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12:59:45 `q001. Explain the difference between x - 2 / x + 4 and (x - 2) / (x + 4).
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RESPONSE --> The parenthesis indicate that the operation within them must be performed first.
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13:05:32 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 --> ok
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13:10:21 `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 --> Again, you must do the operation in paren first.
2^(2+4) = 2^6 = 64 2^2+4= 4+4=8.................................................
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13:11:08 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 --> right
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13:18:03 `q003. What is the numerator of the fraction in the expression x - 3 / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x? What is the denominator? What do you get when you evaluate the expression for x = 2?
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RESPONSE --> The num is x-3. The denom. is [ (2x-5)^2 * 3x+1] - 2+7x.
2-3/ [(2*2-5)^2*3(2)+1] -2+7(2) -1/ [(-1)^2 * 7] - 2+14 -1/-7-12 = 1/19.................................................
ZU֟̎eN㊥V{ Student Name: assignment #001 001. typewriter notation
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20:49:47 `q001. Explain the difference between x - 2 / x + 4 and (x - 2) / (x + 4).
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK.
must do the operation within paren first, then divide..................................................
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20:50:23 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 --> ok
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20:52:48 `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 --> Again, you must follow order of operations and complete the paren expression first.
2^2 +4 = 4+4 = 8 2^ (2+4) = 2^6 = 64.................................................
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20:53:19 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 --> right
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21:03:17 `q003. What is the numerator of the fraction in the expression x - 3 / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x? What is the denominator? What do you get when you evaluate the expression for x = 2?
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RESPONSE --> The numerator is x-3. The denom is [(2x-5)^2 * 3x+1]
2-3/[ (2*2-5)^2 -3(2) +1 ] -2+7(2) -1/[1-5] + 12 -1/8.................................................
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21:11:05 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 --> I don't see why the dvision changed to a fraction for only the 3, as in 3-7??
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21:18:30 `q004. Explain, step by step, how you evaluate the expression (x - 5) ^ 2x-1 + 3 / x-2 for x = 4.
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RESPONSE --> Substituting 4 for x gives:
(4-5) ^2 (4)-1+3 / 4-2 Following order of operations leads me to (4-5) =-1 exponents follow -1^2 = 1 1+4-1+3 / 2 = 7/2.................................................
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21:25:15 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 --> I am confused. I thought this was division but you are solving by changing the denom to a fraction with one of the numerators. I suppose this goes back to the previous problem when you said x isn't part of the denom. Can you clear this up for me?
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F佝\ Student Name: assignment #002 002. Describing Graphs
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21:35:54 `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 --> (0,-4) (3/2,0)
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21:38:33 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 --> ok
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21:42:39 `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. the line has a set slope.
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21:42:50 The graph forms a straight line with no change in steepness.
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RESPONSE --> ok
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21:48:20 `q003. What is the slope of the graph of the preceding two exercises (the function ia y = 3x - 4;slope is rise / run between two points of the graph)?
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RESPONSE --> the slope is 3/1
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21:50:34 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 --> Finally, something I remember how to do!
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21:56:54 `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 --> The graph is increasing at an increasing rate. I don't know about the steepness.
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21:57:38 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 --> ok
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21:59:34 `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 --> The points are (-3,18) (-2,8) (-1,2) and (0.0). The graph is decreasing at a decreasing rate.
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22:00:50 From left to right the graph is decreasing (points (-3,9), (-2,4), (-1,1), (0,0) show y values 9, 4, 1, 0 as we move from left to right ). The magnitudes of the changes in x from 9 to 4 to 1 to 0 decrease, so the steepness is decreasing.
Thus the graph is decreasing, but more and more slowly. We therefore say that the graph is decreasing at a decreasing rate.......!!!!!!!!...................................
RESPONSE --> My points are different than yours.
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22:03:33 `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 --> Increasing at an increasing rate and the steepness does change by increasing.
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22:07:48 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 --> I have the same values as you for my table but my graph doesn't look like what you describe. It actually looks the opposite of your "hill". The hill starts out fairly level and begins to increase as you go. Forgive my ignorance, but I am unsure what plot(s) do.
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22:09:38 `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 --> I am not sure how to read this graph.
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22:12:09 ** 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 --> Thanks for the reminder. My values match yours but I was looking exclusively at the graph for my response.
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22:15:18 `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 --> I think the graph would be increasing as the distance between myself and car would contiue to grow. Increasing at a constant rate.
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22:17:15 ** 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 --> makes sense.
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Student Name: assignment #003 003. PC1 questions
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19:46:57 `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 --> The line from (7,17) to (10, 29) is steeper because the numbers increase. As the number increases the line is longer and steeper.
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19:50:23 The point (3,5) has x coordinate 3 and y coordinate 5. The point (7, 17) has x coordinate 7 and y coordinate 17. To move from (3,5) to (7, 17) we must therefore move 4 units in the x direction and 12 units in the y direction.
Thus between (3,5) and (7,17) the rise is 12 and the run is 4, so the rise/run ratio is 12/4 = 3. Between (7,10) and (10,29) the rise is also 12 but the run is only 3--same rise for less run, therefore more slope. The rise/run ratio here is 12/3 = 4.......!!!!!!!!...................................
RESPONSE --> That makes sense. I am glad you explained it that way. I didn't think to use the simple rise/run formula.
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19:55:23 `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 --> You would set each factor to zero and solve.
x-2=0 add 2 to each side x=2 2x+5=0 subtract 5 and divide by 2 x= -5/2 = -2.5.................................................
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19:56:29 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.......!!!!!!!!...................................
RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK.
ok.................................................
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20:02:05 `q003. For what x values will the expression (3x - 6) * (x + 4) * (x^2 - 4) be zero?
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RESPONSE --> Set each expression to zero and solve:
3x-6 = 0 3x=6 x=2 * x+4 = 0 x=-4 * x^2 -4 = 0 x^2=4 x=2 The solution set is {2,-4,2}. When you check them they all equal zero. Of course, 0 * 0 =0.................................................
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20:02:42 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.**......!!!!!!!!...................................
RESPONSE -->
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20:02:48 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.**......!!!!!!!!...................................
RESPONSE --> right
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20:08:37 `q004. One straight line segment connects the points (3,5) and (7,9) while another connects the points (10,2) and (50,4). From each of the four points a line segment is drawn directly down to the x axis, forming two trapezoids. Which trapezoid has the greater area? Try to justify your answer with something more precise than, for example, 'from a sketch I can see that this one is much bigger so it must have the greater area'.
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RESPONSE --> The trapezoid formed by the points (10,2) and (50,4) seems to have the greater area. The distance between the 2 x coordinates is 40, therefore it would seem that this one should have the greater area.
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20:11:28 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 --> I don't remember the altitude you are referring to from my earlier work but it makes sense. The answer I gave was like yours except for the altitude.
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20:18:22 * `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.......!!!!!!!!...................................
RESPONSE --> I am unsure.
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20:22:07 For x = 1, 2, 3, 4:
The function y = x^2 takes values 1, 4, 9 and 16, increasing more and more for each unit increase in x. This graph therefore increases, as you say, but at an increasing rate. The function y = 1/x takes values 1, 1/2, 1/3 and 1/4, with decimal equivalents 1, .5, .33..., and .25. These values are decreasing, but less and less each time. The decreasing values ensure that the slopes are negative. However, the more gradual the decrease the closer the slope is to zero. The slopes are therefore negative numbers which approach zero. Negative numbers which approach zero are increasing. So the slopes are increasing, and we say that the graph decreases as the slope increases. We could also say that the graph decreases but by less and less each time. So the graph is decreasing at a decreasing rate. For y = `sqrt(x) we get approximate values 1, 1.414, 1.732 and 2. This graph increases but at a decreasing rate.......!!!!!!!!...................................
RESPONSE --> I thought of this but it seemed too simple. Perhaps I am trying to make this harder than it needs to be.
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20:31:13 `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 --> 10%=.10 3 months * .10 =.30
20 * .30 = 6 At the end of three months you would have 30% more than the original twenty, which would equal 26 frogs. 10% * 300 = 30 After 300 months you would have 50 frogs..................................................
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20:37:09 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.......!!!!!!!!...................................
RESPONSE --> I see my error. I simply multiplied 10% by 3 months yielding 30% rather than counting in the increase for each individual month.
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20:47:15 `q007. Calculate 1/x for x = 1, .1, .01 and .001. Describe the pattern you obtain. Why do we say that the values of x are approaching zero? What numbers might we use for x to continue approaching zero? What happens to the values of 1/x as we continue to approach zero? What do you think the graph of y = 1/x vs. x looks for x values between 0 and 1?
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RESPONSE --> The answers are multiples of 10. 1, 10, 100, 1000.
The values of x are becoming smaller and smaller without yet reaching zero. You could next use .0001 and continuing adding a decimal place. The graph forms a straight line..................................................
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20:49:32 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 .......!!!!!!!!...................................
RESPONSE --> I haven't heard the term vertical asymptote before.
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20:51:27 * `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 --> I don't know but I will write down your answer and study it.
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20:51:47
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK.
I don't know but I will write down your answer and study it..................................................
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20:53:08 For t=5, v = 3 t + 9 = (3*5) + 9 = 24. Therefore E = 800 * 24^2 = 460800.
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RESPONSE --> Again, much simpler than I was trying to make it out to be.
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20:53:18
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. ok
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20:55:37 * `q009. Continuing the preceding problem, can you give an expression for E in terms of t?
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RESPONSE --> E = 24/5T
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20:58:32 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).......!!!!!!!!...................................
RESPONSE --> Yes, caluclations done in my head aren't always reliable either, though I am sure more often than those in yours.
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