course Mth 158 I've got a couple of days work in here--I forgot to send some much earlier, but they're all initial problem stuff. ????M??????J????assignment #001001. typewriter notation
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17:02:55 `q001. Explain the difference between x - 2 / x + 4 and (x - 2) / (x + 4).
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RESPONSE --> In x - 2 / x + 4, 2 is divided by x, and then that is subtracted from x and has 4 added to it. In (x - 2) / (x + 4), the groups in the two parenthesis are taken care of first, and then the answer to the top groups is divided by the answer to the second group.
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17:03:33 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 --> Better way to explain it, especially by showing the math behind it.
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17:06:25 `q002. Explain the difference between 2 ^ x + 4 and 2 ^ (x + 4). Then evaluate each expression for x = 2. Note that a ^ b means to raise a to the b power. This process is called exponentiation, and the ^ symbol is used on most calculators, and in most computer algebra systems, to represent exponentiation.
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RESPONSE --> In 2 ^ (x + 4), the 2 is raised by x + 4. In 2 ^ x + 4, the 2 is raised only by x. 2^2 + 4 = 4 + 4 = 8 2^(2 + 4) = 2^(6) = 64
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17:06:33 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 --> Ok.
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17:11:29 `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 = -3 denominator = (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 = 95 / 7
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17:11:56 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 --> Ok, much more detailed than what I wrote.
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17:19:07 `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 --> ((4) - 5) ^ 2(4)-1 + 3 / (4)-2 = ( -1 ) ^ ( 8) -1 + 3 / 4 - 2 = First, take care of what is in the parenthesis. Then handle the multiplication. ( -1) ^ 8 -1 + 3 / 4 -2 = ( -1 ) ^ 7 + 3 / 4 - 2 = Take care of the subtraction in the exponent. The fraction cannot be further simplified. ( -1 ) ^ 7 + 3 / 4 - 2 = -1 + 3 / 4 - 2 = Raise -1 to the seventh power and the next step will be to add and subtract from left to right. -1 + 3 / 4 - 2 = -9 / 4
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17:21:25 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 to multiply the 2 and 4 in the exponent. Also, I made further mistakes because I was assuming there were parenthesis for grouping where there weren't any.
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????w????????y??Student Name: assignment #002 002. Describing Graphs
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17:28:40 `q001. You will frequently need to describe the graphs you have constructed in this course. This exercise is designed to get you used to some of the terminology we use to describe graphs. Please complete this exercise and email your work to the instructor. Problem 1. We make a table for y = 2x + 7 as follows: We construct two columns, and label the first column 'x' and the second 'y'. Put the numbers -3, -2, -1, -, 1, 2, 3 in the 'x' column. We substitute -3 into the expression and get y = 2(-3) + 7 = 1. We substitute -2 and get y = 2(-2) + 7 = 3. Substituting the remaining numbers we get y values 5, 7, 9, 11 and 13. These numbers go into the second column, each next to the x value from which it was obtained. We then graph these points on a set of x-y coordinate axes. Noting that these points lie on a straight line, we then construct the line through the points. Now make a table for and graph the function y = 3x - 4. Identify the intercepts of the graph, i.e., the points where the graph goes through the x and the y axes.
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RESPONSE --> x | y -3 | -13 -2 | -10 -1 | -7 0 | -4 1 | -1 2 | 2 3 | 5 x intercept = (3/4) y intercept = -4
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17:29:23 The graph goes through the x axis when y = 0 and through the y axis when x = 0. The x-intercept is therefore when 0 = 3x - 4, so 4 = 3x and x = 4/3. The y-intercept is when y = 3 * 0 - 4 = -4. Thus the x intercept is at (4/3, 0) and the y intercept is at (0, -4). Your graph should confirm this.
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RESPONSE --> I apparently can't divide. 4 / 3 is NOT 3 / 4.
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17:30:05 `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 --> If by steepness slope is meant, the slope is constant, which is why it is a linear graph.
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17:30:11 The graph forms a straight line with no change in steepness.
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RESPONSE --> ok.
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17:30:27 `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 --> m = 3
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17:30:52 Between any two points of the graph rise / run = 3. For example, when x = 2 we have y = 3 * 2 - 4 = 2 and when x = 8 we have y = 3 * 8 - 4 = 20. Between these points the rise is 20 - 2 = 18 and the run is 8 - 2 = 6 so the slope is rise / run = 18 / 6 = 3. Note that 3 is the coefficient of x in y = 3x - 4. Note the following for reference in subsequent problems: The graph of this function is a straight line. The graph increases as we move from left to right. We therefore say that the graph is increasing, and that it is increasing at constant rate because the steepness of a straight line doesn't change.
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RESPONSE --> Ok.
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17:33:51 `q004. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = 0 and x = 3. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> x | y 0 | 0 1 | 1 2 | 4 3 | 9 the graph is exponential, and decreases from ( - infinity, 0) and increases from (0, infinity)
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17:34:06 Graph points include (0,0), (1,1), (2,4) and (3,9). The y values are 0, 1, 4 and 9, which increase as we move from left to right. The increases between these points are 1, 3 and 5, so the graph not only increases, it increases at an increasing rate.
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RESPONSE --> ok.
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17:36:11 `q005. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = -3 and x = 0. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> Aha, the second part of the question, which I previously answered (for the most part). x | y 0 | 0 -1 | 1 -2 | 4 -3 | 9 the graph is exonential, and is decreasing from (- infinity, 0)
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17:36:18 From left to right the graph is decreasing (points (-3,9), (-2,4), (-1,1), (0,0) show y values 9, 4, 1, 0 as we move from left to right ). The magnitudes of the changes in x from 9 to 4 to 1 to 0 decrease, so the steepness is decreasing. Thus the graph is decreasing, but more and more slowly. We therefore say that the graph is decreasing at a decreasing rate.
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RESPONSE --> Ok.
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17:38:47 `q006. Make a table of y vs. x for y = `sqrt(x). [note: `sqrt(x) means 'the square root of x']. Graph y = `sqrt(x) between x = 0 and x = 3. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> x | y 0 | 0 1 | 1 2 | 1.414214 3 | 1.732051 the graph is increasing at an increasing rate. the slope is not linear.
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17:39:16 If you use x values 0, 1, 2, 3, 4 you will obtain graph points (0,0), (1,1), (2,1.414), (3. 1.732), (4,2). The y value changes by less and less for every succeeding x value. Thus the steepness of the graph is decreasing. The graph would be increasing at a decreasing rate.{}{} If the graph respresents the profile of a hill, the hill starts out very steep but gets easier and easier to climb. You are still climbing but you go up by less with each step, so the rate of increase is decreasing. {}{}If your graph doesn't look like this then you probably are not using a consistent scale for at least one of the axes. If your graph isn't as desribed take another look at your plot and make a note in your response indicating any difficulties.
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RESPONSE --> ah, so it is the opposite of a quadratic graph. Good to know.
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17:42:36 `q007. Make a table of y vs. x for y = 5 * 2^(-x). Graph y = 5 * 2^(-x) between x = 0 and x = 3. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> x | y 0 | 5 1 | 2.5 2 | 1.25 3 | .625 decreasing at a decreasing rate. Steepness changes in a negative manner, and I would say that it is exponential, but I'm not positive.
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17:43:17 ** From basic algebra recall that a^(-b) = 1 / (a^b). So, for example: 2^-2 = 1 / (2^2) = 1/4, so 5 * 2^-2 = 5 * 1/4 = 5/4. 5* 2^-3 = 5 * (1 / 2^3) = 5 * 1/8 = 5/8. Etc. The decimal equivalents of the values for x = 0 to x = 3 will be 5, 2.5, 1.25, .625. These values decrease, but by less and less each time. The graph is therefore decreasing at a decreasing rate. **
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RESPONSE --> But how does the steepness change? Or I'm I just missing the explanation.
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17:44:58 `q008. Suppose you stand still in front of a driveway. A car starts out next to you and moves away from you, traveling faster and faster. If y represents the distance from you to the car and t represents the time in seconds since the car started out, would a graph of y vs. t be increasing or decreasing? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> The graph is increasing at an increasing rate, assuming a constant acceleration, because as the car moves faster, more ground is covered in the same amount of time. I would say that the table would show an increasing trend.
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17:45:10 ** 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 --> Ok.
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????????y?????assignment #001
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???????W????Student Name: assignment #003 003. PC1 questions
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22:57:52 `q001 A straight line connects the points (3, 5) and (7, 17), while another straight line continues on from (7, 17) to the point (10, 29). Which line is steeper and on what basis to you claim your result?
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RESPONSE --> We have two lines, and he have three coordinates. We can use the slope of a line equation to determine which line is the steeper. m = (ysub2 - ysub1)/(xsub2 - xsub1). the coordinate set: m = (17 - 5) / (7 - 3), m = 12 / 4, m = 3. the second coordinate set: m = (29 - 17) / (10 - 7), m = 12 / 3, m = 4. The second line is the steeper, because it's slope is 4, which means that for every 4 you count up the graph, you go right 1. The slope of the first line goes up 3, to the right 1.
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22:58:20 The point (3,5) has x coordinate 3 and y coordinate 5. The point (7, 17) has x coordinate 7 and y coordinate 17. To move from (3,5) to (7, 17) we must therefore move 4 units in the x direction and 12 units in the y direction. Thus between (3,5) and (7,17) the rise is 12 and the run is 4, so the rise/run ratio is 12/4 = 3. Between (7,10) and (10,29) the rise is also 12 but the run is only 3--same rise for less run, therefore more slope. The rise/run ratio here is 12/3 = 4.
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RESPONSE --> ok.
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23:03:14 `q002. The expression (x-2) * (2x+5) is zero when x = 2 and when x = -2.5. Without using a calculator verify this, and explain why these two values of x, and only these two values of x, can make the expression zero.
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RESPONSE --> With this type of problem, we look at each grouping. Whatever makes one group = 0, will make the whole equation = 0, because anything multiplied by a zero = 0. In order to find what value of x makes the equation = 0, we take each grouping and set these ""sub equations"" = 0. And then we solve for x. So: (x - 2) becomes x - 2 = 0. x = 2. (2x + 5) becomes 2x + 5 = 0, 2x = -5 x = -5/2 These two values for x will make the whole equation equal to zero, and only these two will do this because of what happens when you mulitply by a zero--you get zero for your answer.
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23:04:22 If x = 2 then x-2 = 2 - 2 = 0, which makes the product (x -2) * (2x + 5) zero. If x = -2.5 then 2x + 5 = 2 (-2.5) + 5 = -5 + 5 = 0.which makes the product (x -2) * (2x + 5) zero. The only way to product (x-2)(2x+5) can be zero is if either (x -2) or (2x + 5) is zero. Note that (x-2)(2x+5) can be expanded using the Distributive Law to get x(2x+5) - 2(2x+5). Then again using the distributive law we get 2x^2 + 5x - 4x - 10 which simplifies to 2x^2 + x - 10. However this doesn't help us find the x values which make the expression zero. We are better off to look at the factored form.
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RESPONSE --> I'm not sure why the non-factored equation was displayed, unless I missed that I was supposed to find it.
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23:06:45 `q003. For what x values will the expression (3x - 6) * (x + 4) * (x^2 - 4) be zero?
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RESPONSE --> Using the same logic from the previous question: 3x - 6 = 0 3x = 6 x = 2. x + 4 = 0 x = -4 x^2 - 4 = 0 x^2 = 4 (sqrt. of x^2) = (sqrt. of 4) x = 2. the x values that make the equation = 0, x = 2, -4
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23:07:41 In order for the expression to be zero we must have 3x-6 = 0 or x+4=0 or x^2-4=0. 3x-6 = 0 is rearranged to 3x = 6 then to x = 6 / 3 = 2. So when x=2, 3x-6 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero. x+4 = 0 gives us x = -4. So when x=-4, x+4 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero. x^2-4 = 0 is rearranged to x^2 = 4 which has solutions x = + - `sqrt(4) or + - 2. So when x=2 or when x = -2, x^2 - 4 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero. We therefore see that (3x - 6) * (x + 4) * (x^2 - 4) = 0 when x = 2, or -4, or -2. These are the only values of x which can yield zero.**
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RESPONSE --> stupid square roots. I must remember to include the fact that the square root of something can be positive OR negative. if only they were cube roots.
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23:26:01 `q004. One straight line segment connects the points (3,5) and (7,9) while another connects the points (10,2) and (50,4). From each of the four points a line segment is drawn directly down to the x axis, forming two trapezoids. Which trapezoid has the greater area? Try to justify your answer with something more precise than, for example, 'from a sketch I can see that this one is much bigger so it must have the greater area'.
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RESPONSE --> I'm not sure how exactly to solve this problem, but I'll start by finding the slope of the two line segments. For the first one, m = 1. The second one is m = (1 / 20). So far, this doens't have too much relevance to the problem. After drawing out the two trapezoids, I have figures for the 3 sides for each one. trap. X: side 1 = 5, side 2 = 4, side 3 = 9, side 4 = 4. trap. Y: side 1 = 2, side 2 = 40, side 3 = 4, side 4 = 40. the side numbers start on the left, and go counter-clockwise. Side 4 is the same as side 2, due to the reasoning below as to finding the area. trap. X: The area of the rectangle = A = lw, A = 5 * 4 A = 20 The area of triangle = A = (1 / 2)bh, A = (1 / 2 ) * 4 * h. Unfortunately, I don't know how to get the height of a right triangle, so I'll be making some guesses as to the final answer. trap. Y: rectangle = A = 40 * 2 A = 80. triangle = as I explained above, I don't know how to find height. Answer = The second trapezoid most likely has the greater volume, as the area of it's rectangle is 4 times that of the first trapezoid's rectangle, and I don't see how the area of the triangle will greatly affect it. A = The next step will be to break the trapezoids into two shapes--a rectangle and a triangle. I can find the two areas, add them to get the area for the trapezoid.
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23:27:48 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 --> that's a cool way to find the area of a trapezoid A = (sum of altitudes / number of altitudes) * (width of trapezoid). Would have saved me a lot of typing.
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23:39:33 * `q005. Sketch graphs of y = x^2, y = 1/x and y = `sqrt(x) [note: `sqrt(x) means 'the square root of x'] for x > 0. We say that a graph increases if it gets higher as we move toward the right, and if a graph is increasing it has a positive slope. Explain which of the following descriptions is correct for each graph: As we move from left to right the graph increases as its slope increases. As we move from left to right the graph decreases as its slope increases. As we move from left to right the graph increases as its slope decreases. As we move from left to right the graph decreases as its slope decreases.
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RESPONSE --> Joy, tables. Okay, here we go. y = x^2: x | y -3 | 9 -2 | 4 -1 | 1 0 | 0 1 | 1 2 | 4 y = 1/x x | y -3 | (-1 / 3) -1 | -1 0 | undefined 2 | ( 1 / 2) 3 | ( 1 / 3) y = sqrt(x) x | y -3 | -1 | 0 | 0 2 | 1.414 3 | 1.732 I don't think I should put the imaginary x-values in, so I haven't.
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23:42:53 For x = 1, 2, 3, 4: The function y = x^2 takes values 1, 4, 9 and 16, increasing more and more for each unit increase in x. This graph therefore increases, as you say, but at an increasing rate. The function y = 1/x takes values 1, 1/2, 1/3 and 1/4, with decimal equivalents 1, .5, .33..., and .25. These values are decreasing, but less and less each time. The decreasing values ensure that the slopes are negative. However, the more gradual the decrease the closer the slope is to zero. The slopes are therefore negative numbers which approach zero. Negative numbers which approach zero are increasing. So the slopes are increasing, and we say that the graph decreases as the slope increases. We could also say that the graph decreases but by less and less each time. So the graph is decreasing at a decreasing rate. For y = `sqrt(x) we get approximate values 1, 1.414, 1.732 and 2. This graph increases but at a decreasing rate.
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RESPONSE --> I forgot to say which was increasing and whatnot. y = x^2--increasing at increasing rate y = 1 / x--this one's wierd. It decreases at what I would say is an increasing rate, but never approaches x = 0 (the vertical aysmptote). It then decreases at a decreasing rate after the vertical asymptote. y = sqrt(x)--increases at a decreasing rate.
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23:51:42 `q006. If the population of the frogs in your frog pond increased by 10% each month, starting with an initial population of 20 frogs, then how many frogs would you have at the end of each of the first three months (you can count fractional frogs, even if it doesn't appear to you to make sense)? Can you think of a strategy that would allow you to calculate the number of frogs after 300 months (according to this model, which probably wouldn't be valid for that long) without having to do at least 300 calculations?
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RESPONSE --> it's sort of like an interest formula. It's just going to be a bit wierder, because I'm changing the variable. P = increase in population. r = rate, which is 10% t = time, which is in months I = initial population. So it looks like this: P = r * t * I However, this will only give you the first month's increase in population. I know there's a formula to do it for an indefinite period of time, because we figured it out for interest in my pre-calc class. unfortunately, I forgot how we did it. I remember the name for it--the compound interest formula. Now if only I could remember how we got it. However, at the end of three months, you will have: 22 at one month, 24.2 at month two, 26.62 at month three.
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23:52:51 At the end of the first month, the number of frogs in the pond would be (20 * .1) + 20 = 22 frogs. At the end of the second month there would be (22 * .1) + 22 = 24.2 frogs while at the end of the third month there would be (24.2 * .1) + 24.2 = 26.62 frogs. The key to extending the strategy is to notice that multiplying a number by .1 and adding it to the number is really the same as simply multiplying the number by 1.1. 10 * 1.1 = 22; 22 * 1.1 = 24.2; etc.. So after 300 months you will have multiplied by 1.1 a total of 300 times. This would give you 20 * 1.1^300, whatever that equals (a calculator will easily do the arithmetic). A common error is to say that 300 months at 10% per month gives 3,000 percent, so there would be 30 * 20 = 600 frogs after 30 months. That doesn't work because the 10% increase is applied to a greater number of frogs each time. 3000% would just be applied to the initial number, so it doesn't give a big enough answer.
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RESPONSE --> right, okay. I'll have to memorize that--it's just getting too late in the night.
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23:59:20 `q007. Calculate 1/x for x = 1, .1, .01 and .001. Describe the pattern you obtain. Why do we say that the values of x are approaching zero? What numbers might we use for x to continue approaching zero? What happens to the values of 1/x as we continue to approach zero? What do you think the graph of y = 1/x vs. x looks for x values between 0 and 1?
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RESPONSE --> 1/x: 1 / 1 = 1, 1 / .1 = 10, 1 / .01 = 100, 1 / .001 = 1000. As to why x is approaching zero, and not crossing it, is because the horizontal asymptote is y = 0, and the graph cannot cross this line. We know this because the equation falls under the first case for horizontal asymptotes, which says that if the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is y = 0. Not sure if this actually explains what you're after, though. We can continue using smaller and smaller decimals for x, but never 0. Also, depending on what the sign chart says, we can use negative numbers, but I don't think we can use those in the first and fourth quadrants of the graph in this case. The graph of y = x is a straight line with a slope of 1. the graph of y = 1/x is a curved line, decreasing at an increasing rate, in contrast to the constant rise of the first graph.
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00:00:47 If x = .1, for example, 1 / x = 1 / .1 = 10 (note that .1 goes into 1 ten times, since we can count to 1 by .1, getting.1, .2, .3, .4, ... .9, 10. This makes it clear that it takes ten .1's to make 1. So if x = .01, 1/x = 100 Ithink again of counting to 1, this time by .01). If x = .001 then 1/x = 1000, etc.. Note also that we cannot find a number which is equal to 1 / 0. Deceive why this is true, try counting to 1 by 0's. You can count as long as you want and you'll ever get anywhere. The values of 1/x don't just increase, they increase without bound. If we think of x approaching 0 through the values .1, .01, .001, .0001, ..., there is no limit to how big the reciprocals 10, 100, 1000, 10000 etc. can become. The graph becomes steeper and steeper as it approaches the y axis, continuing to do so without bound but never touching the y axis. This is what it means to say that the y axis is a vertical asymptote for the graph .
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RESPONSE --> ok. and I think you mean ""To see"", instead of ""decieve"", but it's actually kind of funny.
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00:03:02 * `q008. At clock time t the velocity of a certain automobile is v = 3 t + 9. At velocity v its energy of motion is E = 800 v^2. What is the energy of the automobile at clock time t = 5?
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RESPONSE --> First substitute 5 for t in the velocity equation, and solve for v. v = 3 (5) + 9, v = 24. Then substitute 24 for v in the second equation, to find energy. E = 800 (24)^2 E = 460,800
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00:03:17 ?
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RESPONSE --> yay, I'm finished.
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00:03:44 For t=5, v = 3 t + 9 = (3*5) + 9 = 24. Therefore E = 800 * 24^2 = 460800.
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RESPONSE --> ah, there popped up the answer. thought that was a little wierd. maybe I'm not finished.
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00:03:57 ?
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RESPONSE --> now it looks as though I'm finished.
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00:05:50 * `q009. Continuing the preceding problem, can you give an expression for E in terms of t?
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RESPONSE --> okay, I think the program is slowing down. to the problem. It should look like this: E = 800 (3t + 9)^2. It's basically one of those f(g(x)) problems.
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00:06:37 Since v = 3 t + 9 the expression would be E = 800 v^2 = 800 ( 3t + 9) ^2. This is the only answer really required here. For further reference, though, note that this expression could also be expanded by applying the Distributive Law:. Since (3t + 9 ) ^ 2 = (3 t + 9 ) * ( 3 t + 9 ) = 3t ( 3t + 9 ) + 9 * (3 t + 9) = 9 t^2 + 27 t + 27 t + 81 = 9 t^2 + 54 t + 81, we get E = 800 ( 9 t^2 + 54 t + 81) = 7200 t^2 + 43320 t + 64800 (check my multiplication because I did that in my head, which isn't always reliable).
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RESPONSE --> ok.
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