course Mth 271 dl|ܡ~Zassignment #001
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21:13:32 `qNote that there are four questions in this assignment. `q001. If your stocks are worth $5000 in mid-March, $5300 in mid-July and $5500 in mid-December, during which period, March-July or July-December was your money growing faster?
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RESPONSE --> July - December is when the money is growing the fastest. 5500 - 5300/12-7 200/5 40(July - December) 5300-5000/7-3 300/4 85(March - July) The graph is steeper between July and December confidence assessment: 2
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21:14:17 The first period (4 months) was shorter than the second (5 months), and the value changed by more during the shorter first period (increase of $300) than during the longer second perios (increase of $200). A greater increase in a shorter period implies a greater rate of change. So the rate was greater during the first period.
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RESPONSE --> Ok, I mixed up my math self critique assessment: 2
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21:18:37 `q002. What were the precise average rates of change during these two periods?
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RESPONSE --> (March - July) 5300-5000/7-3 300/4 85 (July - December) 5500 - 5300/12-7 200/5 40 confidence assessment: 2
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21:19:36 From mid-March thru mid-July is 4 months. A change of $300 in four months gives an average rate of change of $300 / (4 months) = $75/month. From mid-July through mid-December is 5 months, during which the value changes by $200, giving an average rate of $200 / (5 months) = $40 / month. Thus the rate was greater during the first period than during the second.
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RESPONSE --> I am sorry I am doing my multiplication wrong. I dont know what I am thinking self critique assessment: 2
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21:34:34 `q003. If the water in a uniform cylinder is leaking from a hole in the bottom, and if the water depths at clock times t = 10, 40 and 90 seconds are respectively 80 cm, 40 cm and 20 cm, is the depth of the water changing more quickly or less quickly, on the average, between t = 10 sec and t = 40 sec, or between t = 40 sec and t = 90 sec?
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RESPONSE --> t=10 and t=40 40 - 80 / 40 - 10 -40 / 30 -4/3 t=40 and t=90 20 - 40 / 90 - 50 -20 / 40 -2/4 -1/2 The t=10 and 1=40 is changing more quickly than t=40 and t=90 confidence assessment: 2
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21:35:13 Between clock times t = 10 sec and t = 40 sec the depth changes by -40 cm, from 80 cm to 40 cm, in 30 seconds. The average rate is therefore -40 cm / (30 sec) = -1.33 cm/s. Between t = 40 sec and t = 90 sec the change is -20 cm and the time interval is 50 sec so the average rate of change is -20 cm/ 50 sec = -.4 cm/s, approx. The depth is changing more quickly during the first time interval.
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RESPONSE --> I understand self critique assessment: 2
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21:39:04 `q004. How are the two preceding questions actually different versions of the same question? How it is the mathematical reasoning the same in both cases?
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RESPONSE --> They are asking what has the greatest change in each situation. The mathematical reasoning is the same because they both deal with rates of change. confidence assessment: 2
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21:39:38 In each case we were given a quantity that changed with time. We were given the quantities at three different clock times and we were asked to compare the rates of change over the corresponding intervals. We did this by determining the changes in the quantities and the changes in the clock times, and for each interval dividing change in quantity by change in clock time. We could symbolize this process by representing the change in clock time by `dt and the change in the quantity by `dQ. The rate is thus rate = `dQ /`dt.
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RESPONSE --> ok....I see what you mean by mathematical reasoning self critique assessment: 2
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course Mth 271 eۚDO啶ؽassignment #001
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22:27:57 Section 0.1.26 solve x/2-x/3>5
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RESPONSE --> x/2 - x/3 > 5 x/-1>5 x/-1 - 1/1 > 5 - 1 x/1 > 4 x>4 confidence assessment: 3
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22:28:35 It's easiest to avoid denominators where possible. So the preferred first step is to multiply both sides of the original equation by the common denominator 6: 6(x/2) - 6(x/3) = 6 * 5, which gives you3x - 2x = 6 * 5 which gives you x > 6 * 5 which simplifies to x > 30. The interval associated with this solution is 30 < x < infinity, or (30, infinity). To graph you would make an arrow starting at x = 30 and pointing to the right, indicating by an open circle that x = 30 isn't included.**
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RESPONSE --> I see what you are saying self critique assessment: 2
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22:30:43 Section 0.1.28 solve 2x^2+1<9x-3
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RESPONSE --> 2x^2 + 1 < 9x-3 2x^2 -9x + 4 < 0 (2x - 1)(x - 4) < 0 2x-1 < 0 x < 1/2 x-4 <0 x<4 (1/2 < x < 4) confidence assessment: 3
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22:31:09 The given inequality rearranges to give the quadratic 2x^2 - 9 x + 4 < 0. The left-hand side has zeros at x = .5 and x = 4, as we see by factoring [ we get (2x-1)(x-4) = 0 which is true if 2x-1 = 0 or x - 4 = 0, i.e., x = 1/2 or x = 4. ] The left-hand side is a continuous function of x (in fact a quadratic function with a parabola for a graph), and can change sign only by passing thru 0. So on each interval x < 1/2, 1/2 < x < 4, 4 < x the function must have the same sign. Testing an arbitrary point in each interval tells us that only on the middle interval is the function negative, so only on this interval is the inequality true. Note that we can also reason this out from the fact that large negative or positive x the left-hand side is greater than the right because of the higher power. Both intervals contain large positive and large negative x, so the inequality isn't true on either of these intervals. In any case the correct interval is 1/2 < x < 4. ALTERNATE BUT EQUIVALENT EXPLANATION: The way to solve this is to rearrange the equation to get 2 x^2 - 9 x + 4< 0. The expression 2 x^2 - 9 x + 4 is equal to 0 when x = 1/2 or x = 4. These zeros can be found either by factoring the expression to get ( 2x - 1) ( x - 4), which is zero when x = 1/2 or 4, or by substituting into the quadratic formula. You should be able to factor basic quadratics or use the quadratic formula when factoring fails. The function can only be zero at x = 1/2 or x = 4, so the function can only change from positive to negative or negative to positive at these x values. This fact partitions the x axis into the intervals (-infinity, 1/2), (1/2, 4) and (4, infinity). Over each of these intervals the quadratic expression can't change its sign. If x = 0 the quadratic expression 2 x^2 - 9 x + 4 is equal to 4. Therefore the expression is positive on the interval (-infinity, 1/2). The expression changes sign at x = 1/2 and is therefore negative on the interval (1/2, 4). It changes sign again at 4 so is positive on the interval (4, infinity). The solution to the equation is therefore the interval (1/2, 4), or in inequality form 1/2 < x < 4. **
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RESPONSE --> ok self critique assessment: 2
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