course MTH 271 Contains q_a_intial_pbs ( Typewriter Notation, Describing Graphs, and Calculus Initial Problems ) q_a_int_pbs_communication and q_a_rates program SEND file.
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00:04:25 `q001. Explain the difference between x - 2 / x + 4 and (x - 2) / (x + 4).
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RESPONSE --> x - 2 / x + 4 states that the variable x is subtracted by two divided by the variable x plus 4 equals some variable.
Whereas, (x -2) / (x +4) states that the product of x minus two is divided by the product of x plus four..................................................
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00:05:15 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.......!!!!!!!!...................................
RESPONSE --> Acknowledged.
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00:08:46 `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.......!!!!!!!!...................................
RESPONSE --> 2 ^ x + 4 is two raised to x power plus four 2 ^ 2 + 4 4 + 4 = 8
2 ^ (x + 4) is two is raised to the product of x plus four 2 ^ ( 2 + 4 ) 2 ^ 6 = 64.................................................
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00:09:12 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.......!!!!!!!!...................................
RESPONSE --> Acknowledged.
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00:29:22 `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 --> Three is the numerator in the fraction x - 3/[ (2x-5)^2 * 3x +1 ] - 2+ 7x
The denominator is (2x-5)^2 * 3x + 1 When x = 2 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 / 7 - 2 + 14 1 4/7 - 2 + 14 -3/7 + 14 Final answer is 13 4/7.................................................
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00:31:06 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.......!!!!!!!!...................................
RESPONSE --> Acknowledged.
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00:42:02 `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 --> First, plug in 4 for x
( 4 - 5) ^ (2 * 4) - 1 + 3/4 - 2 Then, evaluate all items in parentheses (-1)^(8) - 1 + 3/4 - 2 Then evaluate all exponents 1 - 1 + 3/4 - 2 Do multiplication/division from left to right 1 - 1 + .75 - 2 And then addition / subtraction from left to right 0 + . 75 - 2 . 75 - 2 = - 1.25 The final answer is -1.25 or - 1 1/4 or -5/4.................................................
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00:46:45 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.......!!!!!!!!...................................
RESPONSE --> My mistake was with the exponent, my understanding was that 2 was multiplied by x, because it appeared as 2x instead of 2 * x.
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|cſʖWKDۊ Student Name: assignment #002 002. Describing Graphs
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01:45:01 `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.......!!!!!!!!...................................
RESPONSE --> y = 3x - 4
x y -3 -13 -2 -10 -1 -7 0 -4 1 -1 2 2 3 5 x intercept = ( 4/3, 0 ) y intercept = ( 0, -4 ).................................................
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01:45:26 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.......!!!!!!!!...................................
RESPONSE --> Acknowledged.
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01:45:44 `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 --> No, it is linear.
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01:45:51 The graph forms a straight line with no change in steepness.
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RESPONSE --> Acknowledged.
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01:50:14 `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 --> To find the slope we find the change in y over the change in x which can be translated to:
(y2 - y1) / (x1 - x2) Using data from my table, i will use the points (-3, -13) and (-2, -10) Therefore ( -10 - -13) / ( -2 - -3) equals 3 / 1 or 3. The slope of the graph is 3.................................................
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01:50:40 Between any two points of the graph rise / run = 3.
For example, when x = 2 we have y = 3 * 2 - 4 = 2 and when x = 8 we have y = 3 * 8 - 4 = 20. Between these points the rise is 20 - 2 = 18 and the run is 8 - 2 = 6 so the slope is rise / run = 18 / 6 = 3. Note that 3 is the coefficient of x in y = 3x - 4. Note the following for reference in subsequent problems: The graph of this function is a straight line. The graph increases as we move from left to right. We therefore say that the graph is increasing, and that it is increasing at constant rate because the steepness of a straight line doesn't change.......!!!!!!!!...................................
RESPONSE --> Acknowledged.
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01:54:32 `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?......!!!!!!!!...................................
RESPONSE --> y = x^2
x y 0 0 1 1 2 4 3 9 The graph is increasing The steepness changes exponentially. It is increasing at an increasing rate..................................................
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01:55:17 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.......!!!!!!!!...................................
RESPONSE --> Acknowledged.
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01:57:59 `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?......!!!!!!!!...................................
RESPONSE --> y = x^2
x y -3 9 -2 4 -1 1 0 0 The graph is now decreasing from left to right. The steepness is changing exponentially. The graph is decreasing at a decreasing rate..................................................
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01:59:05 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.......!!!!!!!!...................................
RESPONSE --> Acknowledged.
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02:17:08 `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?......!!!!!!!!...................................
RESPONSE --> y = `sqrt(x)
x y 0 0 1 1 2 `sqrt(2) 3 `sqrt(3) Increasing from left to right The graph curves greatly as x reaches 1, then slowly rises, appearing to be more constant. I would really have to say the graph is increasing at an increasing rate for a moment, then it slowly increases after that.................................................
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02:17:29 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.......!!!!!!!!...................................
RESPONSE --> Acknowledged.
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02:33:45 `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?......!!!!!!!!...................................
RESPONSE --> y = 5 * 2^(-x)
x y 0 5 1 2.5 2 1.25 3 0.625 The graph is decreasing from left to right. The steepness increases exponentially. The graph is decreasing at a decreasing rate..................................................
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02:34:04 ** 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. **......!!!!!!!!...................................
RESPONSE --> Acknowledged.
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02:37:43 `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?......!!!!!!!!...................................
RESPONSE --> The graph of y vs t would be increasing.
It would increase at an increasing rate, because the car is moving faster and faster, but time is staying constant..................................................
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02:37:54 ** 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 --> Acknowledged.
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ր~Ǘ Student Name: assignment #005 005. Calculus
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02:55:34 `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'?......!!!!!!!!...................................
RESPONSE --> We use the slope formula, rise over run which is
(y2 - y1)/(x2-x1) and plug in our points (17 - 5) / (7 -3) = 12/4 = 3 The slope of the line passing through 3,5 and 7,17 is 3 (29 - 17) / (10 - 7) = 12/3 = 4 The slope of the line passing through 7,17 and 10, 29 is 4. It is the steepest on average. We say average because it is not showing the overall curvature of the graph, but only between those points..................................................
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02:55:49 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.......!!!!!!!!...................................
RESPONSE --> Acknowledged.
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03:10:50 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?......!!!!!!!!...................................
RESPONSE -->
x answer 2.1 10 2.01 100 2.001 1000 2.0001 10000 1. They grow larger and larger as x gets closer to 2. 2. Yes, yes. 3. Yes. 4. If infinity could be defined as one number, then my answer is yes, and that number is infinity. 5. Before getting to 2, the graph decreases exponentially to negative infinity on the y axis. The graph will never show at x = 2, because division by zero is undefined, continuing right of x = 2, the graph is slowly coming down from infinity on the y axis at an exponential rate..................................................
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03:11:38 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.......!!!!!!!!...................................
RESPONSE --> Acknowledged, I made a mistake in calling infinity a number.
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10:06:55 `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 --> Since we use the x axis as the side of the trapezoid, we have its height, so we can use the geometric formula to find area for each:
A = 0.5 * (b1 + b2) * h Just by adding and subtracting on the points, one can find the lengths of these segments: to get b1, the y value of the coordinate (3,5) is used because the segment stops on the x-axis. to get b2, i used the same idea, since it would simply be that value minus zero. length of b1 and b2 = 5 and 9 respectively. Next, to get the height, I subtracted the x values and took the absolute value of that number. abs( 3 - 7 ) gives us four, so now I have the length of the third segment, which gives me all information needed to compute the area. 0.5 * ( 5 + 9 ) * 4 0.5 * ( 14 ) * 4 7 * 4 = 28 The Area of the first trapezoid is 28. Next, using the same method I calculated b1 and b2 and the height b1 = 2, b2 = 4, and h = 40 Plugging in those values into the equation: 0.5 * (2 + 4) * 40 0.5 * ( 6) * 40 3 * 40 = 120 The area of the second Trapezoid is 120. We can now see that the second trapezoid has greater area, both visually and mathmatically..................................................
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10:07:38 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.......!!!!!!!!...................................
RESPONSE --> Acknowledged.
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10:13:32 `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 --> Using the function y = x^2 and putting in all x values, we get
Segment 1 x y 2 4 5 25 Segment 2 x y -1 1 7 49 The first segment may not appear to be as steep as the second, but it is the steepest of both. It goes higher in a much shorter distance than Segment 2. Segment 2 takes twice as long to reach its peak, and may cause its slope to be slightly smaller than that of segment 1. This can be proved using the slope formula: Segment 1 = ( 25 - 4) / (5 -2 ) = 21/3 = 7 Segment 2 = ( 49 - 1) / (7 - -1) = 48/8 = 6 Segment 1 is steeper than segment 2..................................................
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10:13:41 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.......!!!!!!!!...................................
RESPONSE --> Acknowledged.
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18:15:24 `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.......!!!!!!!!...................................
RESPONSE --> 1. The graph will be constant, it will be a rising straight line, because the amount of gold rises with time in a constant fashion.
2. The graph will rise, and will appear to curve as more and more gold is added, because you are adding the amount of gold buried last week + 1 more gram for this week. 3. The graph will intially rise quickly, and then taper off to more and more of a straight line because you are halving the amount from the week before..................................................
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18:15:46 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.......!!!!!!!!...................................
RESPONSE --> Acknowledged.
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18:20:13 `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.......!!!!!!!!...................................
RESPONSE --> 1. It would be a rising straight line, because you are adding the same amount each week.
2. It would rise at an increasing rate, because you are burying one more gram + whatever you buried the week before. A line which rises faster and faster. 3. It would rise at an increasing rate, that increases more slowly..................................................
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18:21:25 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.......!!!!!!!!...................................
RESPONSE --> I misinterpreted the question, and therefore answered basically the same as before.
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18:39:42 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 --> The depths are:
x = 30 100 - (2 * 30) + 0.1 (30)^2 100 - 60 + 9 = 49 x = 40 100 - (2 * 40) + 0.1(40)^2 100 - 80 + 16 = 36 x = 60 100 - (2 * 60) + 0.1(60)^2 100 - 120 + 36 = 16 x y 30 49 40 36 60 16 On average, the depth is changing more rapidly during the first time interval, this can be determined by ( 36 - 49) / (40 - 30) = -13/10 = a slope of -1.3 whereas the next segment is (16 - 36) / ( 60 - 40) = -20/20 = a slope of -1.................................................
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18:40:00 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.......!!!!!!!!...................................
RESPONSE --> Acknowledged.
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19:12:17 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 --> It is descending at a rate of 9 cm/s at t = 10 and at a rate of 8 cm/s at t = 20.
The average rate is the two rates added together, 9+8 divided by the number of rates added, 2 17/2 = 8.5 cm/s = average rate. To get how much the water would change on average, you simply multiply 8.5 by 10, to get 85 cm total..................................................
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19:14:33 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.......!!!!!!!!...................................
RESPONSE --> Acknowledged, though I do not understand where the 9.5 cm/s came from. Is that a typo perchance, or is it actually calculated from some set of values in the problem?
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ۜlƛk֥ͩڥgw] Student Name: assignment #001
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19:29:27 `q001. It will be very important in this course for your instructor to see and understand the process of visualization and reasoning you use when you solve problems. This exercise is designed to give you a first experience with these ideas, and your instructor a first look at your work.
Answer the following questions and explain in commonsense terms why your answer makes sense.......!!!!!!!!...................................
RESPONSE --> OK.
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19:31:46 For each question draw a picture to make sense out of the situation, and include a description of the picture.
Samples Sample question and response Question: If a bundle of shingles covers 30 square feet, how many bundles are required to cover a 600 square foot roof? Response: We might draw a picture of a rectangle representing the area, dividing the rectangle into a number of smaller rectangles each representing the area covered by a single bundle. This makes it clear that we are dividing the roof area into 1-bundle areas, and makes it clear why we are going to have to divide. Reasoning this problem out in words, we can say that a single bundle would cover 30 square feet. Two bundles would cover 60 square feet. Three bundles would cover 90 square feet. We could continue in this manner until we reach 600 square feet. However, this would be cumbersome. It is more efficient to use the ideas of multiplication and division. We imagine grouping the 600 square feet into 30 square foot patches. There will be 600 / 30 patches and each will require exactly one bundle. We therefore require 600 / 30 bundles = 20 bundles. {}Your responses might not be as clear as the above, though they might be even more clear. I won't be looking for perfection, though I wouldn't object to it, but for a first effort at visualizing a situation and communicating a reasoning process. This is not something you are used to doing and it might take a few attempts before you can achieve good results, but you will get better every time you try. {}You might be unsure of what to do on a specific question. In such a case specific questions and expressions of confusion are also acceptable responses. Such a response must include your attempts to come up with a picture and reason out an explanation. For example your response might be Sample expression of confusion: I've drawn a picture of a pile of bundles and a roof but I'm not sure how to connect the two. I tried multiplying the number of bundles by the square feet of the roof but I got 18,000, and I know it won't take 18,000 bundles to cover the roof. How do you put the area covered by a bundle together with the roof area to get the number of bundles required? A poor response would be something like 'I don't know how to do #17'. This response reveals nothing of your attempt to understand the question and the situation. Nor does it ask a specific question. Incidentally, you might be tempted to quote rules or formulas about rates and velocities in answering these questions. Don't. This exercise isn't about being able to memorize rules and quote them. It is about expanding your ability to visualize, reason and communicate.......!!!!!!!!...................................
RESPONSE --> OK.
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19:32:59 In your own words briefly summarize the instructions and the intent of this exercise.
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RESPONSE --> This exercise is to allow you to become more familiar with conveying visual mathmatical ideas across, and to try to understand these ideas if problems arise, and to be specific about what is not understood about the problem.
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19:33:37 `q001. If you earn 50 dollars in 5 hours, at what average rate are you earning money, in dollars per hour?
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RESPONSE --> If you divide 50 dollars by 5 hours, you are earning on average 10 dollars an hour.
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19:34:12 If you travel 300 miles in 6 hours, at what average rate are you traveling, in miles per hour?
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RESPONSE --> If you divide 300 miles by 6 hours, you get 50 miles per hour average.
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19:36:10 `q002. If a ball rolling down a grooved track travels 40 centimeters in 5 seconds, at what average rate is the ball moving, in centimeters per second?
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RESPONSE --> Divide 40 centimeters by 5 seconds to get the average rate the ball is moving, which is 8 centimeters per second.
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19:37:39 The preceding three questions illustrate the concept of a rate. In each case, to find the rate we divided the change in some quantity (the number of dollars or the distance, in these examples) by the time required for the change (the number of hours or seconds, in these examples). Explain in your own words what is meant by the idea of a rate.
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RESPONSE --> Rate is the value which is multiplied or divided to give a certain change in a number.
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19:38:21 `q003. If you are earning money at the average rate of 15 dollars per hour, how much do you earn in 6 hours?
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RESPONSE --> We take 15 dollars and multiply that by 6, which gives us 90 dollars total in 6 hours.
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19:38:53 If you are traveling at an average rate of 60 miles per hour, how far do you travel in 9 hours?
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RESPONSE --> We multiply 9 hours by 60 miles per hour, which gives us 540 miles.
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19:39:21 `q004. If a ball travels at and average rate of 13 centimeters per second, how far does it travel in 3 seconds?
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RESPONSE --> We multiply 3 seconds by 13 cm/s to get 39 centimeters traveled in 3 seconds.
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19:40:01 In the preceding three exercises you turned the concept of a rate around. You were given the rate and the change in the clock time, and you calculated the change in the quantity. Explain in your own words how this increases your understanding of the concept of a rate.
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RESPONSE --> It shows that the rate is the independent variable in an equation by which the final answer is derived.
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19:40:54 `q005. How long does it take to earn 100 dollars at an average rate of 4 dollars per hour?
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RESPONSE --> We divide 100 by 4, to give us 25 hours, which is the amount of time required to earn 100 dollars.
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19:41:58 How long does it take to travel 500 miles at an average rate of 25 miles per hour?
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RESPONSE --> We divide 500 miles by 25 miles per hour to get 20 hours to travel.
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19:42:51 `q006. How long does it take a rolling ball to travel 80 centimeters at an average rate of 16 centimeters per second?
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RESPONSE --> We divide 80 centimeters by 16 cm/s to get the time required, which is 5 seconds.
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19:44:33 In the preceding three exercises you again expanded your concept of the idea of a rate. Explain how these problems illustrate the concept of a rate.
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RESPONSE --> The rate is shown through the constant linear rate of the previous problems.
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Ɛ}pŘf]z Student Name: assignment #001 001. Rates
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19:52:14 `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'.......!!!!!!!!...................................
RESPONSE --> Do as instructed in the previous programs and provide a self-critique after each problem, and either save your critique as desired or continue on with the program
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19:52:22 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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19:53:02 `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?......!!!!!!!!...................................
RESPONSE --> 50 divided by 5 gives us 10 dollars per hour. The rate is 10.
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19:54:34 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.......!!!!!!!!...................................
RESPONSE --> The statement ""Time rates in general are found by dividing an accumulated quantity by the time required to accumulate it"" is common sense I didn't even realize.
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19:55:15 `q003.If you make $60,000 per year then how much do you make per month?
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RESPONSE --> 60,000 divided by the amount of months in a year, which is 12, gives us $5000.
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19:55:33 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.......!!!!!!!!...................................
RESPONSE --> OK.
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19:56:37 `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 --> It would be more appropriate to say an average of $5000 per month, because sales can differ, pay rates may increase, or the business must pay fines or taxes of some sort that cause a net loss of money.
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19:56:50 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.......!!!!!!!!...................................
RESPONSE --> OK.
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19:58:08 `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 --> We divide 300 miles by 6 hours, to get 50 miles per hour average. It is average because mileage can and will differ along the travel. Speed limits, ignoring the speed limit, and other factors should be considered.
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19:58:27 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.......!!!!!!!!...................................
RESPONSE --> OK.
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19:59:58 `q006. If you use 60 gallons of gasoline on a 1200 mile trip, then at what average rate are you using gasoline, with respect to miles traveled?
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RESPONSE --> 1200 miles divided by 60 gallons of gasoline gives us an average rate of 20 miles per gallon. This can differ between speeds, amount of gas used in the car, maybe a malfunction, tire pressure, and other various factors.
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20:02:28 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.......!!!!!!!!...................................
RESPONSE --> I did the inverse of the problem, which was miles per gallon, as opposed to gallons per mile, which would have been 60 / 1200 = .05 gallons per mile.
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20:05:54 `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 --> Because the numbers we are dividing by are an average of conditions that we do not know about, or are too complex and insignificant that go ignored.
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20:06:50 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.......!!!!!!!!...................................
RESPONSE --> OK.
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20:41:49 `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 --> By graphing this data, and then finding the slope of the line formed, we can get an average rate for the data.
( 162 - 147) / (50 - 10) = 15/40 which can be reduced to 3/8'ths of a pound per push-up on average. The rate is 3/8 th's per daily pushup.................................................
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20:42:39 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.......!!!!!!!!...................................
RESPONSE --> I intially errored in my notes by writing the wrong value down, which was gave me 5/10 which put me on the wrong track until I re-evaluated the problem, which gave me the correct answer.
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20:47:23 `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 --> Again, I approach this problem by graphing the coordinates:
(10, 171) and (30, 188) and find the slope of the line that connects them. ( 188 - 171) / ( 30 - 10) = 17 / 20 = 0.850 Lifting strength increased at an average rate of 0.850 per pound added shoulder weight..................................................
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20:47:34 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.......!!!!!!!!...................................
RESPONSE --> OK.
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20:50:06 `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 --> Same principle:
( 12, 100) (22, 200) ( 200 - 100) / ( 22 - 12) = 10 The runner was covering distance at an average rate of 10 meters per second..................................................
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20:50:17 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.......!!!!!!!!...................................
RESPONSE --> OK.
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20:56:37 `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 --> Adding the 10 m/s and 9 m/s and then dividing them by two, I get 9.5 m/s which is the average between those two values, and divided 9.5 m/s into 100, which then gives me approximately 10.5 seconds to complete the 100 meter distance.
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20:57:24 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.......!!!!!!!!...................................
RESPONSE --> OK.
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20:58:02 `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 --> Because of the limited amount of data we were given, it was the best way to figure out the average.
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20:59:04 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.......!!!!!!!!...................................
RESPONSE --> The quantities were rates in the last problem, and that is why we average the rates together.
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