course Mth173 I am a police officer with the Town of Abingdon and had to work 70 hours over the last week, that is the reason I am getting this to you a little late. Im not asking for an extension of time for due dates I merely do not want you to think I am not taking the course serious. ??????I???????x?Student Name: assignment #001
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13:34:39 Note 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 --> In order to determine which period my stocks are growing more during I must determine the change between the two periods from March to July the value of my stocks increase $300 dollars in value from July-December my stocks increas $200 dollars in value therefore in 4 months March-July = $300 and in 5 months July-December = $200 therefore my money was growing faster between March and July
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13:34:56 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 --> I understand this.
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13:40:40 `q002. What were the precise average rates of change during these two periods?
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RESPONSE --> March-July $300/4 = avg rate of $75/month July-December $200/5 = avg rate of $40.00/month The reason this only gives us an average rate is due to the fact the amount gained might have fluctuated from month to month during these periods.
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13:41:08 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 understand this.
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13:47:13 `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 --> From the time of 10 seconds to 40 seconds 30 seconds passes and the amount of water lost is 40cm. therefore 40cm/40sec = 1 cm/sec 50second passes during the second interval water changes 20cm during second interval therefore 20cm/50sec =2/5 cm/sec therfore On the average the depth of the water is changing more quickly between t=10sec and t=40sec
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13:48:11 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 made a critical error that might affect me in the future. I failed to note the change in centimeters as a negative value.
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13:49:32 `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 --> Both of the questions are asking me to find two different rates and compare them. For the problems the way to find rate is always the same.
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13:51:09 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 --> I will note down the formula rate = dQ/dt. I have the feeling it will come in handy in the future.
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