtain -406 a - 14 b = 266 396 a + 14 b = -256. Adding the two equations we obtain -10 a = 10, so we have a = -1.
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RESPONSE --> Yes, I was correct.
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13:28:46 `q003. Having obtained a = -1, use either of the equations 58 a + 2 b = -38 198 a + 7 b = -128 to determine the value of b. Check that a = -1 and the value obtained for b are validated by the other equation.
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RESPONSE --> To determine the value of b, I substituted -1 in for a in the first equation: 58*(-1) + 2b = -38 -58 + 2b = -38 -58 (+58) + 2b = -38 (+58) 2b = 20 b = 10 Once I had the values for both a (-1) and b (10) i substituted them in for the values a and b in the second equation to check my answer. 198*(-1) + 7*(10) = -128 -198 + 70 = -128 -128 = -128 a = 1 and b = 10 are validated by the other equation.
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13:29:04 You might have completed this step in your solution to the preceding problem. Substituting a = -1 into the first equation we have 58 * -1 + 2 b = -38, so 2 b = 20 and b = 10.
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RESPONSE --> Yes, I was correct.
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13:32:40 `q004. Having obtained a = -1 and b = 10, determine the value of c by substituting these values for a and b into any of the 3 equations in the original system 2a + 3b + c = 128 60a + 5b + c = 90 200a + 10 b + c = 0. Verify your result by substituting a = -1, b = 10 and the value you obtained for c into another of the original equations.
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RESPONSE --> To determine the value of c, I substituted a = -1 and b = 10 into the first equation (2a + 3b + c = 128). After working the problem, I found that c = 100. I then checked my values for a, b and c to validate the other two equations. They were true.
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13:33:06 Using first equation 2a + 3b + c = 128 we obtain 2 * -1 + 3 * 10 + c = 128, which was some simple arithmetic gives us c = 100. Substituting these values into the second equation we obtain 60 * -1 + 5 * 10 + 100 = 90, or -60 + 50 + 100 = 90, or 90 = 90. We could also substitute the values into the third equation, and will begin obtain an identity. This would completely validate our solution.
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RESPONSE --> Yes, I was correct.
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13:36:15 `q005. If a graph of y vs. x contains the points (1, -2), (3, 5) and (7, 8), as was the case for the graph you sketched in the preceding assignment, then what equation do we get if we substitute the x and y values corresponding to the point (1, -2) into the form y = a x^2 + b x + c?
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RESPONSE --> When I substitute the x and y values of (1, -2) into the form y = a x^2 + b x + c, then I obtained the following equation: -2 = a + b x + c
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13:37:38 We substitute y = -2 and x = 1 to obtain the equation -2 = a * 1^2 + b * 1 + c, or a + b + c = -2.
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RESPONSE --> I was almost right. I forgot to take the x out of the equation: -2 = a + bx + c I knew to do this, I just didn't think.
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13:44:07 `q006. If a graph of y vs. x contains the points (1, -2), (3, 5) and (7, 8), as was the case in the preceding question, then what equations do we get if we substitute the x and y values corresponding to the point (3, 5), then (7, 8) into the form y = a x^2 + b x + c?
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RESPONSE --> When I substitute the x and y values corresponding to the points (3, 5) and (7, 8) into the form of y = a x^2 + b x + c, then I obtained two new equations: 9a + 3b + c = 5 49a + 7b + c = 8
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13:46:31 Using the second point we substitute y = 5 and x = 3 to obtain the equation 5 = a * 3^2 + b * 3 + c, or 9 a + 3 b + c = 5. Using the third point we substitute y = 8 and x = 7 to obtain the equation 8 = a * 7^2 + b * 7 + c, or 49 a + 7 b + c = 7.
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RESPONSE --> Yes, my answers were correct, but i was wondering if my instructor meant to put 8 instead of 7 in the equation: 49a + 7 b + c = 7 ??? since the value of y is 8 for that equation.
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14:01:02 `q007. If a graph of y vs. x contains the points (1, -2), (3, 5) and (7, 8), as was the case in the preceding question, then what system of equations do we get if we substitute the x and y values corresponding to the point (1, -2), (3, 5), and (7, 8), in turn, into the form y = a x^2 + b x + c? What is the solution of this system?
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RESPONSE --> When I substitute the x and y values corresponding to the points (1, -2), (3, 5), and (7, 8) into the form of y = a x^2 + b x + c, then I get this systems of equations: a + b + c = -2 9a + 3b + c = 5 49a + 7b + c = 8
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14:04:57 The system consists of the three equations obtained in the last problem: a + b + c = -2 9 a + 3 b + c = 5 49 a + 7 b + c = 8. This system is solved in the same manner as in the preceding exercise. However the solutions don't come out to be whole numbers. The solution of this system, in decimal form, is approximately a = - 0.45833, b = 5.33333 and c = - 6.875. If you obtained a different system, you should show the system of two equations you obtained when you eliminated c, then indicate what multiple of each equation you put together to eliminate either a or b.
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RESPONSE --> The three equations that I obtained are correct as that of my instructors, but when I tried to solve the system of equations, I didn't get the same values for the variables a, b, and c. I got a = -6.545 b = -22. 681 c = 0.684 I don't know how I got answers so defferent than yours. I can't find where I went wrong. ???
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14:08:35 `q008. Substitute the values you obtained in the preceding problem for a, b and c into the form y = a x^2 + b x + c. What function do you get? What do you get when you substitute x = 1, 3, 5 and 7 into this function?
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RESPONSE --> Since I got three defferent values for a, b and c, than what my instructor got, I didn't get a correct form of equations for y = a x^2 + b x + c. If I substitute x = 1, 3, 5 and 7 into the function, then I get: y = a + b + c y = 9a + 3b + c y = 25a + 5b + c y = 49a + 7b + c
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14:09:52 Substituting the values of a, b and c into the given form we obtain the equation y = - 0.45833 x^2 + 5.33333 x - 6.875. When we substitute 1 into the equation we obtain y = -.45833 * 1^2 + 5.33333 * 1 - 6.875 = -2. When we substitute 3 into the equation we obtain y = -.45833 * 3^2 + 5.33333 * 3 - 6.875 = 5. When we substitute 5 into the equation we obtain y = -.45833 * 5^2 + 5.33333 * 5 - 6.875 = 8.33333. When we substitute 7 into the equation we obtain y = -.45833 * 7^2 + 5.33333 * 7 - 6.875 = 8. Thus the y values we obtain for our x values gives us the points (1, -2), (3, 5) and (7, 8) we used to obtain the formula, plus the point (5, 8.33333).
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RESPONSE --> I can see how my instructor solved the system of equations using the correct values for a, b, and c; although I was not correct in my values. I'm still not sure where I went wrong. ???
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}yz\ Student Name: assignment #003
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14:22:07 `q001. Note that this assignment has 6 questions The function y = a x^2 + b x + c takes the value y = 0 when x = [ -b + `sqrt(b^2 - 4 a c ] / (2 a) or when x = [ -b - `sqrt(b^2 - 4 a c ] / (2 a). For the function y = - 0.45833 x^2 + 5.33333 x - 6.875, which you obtained as a quadratic model of the points (1, -2), (3, 5) and (7, 8) in the preceding assignment, find the values of x for which y = 0. Compare to the estimates you made from the graph through (1,-2), (3, 5) and (7, 8) in Assignment 1.
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RESPONSE --> To find the value of x when y = 0, I used a = 0.45833 b = 5.33333 c = 6.875 and plugged them into the function x = [ -b + 'sqrt(b^2 - 4 a c] / (2 a) I then got that x = -1.483 when y = 0
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14:25:57 For the function y = - 0.45833 x^2 + 5.33333 x - 6.875 we have a = -0.45833, b = 5.33333 and c = -6.875. The quadratic formula therefore tells us that for our function we have y = 0 when x = [-5.33333 + `sqrt(5.33333^2 - 4 * (-0.45833 ) * (-6.875)) ] / ( 2 * (-0.45833)) = 1.47638 and when x = [-5.33333 - `sqrt(5.33333^2 - 4 * (-0.45833 ) * (-6.875)) ] / ( 2 * (-0.45833)) = 10.16006.
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RESPONSE --> I was off by a few points in the answer x = -1.483, but I understand what I did wrong. I left out two negative signs for the values (-0.45833) and (-6.875). I must've overlooked the other equation to get the answer 10.16006.
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14:28:54 `q002. Extend the smooth curve in your sketch to include both points at which y = 0. Estimate the x value at which y takes its maximum value.
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RESPONSE --> When I extend the smooth curve in my sketch to include both points at which y = 0, I estimate that the x value at which y takes its maximum value will be about 2. ????
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14:31:12 Your graph should clearly show how the parabola passes through the x axis at the points where x is approximately 1.5 (corresponding to the more accurate value 1.47638 found in the preceding problem) and where takes is a little more than 10 (corresponding to the more accurate value 10.16006 found in the preceding problem). The graph of the parabola will peak halfway between these x values, at approximately x = 6 (actually closer to x = 5.8), where the y value will be between 8 and 9.
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RESPONSE --> I didn't get the right answer. Since i wasn't exactly sure about my graph in the previous assignment, I'm not sure how my instructor got the conclusion of the graph will peak halfway between x = 1.5 and x = 10, where the y value will be between 8 and 9. ???
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14:35:13 `q003. For the function of the preceding two questions, y will take its maximum value when x is halfway between the two values at which y = 0. Recall that these two values are approximately x = 1.48 and x = 10.16. At what x value will the function take its maximum value? What will be this value? What are the coordinates of the highest point on the graph?
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RESPONSE --> The function will take its maximum value at the x value of about 5.82. The coordinates of the highest point on the graph are (7, 8). ???
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14:37:30 The x value halfway between x = 1.48 and x = 10.16 is the average x = (1.48 + 10.16) / 2 = 5.82. At x = 5.82 we have y = - 0.45833 x^2 + 5.33333 x - 6.875 = -.45833 * 5.82^2 + 5.33333 * 5.82 - 6.875 = 8.64 approx.. Thus the graph of the function will be a parabola whose maximum occurs at its vertex (5.82, 8.64).
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RESPONSE --> I got the value of x right, but not the maximum points of the graph. I understand that the instructor used the function y = a x^2 + b x + c to find the y value of the maximum points.
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14:42:26 `q004. The function y = a x^2 + b x + c has a graph which is a parabola. This parabola will have either a highest point or a lowest point, depending upon whether it opens upward or downward. In either case this highest or lowest point is called the vertex of the parabola. The vertex of a parabola will occur when x = -b / (2a). At what x value, accurate to five significant figures, will the function y = - 0.458333 x^2 + 5.33333 x - 6.875 take its maximum value? Accurate to five significant figures, what is the corresponding y value?
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RESPONSE --> At the x value of 5.81822, the function y = -0.45833 x^2 + 5.33333 x - 6.875 will take its maximum value.
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14:45:15 In the preceding problem we approximated the x value at which the function is maximized by averaging 1.48 and 10.16, the x values at which the function is zero. Here we will use x = -b / (2 a) to obtain x value at which function is maximized: x = -b / (2a) = 5.33333 / (2 * -0.45833) = 5.81818. To find corresponding y value we substitute x = 5.81818 into y = - 0.458333 x^2 + 5.33333 x - 6.875 to obtain y = 8.64024. Thus the vertex of the parabola lies at (5.81818, 8.64024).
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RESPONSE --> I correctly answered the x value at which the function is maximized (5.81818), but not the corresponding y value (8.64024). I understand though, how my instructor reached this answer.
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14:55:50 `q005. As we just saw the vertex of the parabola defined by the function y = - 0.45833 x^2 + 5.33333 x - 6.875 lies at (5.8182, 8.6402). What is the value of x at a point on the parabola which lies 1 unit to the right of the vertex, and what is the value of x at a point on the parabola which lies one unit to the left of the vertex? What is the value of y corresponding to each of these x values? By how much does each of these y values differ from the y value at the vertex, and how could you have determined this number by the equation of the function?
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RESPONSE --> The value of x at a point on the parabola which lies 1 unit to the right of the vertex, will be 6.81818 and the value of x at a point on the parabola which lies one unit to the left of the vertex is 4.81818. The value of y corresponding to the x value which lies 1 unit to the right will be 8.18195 and when the x value lies 1 unit to the left, y will be 8.18187. These y values don't differ much from the y value at the vertex.
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14:59:54 The vertex lies at x = 5.8182, so the x values of points lying one unit to the right and left of the vertex must be x = 6.8182 and x = 4.8182. At these x values we find by substituting into the function y = - 0.458333 x^2 + 5.33333 x - 6.875 that at both points y = 8.1818. Each of these y values differs from the maximum y value, which occurs at the vertex, by -0.4584. This number is familiar. Within roundoff error is identical to to the coefficient of x^2 in the original formula y = - 0.458333 x^2 + 5.33333 x - 6.875. This will always be the case. If we move one unit to the right or left of the vertex of the parabola defined by a quadratic function y = a x^2 + b x + c, the y value always differ from the y value at the vertex by the coefficient a of x^2. Remember this.
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RESPONSE --> i got it right, except my calculator automatically rounds the number off, even by the smallest decimal, therefore it had one of the values for y slightly larger than the other.
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15:14:06 `q006. In the preceding problem we saw an instance of the following rule: The function y = a x^2 + b x + c has a graph which is a parabola. This parabola has a vertex. If we move 1 unit in the x direction from the vertex, moving either 1 unit to the right or to the left, then move vertically a units, we end up at another point on the graph of the parabola. In assignment 2 we obtained the solution a = -1, b = 10, c = 100 for a system of three simultaneous linear equations. If these linear equations had been obtained from 3 points on a graph, we would then have the quadratic model y = -1 x^2 + 10 x + 100 for those points. What would be the coordinates of the vertex of this parabola? What would be the coordinates of the points on the parabola which lie 1 unit to the right and one unit to the left of the vertex? Sketch a graph with these three points, and sketch a parabola through these points. Will this parabola ever touch the x axis?
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RESPONSE --> my answer is that the coordinates of the vertex of this parabola will be (5.00000, 125). If the coordinates of the points in the parabola lied 1 unit to the right it would be (6.00000, 125) and if it lied 1 unit to the left it would be (4.00000, 125). ?????
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15:14:38 02-01-2006 15:14:38 The vertex of the parabola given by y = -1 x^2 + 10 x + 100 will lie at x = -b / (2 a) = -10 / (2 * -1) = 5. At the vertex the y value will therefore be y = -1 x^2 + 10 x + 100 = -1 * 5^2 + 10 * 5 + 100 = 125. It follows that if we move 1 unit in the x direction to the right or left, the y value will change by a = -1. The y value will therefore change from 125 to 124, and we will have the 3 'fundamental points' (4, 124), (5, 125), and (6, 124). Your graph should show a parabola peaking at (5, 125) and descending slightly as we move 1 unit to the right or left of this vertex. The parabola will then descend more and more rapidly, eventually crossing the x-axis both to the left and to the right of the vertex. The points to the right and left show clearly that the parabola descends from its vertex. This is because in this case a = -1, a negative value, which effectively pulls the parabola down on either side of the vertex. Had the value of a been positive, the points one unit to the right and left would lie above the vertex and the parabola would asscend from its vertex.
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NOTES ------->
.................................................R{֗ⶤXǣğ
Student Name: assignment #004
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15:28:00 `q001. Note that this assignment has 4 questions If f(x) = x^2 + 4, then find the values of the following: f(3), f(7) and f(-5). Plot the corresponding points on a graph of y = f(x) vs. x. Give a good description of your graph.
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RESPONSE --> If f(x) = x^2 + 4, then the values of f(3), f(7) and f(-5) would be: f(3) = 3^2 + 4 = 9 + 4 = 13 f(7) = 7^2 + 4 = 49 + 4 = 53 f(-5) = (-5)^2 + 4 = 25 + 4 = 29 My graph, from the left to the right is points (-5, 29), (3, 13) and (7, 53). If followed from the left to the right, the line decreases until it hits the vertex of (3, 13) and then it starts increasing. ???
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15:33:46 f(x) = x^2 + 4. To find f(3) we replace x by 3 to obtain f(3) = 3^2 + 4 = 9 + 4 = 13. Similarly we have f(7) = 7^2 + 4 = 49 + 4 = 53 and f(-5) = (-5)^2 + 9 = 25 + 4 = 29. Graphing f(x) vs. x we will plot the points (3, 13), (7, 53), (-5, 29). The graph of f(x) vs. x will be a parabola passing through these points, since f(x) is seen to be a quadratic function, with a = 1, b = 0 and c = 4. The x coordinate of the vertex is seen to be -b/(2 a) = -0/(2*1) = 0. The y coordinate of the vertex will therefore be f(0) = 0 ^ 2 + 4 = 0 + 4 = 4. Moving along the graph one unit to the right or left of the vertex (0,4) we arrive at the points (1,5) and (-1,5) on the way to the three points we just graphed.
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RESPONSE --> I got most of the answer correct. I understand how f(x) is seen to be a quadratic function, with a = 1, b = 0, and c = 4. I also understand how to solve for the x and y coordinates of the vertex.
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15:45:28 `q002. If f(x) = x^2 + 4, then give the symbolic expression for each of the following: f(a), f(x+2), f(x+h), f(x+h)-f(x) and [ f(x+h) - f(x) ] / h. Expand and/or simplify these expressions as appropriate.
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RESPONSE --> The symbolic expression for each of the following: f(a), f(x+2), f(x+h), f(x+h)-f(x), and [ f(x+h)-f(x) ] / h are: f(a) = a^2 + 4 f(x+2) = (x+2)^2 + 4 = (x+2)(x+2) + 4 f(x+h) = (x+h)^2 + 4 = (x+h)(x+h) + 4 f(x+h-f(x)) = (x+h-f(x))^2 + 4 = (x+h-f(x))(x+h-f(x)) + 4 [f(x+h)-f(x)] / h = ([f(x+h)-f(x)] / h)^2 + 4 = ([f(x+h)-f(x)] / h)([f9x+h)-f(x)] / h) + 4
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15:50:20 If f(x) = x^2 + 4, then the expression f(a) is obtained by replacing x with a: f(a) = a^2 + 4. Similarly to find f(x+2) we replace x with x + 2: f(x+2) = (x + 2)^2 + 4, which we might expand to get (x^2 + 4 x + 4) + 4 or x^2 + 4 x + 8. To find f(x+h) we replace x with x + h to obtain f(x+h) = (x + h)^2 + 4 = x^2 + 2 h x + h^2 + 4. To find f(x+h) - f(x) we use the expressions we found for f(x) and f(x+h): f(x+h) - f(x) = [ x^2 + 2 h x + h^2 + 4 ] - [ x^2 + 4 ] = x^2 + 2 h x + 4 + h^2 - x^2 - 4 = 2 h x + h^2. To find [ f(x+h) - f(x) ] / h we can use the expressions we just obtained to see that [ f(x+h) - f(x) ] / h = [ x^2 + 2 h x + h^2 + 4 - ( x^2 + 4) ] / h = (2 h x + h^2) / h = 2 x + h.
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RESPONSE --> I got my replacements correct, but not the expanding/simplifying. I'm still a little confused about this.
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15:55:56 `q003. If f(x) = 5x + 7, then give the symbolic expression for each of the following: f(x1), f(x2), [ f(x2) - f(x1) ] / ( x2 - x1 ). Note that x1 and x2 stand for subscripted variables (x with subscript 1 and x with subscript 2), not for x * 1 and x * 2. x1 and x2 are simply names for two different values of x. If you aren't clear on what this means please ask the instructor.
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RESPONSE --> the symbolic expressions for f(x1), f(x2) and [ f(x2) - f(x1) ] / (x2 - x1) are: 5*(x1) + 7 5*(x2) + 7 5*[ f(x2) - f(x1) ] / (x2 - x1) + 7 I'm not 100% clear on what the subscripts (x1) and (x2) mean. ???
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16:00:32 Replacing x by the specified quantities we obtain the following: f(x1) = 5 * x1 + 7, f(x2) = 5 * x2 + 7, [ f(x2) - f(x1) ] / ( x2 - x1) = [ 5 * x2 + 7 - ( 5 * x1 + 7) ] / ( x2 - x1) = [ 5 x2 + 7 - 5 x1 - 7 ] / (x2 - x1) = (5 x2 - 5 x1) / ( x2 - x1). We can factor 5 out of the numerator to obtain 5 ( x2 - x1 ) / ( x2 - x1 ) = 5.
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RESPONSE --> I correctly answered the first two expressions, but not the last one. I'm still not sure of the expansion of the expression. ???
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16:04:05 `q004. If f(x) = 5x + 7, then for what value of x is f(x) equal to -3?
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RESPONSE --> If f(x) = 5x + 7, then when f(x) is equal to -3, the value of x is -8 ???
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16:05:58 If f(x) is equal to -3 then we right f(x) = -3, which we translate into the equation 5x + 7 = -3. We easily solve this equation (subtract 7 from both sides then divide both sides by 5) to obtain x = -2.
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RESPONSE --> I didn't get the question correct, but i do understand how to get it now. I was making it much harder than what it really is.
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__________________________________________________________________________________ VM`Ε҆ɍ|Sc Student Name: assignment #005
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10:59:35 `q001. Note that this assignment has 8 questions Evaluate the function y = x^2 for x values -3, -2, -1, 0, 1, 2, and 3. What are your y values?
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RESPONSE --> The y values for the function y = x^2 for the x values -3, -2, -1, 0, 1, 2 and 3 are: y = -9, -4, -1, 0, 1, 4 and 9
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11:01:05 You should have obtained y values 9, 4, 1, 0, 1, 4, 9, in that order.
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RESPONSE --> I miss calculated my values. Instead if saying (-3)^2 I said -3^2. I understand my mistakes.
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11:13:12 `q002. Evaluate the function y = 2^x for x values -3, -2, -1, 0, 1, 2, and 3. What are your y values?
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RESPONSE --> The y values for the function y = 2^x for x values -3, -2, -1, 0, 1, 2 and 3 are: y = 1/8, 1/4, 1/2, 1, 2, 4 and 8
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11:13:32 By velocity exponents, b^-x = 1 / b^x. So for example 2^-2 = 1 / 2^2 = 1/4. Your y values will be 1/8, 1/4, 1/2, 1, 2, 4 and 8. Note that we have used the fact that for any b, b^0 = 1. It is a common error to say that 2^0 is 0. Note that this error would interfere with the pattern or progression of the y values.
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RESPONSE --> I was correct.
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11:19:11 `q003. Evaluate the function y = x^-2 for x values -3, -2, -1, 0, 1, 2, and 3. What are your y values?
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RESPONSE --> The y values for the function y = x^2 for x values -3, -2, -1, 0, 1, 2, and 3 are: y = 9, 4, 1, 0, 1, 4, and 9
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11:21:54 By the laws of exponents, x^-p = 1 / x^p. So x^-2 = 1 / x^2, and your x values should be 1/9, 1/4, and 1. Since 1 / 0^2 = 1 / 0 and division by zero is not defined, the x = 0 value is undefined. The last three values will be 1, 1/4, and 1/9.
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RESPONSE --> I miscalculated. I wasn't clear that x^2 = 1 / x^2. I thought the first symbol was the exponent symbol. I understand the function now.
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11:26:18 `q004. Evaluate the function y = x^3 for x values -3, -2, -1, 0, 1, 2, and 3. What are your y values?
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RESPONSE --> The y values for the function y = x^3 for x values -3, -2, -1, 0, 1, 2 and 3 are: y = -27, -8, -1, 0, 1, 8, and 27
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11:27:58 The y values should be -27, -8, -1, 0, 1, 4, 9.
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RESPONSE --> I didn't correctly answer the last three terms (1, 4, and 9). How does 2^3 = 4 and 3^3 = 9?? I don't understand this.
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11:49:34 `q005. Sketch graphs for y = x^2, y = 2^x, y = x^-2 and y = x^3, using the values you obtained in the preceding four problems. Describe the graph of each function.
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RESPONSE --> The graph for the function y = x^2 is a parabola that at (-3, 9), from left to right, decreases, (-2, 4), (-1, 1) then a vertex of (0, 0) then starts to increase again at (1, 1) and keeps rising (2, 4) and (3, 9). The graph for the second function y = 2^x, from left to right, increases slightly at (-3, 1/8), (-2, 1/4), (-1, 1/2) then a vertex at (0, 1) and then still increases at a more rapid/steep rate at (1, 2), (2, 4) and (3, 8). The graph of the function y = x^2, from left to right, makes a ""m"" shape. It increases (-3, 1/9), (-2, 1/4), (-1, 1) and then decreases to the vertex (0, 0) then starts to increase again to (1, 1) and then decreases again (2, 1/4) and (3, 1/9). The graph of the function y = x^3, from left to right, makes a kind of ""s"" shape, but backwards and tilted on its side. (kind of like an N). It increases (-3, -27), (-2, -8), (-1, -1) and then levels off slightly with point (0, 0) where the line intercepts and then levels slightly with (1, 1) and then increases (2, 4) and (3, 9).
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11:50:55 The graph of y = x^2 is a parabola with its vertex at the origin. It is worth noting that the graph is symmetric with respect to the y-axis. That is, the graph to the left of the y-axis is a mirror image of the graph to the right of the y-axis. The graph of y = 2^x begins at x = -3 with value 1/8, which is relatively close to zero. The graph therefore starts to the left, close to the x-axis. With each succeeding unit of x, with x moving to the right, the y value doubles. This causes the graph to rise more and more quickly as we move from left to right. The graph intercepts the y-axis at y = 1. The graph of y = x^-2 rises more and more rapidly as we approach the y-axis from the left. It might not be clear from the values obtained here that this progression continues, with the y values increasing beyond bound, but this is the case. This behavior is mirrored on the other side of the y-axis, so that the graph rises as we approach the y-axis from either side. In fact the graph rises without bound as we approach the y-axis from either side. The y-axis is therefore a vertical asymptote for this graph. The graph of y = x ^ 3 has negative y values whenever x is negative and positive y values whenever x is positive. As we approach x = 0 from the left, through negative x values, the y values increase toward zero, but the rate of increase slows so that the graph actually levels off for an instant at the point (0,0) before beginning to increase again. To the right of x = 0 the graph increases faster and faster. Be sure to note whether your graph had all these characteristics, and whether your description included these characteristics. Note also any characteristics included in your description that were not included here.
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RESPONSE --> I seem to be correct.
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11:57:26 `q006. Make a table for y = x^2 + 3, using x values -3, -2, -1, 0, 1, 2, 3. How do the y values on the table compare to the y values on the table for y = x^2? How does the graph of y = x^2 + 3 compare to the graph of y = x^2?
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RESPONSE --> The values for y = x^2 + 3 for x values -3, -2, -1, 0, 1, 2, and 3 are: y = 12, 7, 4, 0, 4, 7, 12 These y values compare to the y values of the table for y = x^2, by that 3 is added to the values. Both graphs are a parabola, which the right side mirrors the left side of the graph. The graph for the function y = x^2 + 3 has steeper sides than that of the other function and the parabola is slightly more narrow.
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11:59:21 A list of the y values will include, in order, y = 12, 7, 4, 3, 4, 7, 12. A list for y = x^2 would include, in order, y = 9, 4, 1, 0, 1, 4, 9. The values for y = x^2 + 3 are each 3 units greater than those for the function y = x^2. The graph of y = x^2 + 3 therefore lies 3 units higher at each point than the graph of y = x^2.
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RESPONSE --> I got it correct, all but the x value 0 (zero). it should have been 3, since the function is y = x^2 + 3. I understand this now. I just overlooked it.
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12:06:41 `q007. Make a table for y = (x -1)^3, using x values -3, -2, -1, 0, 1, 2, 3. How did the values on the table compare to the values on the table for y = x^3? Describe the relationship between the graph of y = (x -1)^3 and y = x^3.
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RESPONSE --> The y values for the function y = (x - 1)^3 are: y = -64, -27, -8, -1, 0, 1, and 8 These values compare to those of the function y = x^3 in that they are the same, except one begins with a larger value (-64) and ends with a smaller value (8). Both graphs look similar.
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12:07:09 The values you obtained should have been -64, -27, -8, -1, 0, 1, 8. The values for y = x^3 are -27, -8, -1, 0, 1, 8, 27. The values of y = (x-1)^3 are shifted 1 position to the right relative to the values of y = x^3. The graph of y = (x-1)^3 is similarly shifted 1 unit to the right of the graph of y = x^3.
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RESPONSE --> I was correct.
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12:12:25 `q008. Make a table for y = 3 * 2^x, using x values -3, -2, -1, 0, 1, 2, 3. How do the values on the table compare to the values on the table for y = 2^x? Describe the relationship between the graph of y = 3 * 2^x and y = 2^x.
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RESPONSE --> The y values for the function y = 3 * 2^x and x values of -3, -2, -1, 0, 1, 2, and 3 are: y = 3/8, 3/4, 3/2, 3, 6, 12, and 24 The graphs are similar.
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12:12:49 You should have obtained y values 3/8, 3/4, 3/2, 3, 6, 12 and 24. Comparing these with the values 1/8, 1/4, 1/2, 1, 2, 4, 8 of the function y = 2^x we see that the values are each 3 times as great. The graph of y = 3 * 2^x has an overall shape similar to that of y = 2^x, but each point lies 3 times as far from the x-axis. It is also worth noting that at every point the graph of y = 3 * 2^x is three times as the past that of y = 2^x.
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RESPONSE --> I was correct. I understand this more now.
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