course Mth 173 ??j????????????assignment #004
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08:48:20 query modeling project #2 #5. $200 init investment at 10%. What are the growth rate and growth factor for this function? How long does it take the principle to double? At what time does the principle first reach $300?
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RESPONSE --> the growth rate is .1 which is 10 % and the growth factor is 1.1. to the principles growth rate i used t- year principle which is P(t) = P0 (1+r) ^ t. so at P(t) = $200 (1+.1)^7.28 which is $400 so in 7.28 years the money doubles. the principle reaches $300 in 4.28 years
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08:49:25 ** The growth rate is .10 and the growth factor is 1.10 so the amount function is $200 * 1.10^t. This graph passes through the y axis at (0, 200), increases at an increasing rate and is asymptotic to the negative x axis. For t=0, 1, 2 you get p(t) values $200, $220 and $242--you can already see that the rate is increasing since the increases are $20 and $22. Note that you should know from precalculus the characteristics of the graphs of exponential, polynomial an power functions (give that a quick review if you don't--it will definitely pay off in this course). $400 is double the initial $200. We need to find how long it takes to achieve this. Using trial and error we find that $200 * 1.10^tDoub = $400 if tDoub is a little more than 7. So doubling takes a little more than 7 years. The actual time, accurate to 5 significant figures, is 7.2725 years. If you are using trial and error it will take several tries to attain this accuracy; 7.3 years is a reasonable approximation for trail and error. To reach $300 from the original $200, using amount function $200 * 1.10^t, takes a little over 4.25 years (4.2541 years to 5 significant figures); again this can be found by trial and error. The amount function reaches $600 in a little over 11.5 years (11.527 years), so to get from $300 to $600 takes about 11.5 years - 4.25 years = 7.25 years (actually 7.2725 years to 5 significant figures). **
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RESPONSE --> i understand
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{???x?????????N? assignment #004 ????q???????? Calculus I 07-10-2007
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08:50:36 At what time t is the principle equal to half its t = 20 value? What doubling time is associated with this result?
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RESPONSE --> d
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08:52:29 query modeling project #2 #5. $200 init investment at 10%. What are the growth rate and growth factor for this function? How long does it take the principle to double? At what time does the principle first reach $300?
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RESPONSE --> the growth rate is .1 and the growth factor is 1.1. it takes a little over 7 years. and it reaches 300$ in a little over 4 years.
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08:52:44 ** The growth rate is .10 and the growth factor is 1.10 so the amount function is $200 * 1.10^t. This graph passes through the y axis at (0, 200), increases at an increasing rate and is asymptotic to the negative x axis. For t=0, 1, 2 you get p(t) values $200, $220 and $242--you can already see that the rate is increasing since the increases are $20 and $22. Note that you should know from precalculus the characteristics of the graphs of exponential, polynomial an power functions (give that a quick review if you don't--it will definitely pay off in this course). $400 is double the initial $200. We need to find how long it takes to achieve this. Using trial and error we find that $200 * 1.10^tDoub = $400 if tDoub is a little more than 7. So doubling takes a little more than 7 years. The actual time, accurate to 5 significant figures, is 7.2725 years. If you are using trial and error it will take several tries to attain this accuracy; 7.3 years is a reasonable approximation for trail and error. To reach $300 from the original $200, using amount function $200 * 1.10^t, takes a little over 4.25 years (4.2541 years to 5 significant figures); again this can be found by trial and error. The amount function reaches $600 in a little over 11.5 years (11.527 years), so to get from $300 to $600 takes about 11.5 years - 4.25 years = 7.25 years (actually 7.2725 years to 5 significant figures). **
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RESPONSE --> ok it is fully understood
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09:03:06 At what time t is the principle equal to half its t = 20 value? What doubling time is associated with this result?
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RESPONSE --> t is 20 and when plugged in into the equation the answer is 1345. and it doubles at in 12.7 yrs.
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09:08:16 ** The t = 20 value is $200 * 1.1^20 = $1340, approx. Half the t = 20 value is therefore $1340/2 = $670 approx.. By trial and error or, if you know them, other means we find that the value $670 is reached at t = 12.7, approx.. For example one student found that half of the 20 value is 1345.5/2=672.75, which occurs. between t = 12 and t = 13 (at t = 12 you get $627.69 and at t = 13 you get 690.45).? At 12.75=674.20 so it would probably be about12.72.? This implies that the principle would double in the interval between t = 12.7 yr and t = 20 yr, which is a period of 20 yr - 12.7 yr = 7.3 yr. This is consistent with the doubling time we originally obtained, and reinforces the idea that for an exponential function doubling time is the same no matter when we start or end. **
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RESPONSE --> when i found the value of $200 at t=20 i got $1345
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09:08:31 ** The t = 20 value is $200 * 1.1^20 = $1340, approx. Half the t = 20 value is therefore $1340/2 = $670 approx.. By trial and error or, if you know them, other means we find that the value $670 is reached at t = 12.7, approx.. For example one student found that half of the 20 value is 1345.5/2=672.75, which occurs. between t = 12 and t = 13 (at t = 12 you get $627.69 and at t = 13 you get 690.45).? At 12.75=674.20 so it would probably be about12.72.? This implies that the principle would double in the interval between t = 12.7 yr and t = 20 yr, which is a period of 20 yr - 12.7 yr = 7.3 yr. This is consistent with the doubling time we originally obtained, and reinforces the idea that for an exponential function doubling time is the same no matter when we start or end. **
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RESPONSE --> ok
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09:14:29 query #8. Sketch principle vs. time for the first four years with rates 10%, 20%, 30%, 40%
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RESPONSE --> for the first rate at for years the valuese are 1.1 1.21 1.33 and 1.46 and the same equation is applied to the oter rates. so the doubling time decreases, but it become less every time there is a 10 % increase
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09:14:35 ** We find that for the first interest rate, 10%, we have amount = 1(1.10)^t. For the first 4 years the values at t=1,2,3,4 years are 1.10, 1.21, 1.33 and 1.46. By trial and error we find that it will take 7.27 years for the amount to double. for the second interest rate, 20%, we have amount = 1(1.20)^t. For the first 4 years the values at t=1,2,3,4 years are 1.20, 1.44, 1.73 and 2.07. By trial and error we find that it will take 3.80 years for the amount to double. Similar calculations tell us that for interest rate 30% we have $286 after 4 years and require 2.64 years to double, and for interest rate 40% we have $384 after 4 years and require 2.06 years to double. The final 4-year amount increases by more and more with each 10% increase in interest rate. The doubling time decreases, but by less and less with each 10% increase in interest rate. **
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RESPONSE --> ok
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09:14:41 ** We find that for the first interest rate, 10%, we have amount = 1(1.10)^t. For the first 4 years the values at t=1,2,3,4 years are 1.10, 1.21, 1.33 and 1.46. By trial and error we find that it will take 7.27 years for the amount to double. for the second interest rate, 20%, we have amount = 1(1.20)^t. For the first 4 years the values at t=1,2,3,4 years are 1.20, 1.44, 1.73 and 2.07. By trial and error we find that it will take 3.80 years for the amount to double. Similar calculations tell us that for interest rate 30% we have $286 after 4 years and require 2.64 years to double, and for interest rate 40% we have $384 after 4 years and require 2.06 years to double. The final 4-year amount increases by more and more with each 10% increase in interest rate. The doubling time decreases, but by less and less with each 10% increase in interest rate. **
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RESPONSE --> ok
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09:14:45 query #11. equation for doubling time
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RESPONSE --> dafds
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09:16:49 query #11. equation for doubling time
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RESPONSE --> the equation for doubling time is P0 * (1+r)^t= 2 P0 becaues the orginial eqution is P0 * (1+r)^t but i am trying to find double the time.
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09:16:59 ** the basic equation says that the amount at clock time t, which is P0 * (1+r)^t, is double the original amount P0. The resulting equation is therefore P0 * (1+r)^t = 2 P0. Note that this simplifies to (1 + r)^ t = 2, and that this result depends only on the interest rate, not on the initial amount P0. **
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RESPONSE --> ok
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09:17:12 Write the equation you would solve to determine the doubling time 'doublingTime, starting at t = 2, for a $5000 investment at 8%.
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RESPONSE --> ok
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09:18:00 ** the basic equation says that the amount at clock time t, which is P0 * (1+r)^t, is double the original amount P0. The resulting equation is therefore P0 * (1+r)^t = 2 P0. Note that this simplifies to (1 + r)^ t = 2, and that this result depends only on the interest rate, not on the initial amount P0. **
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RESPONSE --> i do understand the equation can simplify to be (1+ r)^t = 2
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09:37:32 Write the equation you would solve to determine the doubling time 'doublingTime, starting at t = 2, for a $5000 investment at 8%.
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RESPONSE --> well the equation i used was P(2 + 'doublingTime) = 2 P(1) so when the valuse are plugged in the equation looks like 5000*1.08^(2+doubling time) = 2*(5000 * 1.08 ^2) after applying the prop. the value came out to be 1.08^2* ^dt = 2*1.08^2 then both sides are divided by 1.08 and the doubling time is 2
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09:38:00 **dividing the equation $5000 * 1.08 ^ ( 2 + doubling time) = 2 * [$5000 * 1.08 ^2] by $5000 we get 1.08 ^ ( 2 + doubling time) = 2 * 1.08 ^2]. This can be written as 1.08^2 * 1.08^doublingtime = 2 * 1.08^2. Dividing both sides by 1.08^2 we obtain 1.08^doublingtime = 2. We can then use trial and error to find the doubling time that works. We get something like 9 years. **
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RESPONSE --> ok
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09:38:07 Desribe how on your graph how you obtained an estimate of the doubling time.
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RESPONSE --> ok
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09:38:26 **dividing the equation $5000 * 1.08 ^ ( 2 + doubling time) = 2 * [$5000 * 1.08 ^2] by $5000 we get 1.08 ^ ( 2 + doubling time) = 2 * 1.08 ^2]. This can be written as 1.08^2 * 1.08^doublingtime = 2 * 1.08^2. Dividing both sides by 1.08^2 we obtain 1.08^doublingtime = 2. We can then use trial and error to find the doubling time that works. We get something like 9 years. **
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RESPONSE --> i see that the doubling time takes somewhere around 9 years
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09:40:03 Desribe how on your graph how you obtained an estimate of the doubling time.
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RESPONSE --> first i put the double amount and put it on the vert. axis then i moved to the horiz. and clock time was put there.
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09:40:10 **In this case you would find the double of the initial amount, $10000, on the vertical axis, then move straight over to the graph, then straight down to the horizontal axis. The interval from t = 0 to the clock time found on the horizontal axis is the doubling time. **
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RESPONSE --> ok
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09:40:37 **In this case you would find the double of the initial amount, $10000, on the vertical axis, then move straight over to the graph, then straight down to the horizontal axis. The interval from t = 0 to the clock time found on the horizontal axis is the doubling time. **
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RESPONSE --> ok
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12:49:46 Growth rate and growth factor: Describe the difference between growth rate and growth factor and give a short example of how each might be used
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RESPONSE --> fdadf
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n??????????? assignment #005 ????q???????? Calculus I 07-10-2007
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12:51:57 ** Specific statements: When multiplied by a quantity the growth rate tells us how much the quantity will change over a single period. When multiplied by the quantity the growth factor gives us the new quantity at the end of the next period. **
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RESPONSE --> ok
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12:52:08 Growth rate and growth factor: Describe the difference between growth rate and growth factor and give a short example of how each might be used
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RESPONSE --> growth rate tells how much something will change over a period of time. and growth factor predicts how much changed in the future of a quantity
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12:52:15 Class notes #05 trapezoidal representation. Explain why the slope of a depth vs. time trapezoid represents the average rate of change of the depth with respect to the time during the time interval represented
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RESPONSE --> ok
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12:52:34 ** Specific statements: When multiplied by a quantity the growth rate tells us how much the quantity will change over a single period. When multiplied by the quantity the growth factor gives us the new quantity at the end of the next period. **
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RESPONSE --> ok
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12:54:04 Class notes #05 trapezoidal representation. Explain why the slope of a depth vs. time trapezoid represents the average rate of change of the depth with respect to the time during the time interval represented
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RESPONSE --> when figuring rise over run it also translates as change in depth/ change in time
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12:54:16 ** GOOD ANSWER BY STUDENT WITH INSTRUCTOR COMMENTS: The slope of the trapezoids will indicate rise over run or the slope will represent a change in depth / time interval thus an average rate of change of depth with respect to time INSTRUCTOR COMMENTS: More detail follows: ** To explain the meaning of the slope you have to reason the question out in terms of rise and run and slope. For this example rise represents change in depth and run represent change in clock time; rise / run therefore represents change in depth divided by change in clock time, which is the average rate of change. **
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RESPONSE --> ok
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12:54:25 Explain why the area of a rate vs. time trapezoid for a given time interval represents the change in the quantity corresponding to that time interval.
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RESPONSE --> asdf
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12:54:42 ** GOOD ANSWER BY STUDENT WITH INSTRUCTOR COMMENTS: The slope of the trapezoids will indicate rise over run or the slope will represent a change in depth / time interval thus an average rate of change of depth with respect to time INSTRUCTOR COMMENTS: More detail follows: ** To explain the meaning of the slope you have to reason the question out in terms of rise and run and slope. For this example rise represents change in depth and run represent change in clock time; rise / run therefore represents change in depth divided by change in clock time, which is the average rate of change. **
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RESPONSE --> ok
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12:57:07 **STUDENT RESPONSE WITH INSTRUCTOR COMMENTS: The area of a rate vs. time graph rep. the change in quantity. Calculating the area under the graph is basically integration The accumulated area of all the trapezoids for a range will give us thetotal change in quantity. The more trapezoids used the more accurate the approx. INSTRUCTOR COMMENTS: All very good but the other key point is that the average altitude represents the average rate, which when multiplied by the width which represents time interval gives the change in quantity You have to reason this out in terms of altitudes, widths and areas. For the rate of depth change example altitude represents rate of depth change so average altitude represents average rate of depth change, and width represents change in clock time. average altitude * width therefore represents ave rate of depth change * duration of time interval = change in depth. For the rate of change of a quantity other than depth, the reasoning is identical except you'll be talking about something besides depth. **
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RESPONSE --> asdf
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12:57:17 Explain why the area of a rate vs. time trapezoid for a given time interval represents the change in the quantity corresponding to that time interval.
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RESPONSE --> because there is a change of a quantity found so therefore average can also be foound
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12:57:25 **STUDENT RESPONSE WITH INSTRUCTOR COMMENTS: The area of a rate vs. time graph rep. the change in quantity. Calculating the area under the graph is basically integration The accumulated area of all the trapezoids for a range will give us thetotal change in quantity. The more trapezoids used the more accurate the approx. INSTRUCTOR COMMENTS: All very good but the other key point is that the average altitude represents the average rate, which when multiplied by the width which represents time interval gives the change in quantity You have to reason this out in terms of altitudes, widths and areas. For the rate of depth change example altitude represents rate of depth change so average altitude represents average rate of depth change, and width represents change in clock time. average altitude * width therefore represents ave rate of depth change * duration of time interval = change in depth. For the rate of change of a quantity other than depth, the reasoning is identical except you'll be talking about something besides depth. **
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RESPONSE --> ok
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12:57:29 ??? #17. At 10:00 a.m. a certain individual has 550 mg of penicillin in her bloodstream. Every hour, 11% of the penicillin present at the beginning of the hour is removed by the end of the hour. What is the function Q(t)?
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RESPONSE --> dasf
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12:59:47 ** Every hour 11% or .11 of the total is lost so the growth rate is -.11 and the growth factor is 1 + r = 1 + (-.11) = .89 and we have Q(t) = Q0 * (1 + r)^t = 550 mg (.89)^t or Q(t)=550(.89)^t? **
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RESPONSE --> the growth rate shows is -.11 but i do not understand where .89 is coming from
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12:59:52 ** Every hour 11% or .11 of the total is lost so the growth rate is -.11 and the growth factor is 1 + r = 1 + (-.11) = .89 and we have Q(t) = Q0 * (1 + r)^t = 550 mg (.89)^t or Q(t)=550(.89)^t? **
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RESPONSE --> ok
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13:00:01 How much antibiotic is present at 3:00 p.m.?
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RESPONSE --> ok
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13:02:21 How much antibiotic is present at 3:00 p.m.?
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RESPONSE --> using the equation Q(t) = 550 mg*.89^t when 5 is plugged in because it is 5hrs after the intial time Q(5) = 550 * .89^5=307.123
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13:02:24 ** 3:00 p.m. is 5 hours after the initial time so at that time there will be Q(5) = 550 mg * .89^5 = 307.123mg in the blood **
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RESPONSE --> ok
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13:02:29 Describe your graph and explain how it was used to estimate half-life.
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RESPONSE --> adsf
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13:02:39 ** 3:00 p.m. is 5 hours after the initial time so at that time there will be Q(5) = 550 mg * .89^5 = 307.123mg in the blood **
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RESPONSE --> ok
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13:04:56 Describe your graph and explain how it was used to estimate half-life.
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RESPONSE --> starting anywhere on the graph i went out over the graph horiz and until i hit the line and when i go down to the ver. line and the dis. from the first to the sec point is the half life.
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13:04:58 ** Starting from any point on the graph we first project horizontally and vertically to the coordinate axes to obtain the coordinates of that point. The vertical coordinate represents the quantity Q(t), so we find the point on the vertical axis which is half as high as the vertical coordinate of our starting point. We then draw a horizontal line directly over to the graph, and project this point down. The horizontal distance from the first point to the second will be the distance on the t axis between the two vertical projection lines. **
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RESPONSE --> ok
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13:05:04 What is the equation to find the half-life?? What is its most simplified form?
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RESPONSE --> asdf
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13:05:17 ** Starting from any point on the graph we first project horizontally and vertically to the coordinate axes to obtain the coordinates of that point. The vertical coordinate represents the quantity Q(t), so we find the point on the vertical axis which is half as high as the vertical coordinate of our starting point. We then draw a horizontal line directly over to the graph, and project this point down. The horizontal distance from the first point to the second will be the distance on the t axis between the two vertical projection lines. **
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RESPONSE --> ok
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13:06:57 What is the equation to find the half-life?? What is its most simplified form?
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RESPONSE --> the equation is Q(doubling time) = 1/2 Q the simpler equation is .89^dt = 5
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13:07:00 ** Q(doublingTime) = 1/2 Q(0)or 550 mg * .89^doublingTIme = .5 * 550 mg. Dividing thru by the 550 mg we have .89^doublingTime = .5. We can use trial and error to find an approximate value for doublingTIme (later we use logarithms to get the accurate solution). **
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RESPONSE --> ok
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13:07:08 #19. For the function Q(t) = Q0 (1.1^ t), a value of t such that Q(t) lies between .05 Q0 and .1 Q0. For what values of t did Q(t) lie between .005 Q0 and .01 Q0?
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RESPONSE --> hgv
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13:07:19 ** Q(doublingTime) = 1/2 Q(0)or 550 mg * .89^doublingTIme = .5 * 550 mg. Dividing thru by the 550 mg we have .89^doublingTime = .5. We can use trial and error to find an approximate value for doublingTIme (later we use logarithms to get the accurate solution). **
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RESPONSE --> ok
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13:13:26 ** Any value between about t = -24.2 and t = -31.4 will result in Q(t) between .05 Q0 and .1 Q0. Note that these values must be negative, since positive powers of 1.1 are all greater than 1, resulting in values of Q which are greater than Q0. Solving Q(t) = .05 Q0 we rewrite this as Q0 * 1.1^t = .05 Q0. Dividing both sides by Q0 we get 1.1^t = .05. We can use trial and error (or if you know how to use them logarithms) to approximate the solution. We get t = -31.4 approx. Solving Q(t) = .1 Q0 we rewrite this as Q0 * 1.1^t = .1 Q0. Dividing both sides by Q0 we get 1.1^t = .1. We can use trial and error (or if you know how to use them logarithms) to approximate the solution. We get t = -24.2 approx. (The solution for .005 Q0 is about -55.6, for .01 is about -48.3 For this solution any value between about t = -48.3 and t = -55.6 will work). **
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RESPONSE --> jkhb
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13:13:31 explain why the negative t axis is a horizontal asymptote for this function.
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RESPONSE --> sdaf
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13:13:40 #19. For the function Q(t) = Q0 (1.1^ t), a value of t such that Q(t) lies between .05 Q0 and .1 Q0. For what values of t did Q(t) lie between .005 Q0 and .01 Q0?
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RESPONSE --> the equation is wrote out as Q0 * 1.1^t = .05 Q0 and then both sides are divided by Q0 and tht leaves me with 1.1^t= .05 and when solved i get -31.4 and the same applies for Q0 = .1 and the asnwer is -24.2
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13:13:46 ** Any value between about t = -24.2 and t = -31.4 will result in Q(t) between .05 Q0 and .1 Q0. Note that these values must be negative, since positive powers of 1.1 are all greater than 1, resulting in values of Q which are greater than Q0. Solving Q(t) = .05 Q0 we rewrite this as Q0 * 1.1^t = .05 Q0. Dividing both sides by Q0 we get 1.1^t = .05. We can use trial and error (or if you know how to use them logarithms) to approximate the solution. We get t = -31.4 approx. Solving Q(t) = .1 Q0 we rewrite this as Q0 * 1.1^t = .1 Q0. Dividing both sides by Q0 we get 1.1^t = .1. We can use trial and error (or if you know how to use them logarithms) to approximate the solution. We get t = -24.2 approx. (The solution for .005 Q0 is about -55.6, for .01 is about -48.3 For this solution any value between about t = -48.3 and t = -55.6 will work). **
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RESPONSE --> ok
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13:14:18 explain why the negative t axis is a horizontal asymptote for this function.
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RESPONSE --> i have no idea
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13:15:09 ** The value of 1.1^t increases for increasing t; as t approaches infinity 1.1^t also approaches infinity. Since 1.1^-t = 1 / 1.1^t, we see that for increasingly large negative values of t the value of 1.1^t will be smaller and smaller, and would in fact approach zero. **
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RESPONSE --> ok
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13:15:12 #22. What value of b would we use to express various functions in the form y = A b^x? What is b for the function y = 12 ( e^(-.5x) )?
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RESPONSE --> sdaf
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13:15:19 ** The value of 1.1^t increases for increasing t; as t approaches infinity 1.1^t also approaches infinity. Since 1.1^-t = 1 / 1.1^t, we see that for increasingly large negative values of t the value of 1.1^t will be smaller and smaller, and would in fact approach zero. **
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RESPONSE --> now i understand because when approaches infinity it approaches 0 but never to be 0
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13:17:39 #22. What value of b would we use to express various functions in the form y = A b^x? What is b for the function y = 12 ( e^(-.5x) )?
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RESPONSE --> b for the function is set up as y = 12 ( e^(-.5x) ) so b is 61
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13:17:42 ** 12 e^(-.5 x) = 12 (e^-.5)^x = 12 * .61^x, approx. So this function is of the form y = A b^x for b = .61 approx.. **
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RESPONSE --> ok
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13:17:47 what is b for the function y = .007 ( e^(.71 x) )?
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RESPONSE --> dafs
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13:18:00 ** 12 e^(-.5 x) = 12 (e^-.5)^x = 12 * .61^x, approx. So this function is of the form y = A b^x for b = .61 approx.. **
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RESPONSE --> ok
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13:19:41 what is b for the function y = .007 ( e^(.71 x) )?
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RESPONSE --> i have no i dea
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13:19:45 ** 12 e^(.71 x) = 12 (e^.71)^x = 12 * 2.04^x, approx. So this function is of the form y = A b^x for b = 2.041 approx.. **
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RESPONSE --> ok
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13:19:51 ** 12 e^(.71 x) = 12 (e^.71)^x = 12 * 2.04^x, approx. So this function is of the form y = A b^x for b = 2.041 approx.. **
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RESPONSE --> ok
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13:19:55 what is b for the function y = -13 ( e^(3.9 x) )?
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RESPONSE --> dfa
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13:20:26 what is b for the function y = -13 ( e^(3.9 x) )?
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RESPONSE --> the function is 49 when the equation is set up as 12 e^(3.9 x) = 12 (e^3.9)^x = 12 * 49.4^x
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13:20:31 ** 12 e^(3.9 x) = 12 (e^3.9)^x = 12 * 49.4^x, approx. So this function is of the form y = A b^x for b = 49.4 approx.. **
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RESPONSE --> ok
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13:20:35 List these functions, each in the form y = A b^x.
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RESPONSE --> dsaf
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13:20:39 ** 12 e^(3.9 x) = 12 (e^3.9)^x = 12 * 49.4^x, approx. So this function is of the form y = A b^x for b = 49.4 approx.. **
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RESPONSE --> ok
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13:21:20 ** The functions are y=12(.6065^x) y=.007(2.03399^x) and y=-13(49.40244^x) **
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RESPONSE --> ok
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13:21:31 List these functions, each in the form y = A b^x.
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RESPONSE --> y=12(.6065^x) y=.007(2.03399^x) y=-13(49.40244^x
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13:21:44 ** The functions are y=12(.6065^x) y=.007(2.03399^x) and y=-13(49.40244^x) **
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RESPONSE --> ok
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13:21:49 query text problem 1.1 #24 dolphin energy prop cube of vel
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RESPONSE --> dsaf
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13:22:42 query text problem 1.1 #24 dolphin energy prop cube of vel
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RESPONSE --> dont kno where the problem is coming from
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13:22:47 ** A proportionality to the cube would be E = k v^3. **
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RESPONSE --> ok
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13:22:52 query text problem 1.1 #27 temperature function H = f(t), meaning of H(30)=10, interpret vertical and horizontal intercepts
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RESPONSE --> fd
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13:23:00 ** A proportionality to the cube would be E = k v^3. **
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RESPONSE --> ok
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13:24:36 ** The interpretation would be that the vertical intercept represents the temperature at clock time t = 0, while the horizontal intercept represents the clock time at which the temperature reaches zero. **
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RESPONSE --> ok
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13:24:41 what is the meaning of the equation H(30) = 10?
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RESPONSE --> gfs
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13:24:48 query text problem 1.1 #27 temperature function H = f(t), meaning of H(30)=10, interpret vertical and horizontal intercepts
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RESPONSE --> t = 0 temp and h is time when temp is 0
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13:24:52 ** The interpretation would be that the vertical intercept represents the temperature at clock time t = 0, while the horizontal intercept represents the clock time at which the temperature reaches zero. **
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RESPONSE --> ok
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13:25:18 ** This means that when clock time t is 30, the temperature H is 10. **
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RESPONSE --> ok
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13:25:22 What is the meaning of the vertical intercept?
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RESPONSE --> dfa
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13:25:29 what is the meaning of the equation H(30) = 10?
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RESPONSE --> when clock time is 30 h equals 10
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13:25:35 ** This means that when clock time t is 30, the temperature H is 10. **
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RESPONSE --> ok
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13:26:07 What is the meaning of the vertical intercept?
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RESPONSE --> the temp at t = 0
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13:26:16 ** This is the value of H when t = 0--i.e., the temperature at clock time 0. **
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RESPONSE --> ok
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13:26:23 What is the meaning of the horizontal intercept?
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RESPONSE --> dfa
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13:26:28 ** This is the value of H when t = 0--i.e., the temperature at clock time 0. **
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RESPONSE --> ok
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13:26:53 What is the meaning of the horizontal intercept?
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RESPONSE --> clock time when temp is 0
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13:26:57 ** This is the t value when H = 0--the clock time when temperature reaches 0 **
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RESPONSE --> ok
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13:27:06 query text problem 1.1.32. Water freezes 0 C, 32 F; boils 100 C, 212 F. Give your solution to problem 1.1.32.
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RESPONSE --> fdsa
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13:27:18 ** This is the t value when H = 0--the clock time when temperature reaches 0 **
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RESPONSE --> ok
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13:29:00 ** The graph contains points (0, 32) and (100, 212). The slope is therefore (212-32) / (100-0) = 1.8. The y-intercept is 32 so the equation of the line is y = 1.8 x + 32, or using F and C F = 1.8 C + 32. To find the Fahrenheit temp corresponding to 20 C we substitute C = 20 into F = 1.8 C + 32 to get F = 1.8 * 20 + 32 = 36 + 32 = 68 The two temperatures will be equal when F = C. Substituting C for F in F = 1.8 C + 32 we get C = 1.8 C + 32. Subtracting 1.8 C from both sides we have -.8 C = 32 or C = 32 / (-.8) = -40. The scales read the same at -40 degrees. **
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RESPONSE --> ok
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13:29:22 query text problem 1.1.32. Water freezes 0 C, 32 F; boils 100 C, 212 F. Give your solution to problem 1.1.32.
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RESPONSE --> when the slope is found by finding (212-32) / (100-0) = 1.8. it can be plugged into slope int. form and the equ. is 1.8 C + 32
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13:29:25 ** The graph contains points (0, 32) and (100, 212). The slope is therefore (212-32) / (100-0) = 1.8. The y-intercept is 32 so the equation of the line is y = 1.8 x + 32, or using F and C F = 1.8 C + 32. To find the Fahrenheit temp corresponding to 20 C we substitute C = 20 into F = 1.8 C + 32 to get F = 1.8 * 20 + 32 = 36 + 32 = 68 The two temperatures will be equal when F = C. Substituting C for F in F = 1.8 C + 32 we get C = 1.8 C + 32. Subtracting 1.8 C from both sides we have -.8 C = 32 or C = 32 / (-.8) = -40. The scales read the same at -40 degrees. **
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RESPONSE --> ok
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