course Mth 271 HZZƫې~ڤassignment #001
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09:17:18 `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
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09:19:12 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 found it
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09:23:16 `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 --> The send file contains all of my responses. Each response is added, and nothing is written over. If for some reason the program dissapears or is deleted the send file will still be intact and contain all my responses.
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09:35:25 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 --> On this page is a link to my portfolio, a submit work form, a discussion form, a test taken form, help with programs, and a proctor information form.
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09:42:20 `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 --> It is a good idea in all situations to back up work on an outside source. I have a USB drive just for this.
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ɗTٸ~x Student Name: assignment #001 001. typewriter notation
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22:26:08 `q001. Explain the difference between x - 2 / x + 4 and (x - 2) / (x + 4).
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RESPONSE --> This is two totally different problems. The first has the middle as 2 over x. I would read it as x minus (two over X) plus four. the second has order as what is in ( ) should be done first. this is x minus 2 should be divided by x plus 4.
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22:27:46 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 --> you stated it better than I did,
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22:33:11 `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 --> the first one is 2 raised to the power of x plus 4. 2^2 +4= 4+4= 8 is the first answer the second is 2 raised to the power of x plus 4 2^(2+4)= 2^6= 64 is the second answer
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22:34:30 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 --> cool, I need to show my work a little bit differently
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22:58:14 `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 --> numerator is three denominator is the expression in the brackets [ (2x-5)^2 * 3x+1] 2 - 3 / [ (2*2-5)^2 *(3*2) +1] -2 + 7*2= 2 - 3 / [ (4-5)^2 * 6 +1 ] -2 + 14 = 2 - 3/ [ -1^2 * 6 + 1] -2 +14 = 2 - 3/ [ 1*6 + 1] -2 +14= 2 - 3/ [6+1] -2 +14= 2 - 3/ [7] -2 +14= 2 - 3/7 -2 +14 = -3/7 + 14= 13 4/7
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22:59:48 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 --> would you rather see 95/7 or 13 4/7 ?
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Θ}Մ^he Student Name: assignment #001 001. typewriter notation
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16:57:27 `q001. Explain the difference between x - 2 / x + 4 and (x - 2) / (x + 4).
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RESPONSE -->
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16:57:36 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 -->
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16:57:50 `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 -->
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16:58:00 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 -->
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18:16:11 `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 --> numerator is 3 denominator is brackets of [ (2x-5)^2 * 3x +1] to evaluate this expression first evaluate what is in the ()'s then out to the brackets. First multiply 2 by 2 then subtract 5 this will give you -1 raise -1 to the 2nd power which is 1. multiply 3 by 2 then multiply this answer by the first result of 1 add this result to 1 which gives you a denominator of 7. so at this point we have 2 - 3/7 -2 + 7x the 2's will cancel each other out, multiply 2 by 7 then subtract 3/7 to give a final answer of 13 and 3/7 or 94/7 or 13.5714285714
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18:19:13 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 may not have explained it correctly, but still came up with that answer. Hopefully I will get the hang of this.
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18:49:32 `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 --> start with the ()'s 4 - 5= -1 Since there is no ()'s after the ^ i will assume 2 is the exponent. -1 squared is 1. multiply 1 by 4 then subtract 1 so now we would have 3 + 3/4 - 2 re-align the problem 3-2+3/4 which equals 1 and 3/4 or 7/4 or 1.75 this is my calculation unless the extra spaces seperate the problem, then I'm wrong.
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18:50:42 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 --> yeah, spaces mean nothing. exponents come first. so far so good.
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oH}ǿї Student Name: assignment #005 005. Calculus
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20:32:26 `q001. The graph of a certain function is a smooth curve passing through the points (3, 5), (7, 17) and (10, 29). Between which two points do you think the graph is steeper, on the average? Why do we say 'on the average'?
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RESPONSE --> to determine which is steeper, we must determine the slope of each set. to do that you divide the rise by the run. I guess that is why we say on average. the actual line crosses many plots and you use the beginning and ending to find the slope of a particular set. slope = rise divided by run which is change in y divided by the change in x or (y2 - y1) / (x2-x1) for the first set that is (17-5)/(7-3 )= 12/4 = 3 for the second set (29-17)/(10-7) = 12/3 = 4 the slope between (7, 17) and (10, 29) is steeper on average.
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20:33:29 Slope = rise / run. Between points (7, 17) and (10, 29) we get rise / run = (29 - 17) / (10 - 7) =12 / 3 = 4. The slope between points (3, 5) and (7, 17) is 3 / 1. (17 - 5) / (7 -3) = 12 / 4 = 3. The segment with slope 4 is the steeper. The graph being a smooth curve, slopes may vary from point to point. The slope obtained over the interval is a specific type of average of the slopes of all points between the endpoints.
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RESPONSE --> ok
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21:00:13 2. Answer without using a calculator: As x takes the values 2.1, 2.01, 2.001 and 2.0001, what values are taken by the expression 1 / (x - 2)? 1. As the process continues, with x getting closer and closer to 2, what happens to the values of 1 / (x-2)? 2. Will the value ever exceed a billion? Will it ever exceed one trillion billions? 3. Will it ever exceed the number of particles in the known universe? 4. Is there any number it will never exceed? 5. What does the graph of y = 1 / (x-2) look like in the vicinity of x = 2?
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RESPONSE --> to solve, subtract 2 from 2.1 which is .1 then divide 1 by this value 1/.1 = 10 1/.01 = 100. 1/.001 = 1000. 1 as x continues closer to 2, the value of the equation increases to the next tenth, oh ok i forgot the correct mathmatical term but the next one would be 10 thousand, then 100 thousand, then 1 million. 2. yes it can exceed a billion and it can exceed one trillion billions. to exceed a billion x would be 2.000000001 just keep extending the number in the same way to reach the next amount. 3. yes, the value of the equation could go on to infinity. 4. I don't think there is any # that it could exceed. If the values continued in the pattern of expanding by a "".0"" then the value would increase to infinity. 5. y would equal zero so the plot woud be (2 , 0)
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21:04:57 For x = 2.1, 2.01, 2.001, 2.0001 we see that x -2 = .1, .01, .001, .0001. Thus 1/(x -2) takes respective values 10, 100, 1000, 10,000. It is important to note that x is changing by smaller and smaller increments as it approaches 2, while the value of the function is changing by greater and greater amounts. As x gets closer in closer to 2, it will reach the values 2.00001, 2.0000001, etc.. Since we can put as many zeros as we want in .000...001 the reciprocal 100...000 can be as large as we desire. Given any number, we can exceed it. Note that the function is simply not defined for x = 2. We cannot divide 1 by 0 (try counting to 1 by 0's..You never get anywhere. It can't be done. You can count to 1 by .1's--.1, .2, .3, ..., .9, 1. You get 10. You can do similar thing for .01, .001, etc., but you just can't do it for 0). As x approaches 2 the graph approaches the vertical line x = 2; the graph itself is never vertical. That is, the graph will have a vertical asymptote at the line x = 2. As x approaches 2, therefore, 1 / (x-2) will exceed all bounds. Note that if x approaches 2 through the values 1.9, 1.99, ..., the function gives us -10, -100, etc.. So we can see that on one side of x = 2 the graph will approach +infinity, on the other it will be negative and approach -infinity.
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RESPONSE --> I was thinking 1/0 is 0, ok i forgot that. it has been a while for me. it is 1 times 0 is zero. It's amazing to me that I remember some of this other, then forget something that simple.
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21:39:27 `q003. One straight line segment connects the points (3,5) and (7,9) while another connects the points (10,2) and (50,4). From each of the four points a line segment is drawn directly down to the x axis, forming two trapezoids. Which trapezoid has the greater area? Try to justify your answer with something more precise than, for example, 'from a sketch I can see that this one is much bigger so it must have the greater area'.
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RESPONSE --> i graphed the 2 trapezoids to be able to see them visually - then i would need to use the formula to find the area. I believe the formula is .5h times the bases added together. the height is actually the ""length"" of the trapezoid along the x axis and each base is the ""length"" of the bases up the y axis. since the bases start out at zero on the y axis, the ""length"" of the bases is the actual plot points. the first one of 5 and 9 and the second is 2 and 4. to find the heights along the x axis I would need to subtract the first plot point from the second plot point. the first trapezoid would be 7 minus 3 = 4 the second trapezoid would be 50 minus 10 = 40 so to calculate the first area it would be: .5(4)(5+9) = 2(14) = 28 the second one would be: .5(40)(2+4) = 20(6) = 120 the second trapezoid with the points of (10,2) and (50,4) is mathmatically larger by 92.
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21:42:38 Your sketch should show that while the first trapezoid averages a little more than double the altitude of the second, the second is clearly much more than twice as wide and hence has the greater area. To justify this a little more precisely, the first trapezoid, which runs from x = 3 to x = 7, is 4 units wide while the second runs from x = 10 and to x = 50 and hence has a width of 40 units. The altitudes of the first trapezoid are 5 and 9,so the average altitude of the first is 7. The average altitude of the second is the average of the altitudes 2 and 4, or 3. So the first trapezoid is over twice as high, on the average, as the first. However the second is 10 times as wide, so the second trapezoid must have the greater area. This is all the reasoning we need to answer the question. We could of course multiply average altitude by width for each trapezoid, obtaining area 7 * 4 = 28 for the first and 3 * 40 = 120 for the second. However if all we need to know is which trapezoid has a greater area, we need not bother with this step.
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RESPONSE --> ok, I just assumed you wanted proof, when in doubt, I hope that is ok.
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22:06:14 `q004. If f(x) = x^2 (meaning 'x raised to the power 2') then which is steeper, the line segment connecting the x = 2 and x = 5 points on the graph of f(x), or the line segment connecting the x = -1 and x = 7 points on the same graph? Explain the basisof your reasoning.
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RESPONSE --> the first plots of x= 2 and x= 5 only expands 3 units wide as it climbs 21 units up the slope, but with the second plots of x=-1 and x=7 this expands across 8 units wide as it climbs 48 units high. Because the second segment is crossing more units across the x axis the first segment will be steeper. I know what I mean, but it is hard for me to express. If you have two segments, the one that expands the least across the x axis will be the steepest one.
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22:06:33 The line segment connecting x = 2 and the x = 5 points is steeper: Since f(x) = x^2, x = 2 gives y = 4 and x = 5 gives y = 25. The slope between the points is rise / run = (25 - 4) / (5 - 2) = 21 / 3 = 7. The line segment connecting the x = -1 point (-1,1) and the x = 7 point (7,49) has a slope of (49 - 1) / (7 - -1) = 48 / 8 = 6. The slope of the first segment is greater.
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RESPONSE --> ok
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22:14:47 `q005. Suppose that every week of the current millenium you go to the jewler and obtain a certain number of grams of pure gold, which you then place in an old sock and bury in your backyard. Assume that buried gold lasts a long, long time ( this is so), that the the gold remains undisturbed (maybe, maybe not so), that no other source adds gold to your backyard (probably so), and that there was no gold in your yard before.. 1. If you construct a graph of y = the number of grams of gold in your backyard vs. t = the number of weeks since Jan. 1, 2000, with the y axis pointing up and the t axis pointing to the right, will the points on your graph lie on a level straight line, a rising straight line, a falling straight line, a line which rises faster and faster, a line which rises but more and more slowly, a line which falls faster and faster, or a line which falls but more and more slowly? 2. Answer the same question assuming that every week you bury 1 more gram than you did the previous week. {}3. Answer the same question assuming that every week you bury half the amount you did the previous week.
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RESPONSE --> if you went to the jeweler and obtained the exact same amount every week, without missing any weeks, you would have a rising straight line. the line would increase the exact same amount each time the slope would be consistently the same. Each increase of the number of weeks would be exactly the same: 1,2,3,4,5,6 etc and the amount of gold would increase at exactly the same rate . 16oz, 32 oz, 48 oz, 64oz etc.
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22:15:53 1. If it's the same amount each week it would be a straight line. 2. Buying gold every week, the amount of gold will always increase. Since you buy more each week the rate of increase will keep increasing. So the graph will increase, and at an increasing rate. 3. Buying gold every week, the amount of gold won't ever decrease. Since you buy less each week the rate of increase will just keep falling. So the graph will increase, but at a decreasing rate. This graph will in fact approach a horizontal asymptote, since we have a geometric progression which implies an exponential function.
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RESPONSE --> ok
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22:29:03 `q006. Suppose that every week you go to the jewler and obtain a certain number of grams of pure gold, which you then place in an old sock and bury in your backyard. Assume that buried gold lasts a long, long time, that the the gold remains undisturbed, and that no other source adds gold to your backyard. 1. If you graph the rate at which gold is accumulating from week to week vs. tne number of weeks since Jan 1, 2000, will the points on your graph lie on a level straight line, a rising straight line, a falling straight line, a line which rises faster and faster, a line which rises but more and more slowly, a line which falls faster and faster, or a line which falls but more and more slowly? 2. Answer the same question assuming that every week you bury 1 more gram than you did the previous week. 3. Answer the same question assuming that every week you bury half the amount you did the previous week.
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RESPONSE --> 2. ok. if i bought an ounce more each week, the line would be rising faster and faster. the graph would have a steeper slope because the # of ounces would be increasing exceedingly faster than the number of weeks. the accumulation would be quicker. 3. if you bought half the amount as you did from the previous week, the line would still rise but more and more slowly. This is because the weeks increase at a steady amount but the gold amount only increases half the amount of the last increase. This will cause the slope to be notably less.
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22:32:52 This set of questions is different from the preceding set. This question now asks about a graph of rate vs. time, whereas the last was about the graph of quantity vs. time. Question 1: This question concerns the graph of the rate at which gold accumulates, which in this case, since you buy the same amount eact week, is constant. The graph would be a horizontal straight line. Question 2: Each week you buy one more gram than the week before, so the rate goes up each week by 1 gram per week. You thus get a risingstraight line because the increase in the rate is the same from one week to the next. Question 3. Since half the previous amount will be half of a declining amount, the rate will decrease while remaining positive, so the graph remains positive as it decreases more and more slowly. The rate approaches but never reaches zero.
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RESPONSE --> i completely missed those, but I understand the concept. Apparently I didn't fully read the questions. I was still thinking of graphing what you have total not just the intake. My mistake
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22:56:01 7. If the depth of water in a container is given, in centimeters, by 100 - 2 t + .01 t^2, where t is clock time in seconds, then what are the depths at clock times t = 30, t = 40 and t = 60? On the average is depth changing more rapidly during the first time interval or the second?
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RESPONSE --> I'm not exactly sure on this one, but here is my best try: solve each equation for t for t = 30 100-2(30) + .01(30^2)=100 - 60 + .01(900)=40 +9=49 for t = 40 100-2(40) + .01(40^2)=100 - 80 + .01(1600)=20+16=36 for t = 60 100-2(60) + .01(60^2)=100-120 + .01(3600)=36-20=16 to me, if I did the calculations correctly the depth is changing more rapidly during the second time intreval. the difference would be 20 in the second interval and only 13 in the first. I would assume the water is being poured out.
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22:58:13 At t = 30 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 30 + .01 * 30^2 = 49. At t = 40 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 40 + .01 * 40^2 = 36. At t = 60 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 60 + .01 * 60^2 = 16. 49 cm - 36 cm = 13 cm change in 10 sec or 1.3 cm/s on the average. 36 cm - 16 cm = 20 cm change in 20 sec or 1.0 cm/s on the average.
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RESPONSE --> I didn't take into consideration the amount of time of the interval of 10 seconds in the first one and the 20 seconds in the second one.
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23:11:46 8. If the rate at which water descends in a container is given, in cm/s, by 10 - .1 t, where t is clock time in seconds, then at what rate is water descending when t = 10, and at what rate is it descending when t = 20? How much would you therefore expect the water level to change during this 10-second interval?
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RESPONSE --> Would it be the average between the two multiplied by the 10 second intreval between the two. solve each t = 10 10 - .1(10) = 10-1 = 9 t = 20 10 - .1(20) = 10 - 2 = 8 the average rate of the two is (9 + 8)/2 = 8.5 multiplied by the number of seconds in the intreval (10) = 85 cm
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23:13:01 At t = 10 sec the rate function gives us 10 - .1 * 10 = 10 - 1 = 9, meaning a rate of 9 cm / sec. At t = 20 sec the rate function gives us 10 - .1 * 20 = 10 - 2 = 8, meaning a rate of 8 cm / sec. The rate never goes below 8 cm/s, so in 10 sec the change wouldn't be less than 80 cm. The rate never goes above 9 cm/s, so in 10 sec the change wouldn't be greater than 90 cm. Any answer that isn't between 80 cm and 90 cm doesn't fit the given conditions.. The rate change is a linear function of t. Therefore the average rate is the average of the two rates, or 9.5 cm/s. The average of the rates is 8.5 cm/sec. In 10 sec that would imply a change of 85 cm.
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RESPONSE --> ok
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nƜcvм١ӂZ Student Name: assignment #001 001. typewriter notation ɀQԕӊܱ Student Name: assignment #005 005. Calculus ϔүօazs Student Name: assignment #001 001. Rates
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19:51:51 `q001. You should copy and paste these instructions to a word processor for reference. However you can always view them, as well as everything else that has appeared in this box, by clicking the 'Display Everything' button. 1. For the next question or answer, you click on 'Next Question / Answer' button above the box at top left until a question has been posed. Once a question has been posed you are to answer before you click again on this button. 2. Before clicking for an answer, type your best answer to the current question into the box to the right, then clip on the 'Enter Answer' button. 3. After entering your answer you will click on 'Next Question / Answer' to view the answer to the question. Do not tamper with the information displayed in the left-hand box. 4. If your answer was incorrect, incomplete or would otherwise require revision, you will enter a self-critique. If you learned something from the answer, you need to restate it in your own words in order to reinforce your learning. If there is something you feel you should note for future reference, you should make a note in your own words. Go to the response box (the right-hand box) and type in a self-critique and/or notes, as appropriate. Do not copy and paste anything from the left-hand box, since that information will be saved in any case. 5. If you wish to save your response to your Notes file you may choose to click on the 'Save As Notes' button rather than the 'Enter Answer' button. Doing so will save your work for your future reference. Your work will be saved in a Notes file in the c:\vhmthphy folder. The title of the Notes file will also include the name you gave when you started the program. 6. After clicking either the 'Enter Response' or the 'Save as Notes' button, click on 'Next Question / Answer' and proceed in a similar manner. In the right-hand box briefly describe your understanding of these instructions, then click 'Enter Answer'.
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RESPONSE --> I understand the instructions
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19:52:01 Your answer has been noted. Enter 'ok' in the Response Box and click on Enter Response, then click on Next Question/Answer for the first real question.
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RESPONSE --> ok
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19:53:09 `q002. Note that there are 10 questions in this assignment. The questions are of increasing difficulty--the first questions are fairly easy but later questions are very tricky. The main purposes of these exercises are to refine your thinking about rates, and to see how you process challenging information. Continue as far as you can until you are completely lost. Students who are prepared for the highest-level math courses might not ever get lost. If you make $50 in 5 hr, then at what rate are you earning money?
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RESPONSE --> $10 per hour 50/5 = 10
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19:53:41 The rate at which you are earning money is the number of dollars per hour you are earning. You are earning money at the rate of 50 dollars / (5 hours) = 10 dollars / hour. It is very likely that you immediately came up with the $10 / hour because almosteveryone is familiar with the concept of the pay rate, the number of dollars per hour. Note carefully that the pay rate is found by dividing the quantity earned by the time required to earn it. Time rates in general are found by dividing an accumulated quantity by the time required to accumulate it. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> ok
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19:56:07 `q003.If you make $60,000 per year then how much do you make per month?
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RESPONSE --> that would be $5,000 per month 60,000 for 12 months 60,000/12 = 5,000
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19:56:19 Most people will very quickly see that we need to divide $60,000 by 12 months, giving us 60,000 dollars / (12 months) = 5000 dollars / month. Note that again we have found a time rate, dividing the accumulated quantity by the time required to accumulate it. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> ok
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19:59:44 `q004. Suppose that the $60,000 is made in a year by a small business. Would be more appropriate to say that the business makes $5000 per month, or that the business makes an average of $5000 per month?
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RESPONSE --> You should say the business makes an average of $12,000 per month because sales for a business normally fluctuates from month to month.
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20:00:07 Small businesses do not usually make the same amount of money every month. The amount made depends on the demand for the services or commodities provided by the business, and there are often seasonal fluctuations in addition to other market fluctuations. It is almost certain that a small business making $60,000 per year will make more than $5000 in some months and less than $5000 in others. Therefore it is much more appropriate to say that the business makes and average of $5000 per month. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> ok
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20:03:03 `q005. If you travel 300 miles in 6 hours, at what average rate are you covering distance, and why do we say average rate instead of just plain rate?
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RESPONSE --> to determine miles per hour we divide the 300 miles by the number of hours (6) 300/6 = 50 mph. We call this an average rate because it is not feasible to drive 300 miles at exactly 50 mph due to traffic, stop lights, ect.
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20:03:31 The average rate is 50 miles per hour, or 50 miles / hour. This is obtained by dividing the accumulated quantity, the 300 miles, by the time required to accumulate it, obtaining ave rate = 300 miles / ( 6 hours) = 50 miles / hour. Note that the rate at which distance is covered is called speed. The car has an average speed of 50 miles/hour. We say 'average rate' in this case because it is almost certain that slight changes in pressure on the accelerator, traffic conditions and other factors ensure that the speed will sometimes be greater than 50 miles/hour and sometimes less than 50 miles/hour; the 50 miles/hour we obtain from the given information is clearly and overall average of the velocities. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> ok
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20:04:59 `q006. If you use 60 gallons of gasoline on a 1200 mile trip, then at what average rate are you using gasoline, with respect to miles traveled?
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RESPONSE --> to determine the miles per gallon we divide 1200 miles by the 60 gallons 1200/60 = 20 miles per gallons
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20:07:40 The rate of change of one quantity with respect to another is the change in the first quantity, divided by the change in the second. As in previous examples, we found the rate at which money was made with respect to time by dividing the amount of money made by the time required to make it. By analogy, the rate at which we use fuel with respect to miles traveled is the change in the amount of fuel divided by the number of miles traveled. In this case we use 60 gallons of fuel in 1200 miles, so the average rate it 60 gal / (1200 miles) = .05 gallons / mile. Note that this question didn't ask for miles per gallon. Miles per gallon is an appropriate and common calculation, but it measures the rate at which miles are covered with respect to the amount of fuel used. Be sure you see the difference. Note that in this problem we again have here an example of a rate, but unlike previous instances this rate is not calculated with respect to time. This rate is calculated with respect to the amount of fuel used. We divide the accumulated quantity, in this case miles, by the amount of fuel required to cover those miles. Note that again we call the result of this problem an average rate because there are always at least subtle differences in driving conditions that require the use of more fuel on some miles than on others. It's very important to understand the phrase 'with respect to'. Whether the calculation makes sense or not, it is defined by the order of the terms. In this case gallons / mile tells you how many gallons you are burning, on the average, per mile. This concept is not as familiar as miles / gallon, but except for familiarity it's technically no more difficult. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> Here we go again when I read but don't read. I do see that I need to be very careful and not make assumptions. I do understand how to do this calculation as well.
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20:16:03 `q007. The word 'average' generally connotes something like adding two quantities and dividing by 2, or adding several quantities and dividing by the number of quantities we added. Why is it that we are calculating average rates but we aren't adding anything?
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RESPONSE --> We are determing the average of two quantities in relation to each other. this calculation determines the miles per gallon for one trip. so essentially you could say that the miles are an average for 1 trip, and the gallons are the average for 1 trip. a better more reliable average would be over several trips.
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20:16:40 The word 'average' in the context of the dollars / month, miles / gallon types of questions we have been answering was used because we expect that in different months different amounts were earned, or that over different parts of the trip the gas mileage might have varied, but that if we knew all the individual quantities (e.g., the dollars earned each month, the number of gallons used with each mile) and averaged them in the usual manner, we would get the .05 gallons / mile, or the $5000 / month. In a sense we have already added up all the dollars earned in each month, or the miles traveled on each gallon, and we have obtained the total $60,000 or 1200 miles. Thus when we divide by the number of months or the number of gallons, we are in fact calculating an average rate. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> ok
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20:27:36 `q008. In a study of how lifting strength is influenced by various ways of training, a study group was divided into 2 subgroups of equally matched individuals. The first group did 10 pushups per day for a year and the second group did 50 pushups per day for year. At the end of the year to lifting strength of the first group averaged 147 pounds, while that of the second group averaged 162 pounds. At what average rate did lifting strength increase per daily pushup?
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RESPONSE --> ok, since I don't know the beginning lifting strength, but i do know both groups started equally. I would make the assumption that the difference in the number of pushups per day is directly related to the difference in the average lifting pounds at the end. the second group did 40 more pushups per day and at the end had increased to 15 more average pounds than the first group. so I would assume to find the average lifting strength increase per daily pushup I would divied the difference in the average pounds (15) by the difference in the number of pushups (40). 15/40 =.375 that would mean a .375 average lifting increase per daily pushup.
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20:27:45 The second group had 15 pounds more lifting strength as a result of doing 40 more daily pushups than the first. The desired rate is therefore 15 pounds / 40 pushups = .375 pounds / pushup. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> ok
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20:41:12 `q009. In another part of the study, participants all did 30 pushups per day, but one group did pushups with a 10-pound weight on their shoulders while the other used a 30-pound weight. At the end of the study, the first group had an average lifting strength of 171 pounds, while the second had an average lifting strength of 188 pounds. At what average rate did lifting strength increase with respect to the added shoulder weight?
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RESPONSE --> now, in this question do we start even again? I am going to assume so. i really feel like I'm missing something here, but this problem looks the same to me except we are increasing the weight instead of reps. so the difference in weight is 20 pounds (30-20) and the difference in the average lifting strenth at the end is 17 pounds (188-171). the strength increased 17 pounds per the extra 20 pounds of added weight on the shoulders. The way this problem is worded I would assume we do not have to find strength per pound added.
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20:42:36 The difference in lifting strength was 17 pounds, as a result of a 20 pound difference in added weight. The average rate at which strength increases with respect added weight would therefore be 17 lifting pounds / (20 added pounds) = .85 lifting pounds / added pound. The strength advantage was .85 lifting pounds per pound of added weight, on the average. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> ok, so I did need to find it pound per pound. I knew that was 17/20 = .85
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20:50:43 `q010. During a race, a runner passes the 100-meter mark 12 seconds after the start and the 200-meter mark 22 seconds after the start. At what average rate was the runner covering distance between those two positions?
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RESPONSE --> to find the avg rate covering the distance between the 100 meter mark and the 200 meter mark, we must determine the amount of distance covered and the amount of time to cover this second distance. first we must determine the distance, so we would subtract the first mark from the second mark. 200-100 = 100 meters covered in the second segment. now, we must determine the amount of time for the second segment. we would subtract the first mark time from the second mark time to get the second segment time. 22-12 = 10 seconds. so during the second segment 100 meters were covered in 10 seconds so that is 10 meters per second (100/10 = 10)
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20:50:50 The runner traveled 100 meters between the two positions, and required 10 seconds to do so. The average rate at which the runner was covering distance was therefore 100 meters / (10 seconds) = 10 meters / second. Again this is an average rate; at different positions in his stride the runner would clearly be traveling at slightly different speeds. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> ok
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20:56:59 `q011. During a race, a runner passes the 100-meter mark moving at 10 meters / second, and the 200-meter mark moving at 9 meters / second. What is your best estimate of how long it takes the runner to cover the 100 meter distance?
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RESPONSE --> to me i would average the meters per second, which would be (10+9)/2 =9.5 then multiply the average by the distance. (9.5 *100 = 950) so my best estimate would be 950 seconds.
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20:59:30 At 10 meters/sec, the runner would require 10 seconds to travel 100 meters. However the runner seems to be slowing, and will therefore require more than 10 seconds to travel the 100 meters. We don't know what the runner's average speed is, we only know that it goes from 10 m/s to 9 m/s. The simplest estimate we could make would be that the average speed is the average of 10 m/s and 9 m/s, or (10 m/s + 9 m/s ) / 2 = 9.5 m/s. Taking this approximation as the average rate, the time required to travel 100 meters will be (100 meters) / (9.5 m/s) = 10.5 sec, approx.. Note that simply averaging the 10 m/s and the 9 m/s might not be the best way to approximate the average rate--for example we if we knew enough about the situation we might expect that this runner would maintain the 10 m/s for most of the remaining 100 meters, and simply tire during the last few seconds. However we were not given this information, and we don't add extraneous assumptions without good cause. So the approximation we used here is pretty close to the best we can do with the given information. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> ok, i missed this one.but i understand it now.
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21:01:53 `q012. We just averaged two quantities, adding them in dividing by 2, to find an average rate. We didn't do that before. Why we do it now?
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RESPONSE --> before we were given the exact amount of time it took to cover the first 100 meters and the total 200 meters. now we were just given speed at the 100 meter mark and the 200 meter mark with no notation of how much time it took to cover the distance.
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21:02:05 In previous examples the quantities weren't rates. We were given the amount of change of some accumulating quantity, and the change in time or in some other quantity on which the first was dependent (e.g., dollars and months, miles and gallons). Here we are given 2 rates, 10 m/s and 9 m/s, in a situation where we need an average rate in order to answer a question. Within this context, averaging the 2 rates was an appropriate tactic. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> ok
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