course Mth173 Please email me to let me know if you are getting these. THANKS! 湠庉k鯥噪篹鈱g治┞禎櫧俛ssignment #005
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16:15:14 `qNote that there are 9 questions in this assignment. `q001. We see that the water depth vs. clock time system likely behaves in a much more predictable detailed manner than the stock market. So we will focus for a while on this system. An accurate graph of the water depth vs. clock time will be a smooth curve. Does this curve suggest a constantly changing rate of depth change or a constant rate of depth change? What is in about the curve at a point that tells you the rate of depth change?
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RESPONSE --> It is constantly changing. The slope confidence assessment: 3
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16:15:21 The steepness of the curve is continually changing. Since it is the slope of the curve then indicates the rate of depth change, the depth vs. clock time curve represents a constantly changing rate of depth change.
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RESPONSE --> ok self critique assessment: 3
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16:18:17 `q002. As you will see, or perhaps have already seen, it is possible to represent the behavior of the system by a quadratic function of the form y = a t^2 + b t + c, where y is used to represent depth and t represents clock time. If we know the precise depths at three different clock times there is a unique quadratic function that fits those three points, in the sense that the graph of this function passes precisely through the three points. Furthermore if the cylinder and the hole in the bottom are both uniform the quadratic model will predict the depth at all in-between clock times with great accuracy. Suppose that another system of the same type has quadratic model y = y(t) = .01 t^2 - 2 t + 90, where y is the depth in cm and t the clock time in seconds. What are the depths for this system at t = 10, t = 40 and t = 90?
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RESPONSE --> y(10)=.01(10^2) +2(10)+90=71 confidence assessment: 2
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16:18:28 At t=10 the depth is y(10) = .01(10^2) + 2(10) + 90 = 1 - 20 + 90 = 71, representing a depth of 71 cm. At t=20 the depth is y(20) = .01(20^2) - 2(20) + 90 = 4 - 40 + 90 = 54, representing a depth of 54 cm. At t=90 the depth is y(90) = .01(90^2) - 2(90) + 90 = 81 - 180 + 90 = -9, representing a depth of -9 cm.
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RESPONSE --> ok self critique assessment: 2
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16:21:39 `q003. For the preceding situation, what are the average rates which the depth changes over each of the two-time intervals?
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RESPONSE --> between 10 and 20 is a difference of -17cm so the avg. change is -1.7cm/s. between 54 and 9 is a difference of -63cm and avg. change of -1.26cm. confidence assessment: 3
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16:21:54 From 71 cm to 54 cm is a change of 54 cm - 71 cm = -17 cm; this change takes place between t = 10 sec and t = 20 sec, so the change in clock time is 20 sec - 10 sec = 10 sec. The average rate of change between these to clock times is therefore ave rate = change in depth / change in clock time = -17 cm / 10 sec = -1.7 cm/s. From 54 cm to -9 cm is a change of -9 cm - 54 cm = -63 cm; this change takes place between t = 40 sec and t = 90 sec, so the change in clock time is a9 0 sec - 40 sec = 50 sec. The average rate of change between these to clock times is therefore ave rate = change in depth / change in clock time = -63 cm / 50 sec = -1.26 cm/s.
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RESPONSE --> ok self critique assessment: 3
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16:25:50 `q004. What is the average rate at which the depth changes between t = 10 and t = 11, and what is the average rate at which the depth changes between t = 10 and t = 10.1?
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RESPONSE --> depth for 10 is 71cm and for 11 is 69.21cm. Avg. rate change from 10-11 is 69.21-71/(11-10)=-1.79cm/s between 10 and10.1 is -1.799 confidence assessment: 3
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16:26:03 At t=10 the depth is y(10) = .01(10^2) - 2(10) + 90 = 1 - 20 + 90 = 71, representing a depth of 71 cm. At t=11 the depth is y(11) = .01(11^2) - 2(11) + 90 = 1.21 - 22 + 90 = 69.21, representing a depth of 69.21 cm. The average rate of depth change between t=10 and t = 11 is therefore change in depth / change in clock time = ( 69.21 - 71) cm / [ (11 - 10) sec ] = -1.79 cm/s. At t=10.1 the depth is y(10.1) = .01(10.1^2) - 2(10.1) + 90 = 1.0201 - 20.2 + 90 = 70.8201, representing a depth of 70.8201 cm. The average rate of depth change between t=10 and t = 10.1 is therefore change in depth / change in clock time = ( 70.8201 - 71) cm / [ (10.1 - 10) sec ] = -1.799 cm/s. We see that for the interval from t = 10 sec to t = 20 sec, then t = 10 s to t = 11 s, then from t = 10 s to t = 10.1 s the progression of average rates is -1.7 cm/s, -1.79 cm/s, -1.799 cm/s. It is important to note that rounding off could have hidden this progression. For example if the 70.8201 cm had been rounded off to 70.82 cm, the last result would have been -1.8 cm and the interpretation of the progression would change. When dealing with small differences it is important not around off too soon.
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RESPONSE --> ok self critique assessment: 2
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16:28:20 `q005. What do you think is the precise rate at which depth is changing at the instant t = 10?
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RESPONSE --> increases from -1.7 to -1.79 to -1.799 which would make it -1.8 at when t=10 confidence assessment: 3
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16:28:28 The progression -1.7 cm/s, -1.79 cm/s, -1.799 cm/s corresponds to time intervals of `dt = 10, 1, and .1 sec, with all intervals starting at the instant t = 10 sec. That is, we have shorter and shorter intervals starting at t = 10 sec. We therefore expect that the progression might well continue with -1.7999 cm/s, -1.79999 cm/s, etc.. We see that these numbers approach more and more closely to -1.8, and that there is no limit to how closely they approach. It therefore makes sense that at the instant t = 10, the rate is exactly -1.8.
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RESPONSE --> ok self critique assessment: 3
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16:31:08 `q006. In symbols, what are the depths at clock time t = t1 and at clock time t = t1 + `dt, where `dt is the time interval between the two clock times?
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RESPONSE --> y=.01 t1^2 - 2 t1+90 and t1+'dt=.01(t1 + 'dt)^2 - 2(t1 + 'dt) confidence assessment: 3
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16:31:14 At clock time t = t1 the depth is y(t1) = .01 t1^2 - 2 t1 + 90 and at clock time t = t1 + `dt the depth is y(t1 + `dt) = .01 (t1 + `dt)^2 - 2 (t1 + `dt) + 90.
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RESPONSE --> ok self critique assessment: 3
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16:32:42 `q007. What is the change in depth between these clock times?
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RESPONSE --> .01 t1^2 + .02 t1 `dt + .01`dt^2 - 2 t1 - 2 `dt + 90 - .01 t1^2 + 2 t1 - 90) therefore = .02 t1 `dt + - 2 `dt + .01 `dt^2. confidence assessment: 3
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16:32:49 The change in depth is .01 (t1 + `dt)^2 - 2 (t1 + `dt) + 90 - (.01 t1^2 - 2 t1 + 90) = .01 (t1^2 + 2 t1 `dt + `dt^2) - 2 t1 - 2 `dt + 90 - (.01 t1^2 - 2 t1 + 90) = .01 t1^2 + .02 t1 `dt + .01`dt^2 - 2 t1 - 2 `dt + 90 - .01 t1^2 + 2 t1 - 90) = .02 t1 `dt + - 2 `dt + .01 `dt^2.
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RESPONSE --> ok self critique assessment: 3
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16:33:31 `q008. What is the average rate at which depth changes between these clock time?
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RESPONSE --> .02 t1 - 2 + .01 `dt. confidence assessment: 2
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16:33:38 The average rate is ave rate = change in depth / change in clock time = ( .02 t1 `dt + - 2 `dt + .01 `dt^2 ) / `dt = .02 t1 - 2 + .01 `dt. Note that as `dt shrinks to 0 this expression approaches .02 t1 - 2.
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RESPONSE --> ok self critique assessment: 2
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16:35:15 `q009. What is the value of .02 t1 - 2 at t1 = 10 and how is this consistent with preceding results?
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RESPONSE --> The value at t1 = 10 is .02 * 10 - 2 = .2 - 2 = -1.8. the rate of 10 confidence assessment: 3
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16:35:27 At t1 = 10 we get .02 * 10 - 2 = .2 - 2 = -1.8. This is the rate we conjectured for t = 10.
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RESPONSE --> ok self critique assessment: 3
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course Mth173 Rυ{~嬨儐椒背涮}姧迮捔\齽F巃ssignment #005
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17:45:55 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 is change change over a period. growth factor is the result of each period.
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17:46:16 ** 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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17:47:43 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 --> rise over run is the depth and time which is also the slope. slope is equal to the avg. rate of change
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17:47:49 ** 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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17:50:22 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 --> the area is integration. Rise is = to depth change and Run is = to time
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17:50:26 **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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17:52:51 ��� #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 --> growth rate is -.11 and factor is = 1 + (-.11) = .89. therefore is = 550mg (.89)^t
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17:52:57 ** 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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17:53:32 How much antibiotic is present at 3:00 p.m.?
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RESPONSE --> = 550mg * .89^5 = 307.123mg
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17:57:18 Describe your graph and explain how it was used to estimate half-life.
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RESPONSE --> Find the starting point on the graph the find where it is approx. half that then move vertically and horizontally to time and amount.
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17:57:25 ** 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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17:59:40 What is the equation to find the half-life?� What is its most simplified form?
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RESPONSE --> 550 mg * .89^dt = .5 * 550 mg.
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17:59:45 ** 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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18:01:11 #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 --> NO IDEA
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18:01:16 ** 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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18:03:44 explain why the negative t axis is a horizontal asymptote for this function.
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RESPONSE --> as t gets bigger 1.1^t does also. so as neg. #'s increase 1.1^t will get closer and closer to 0.
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18:03:55 ** 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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18:06:16 #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 --> not sure about the first question but the secound is = 12 (e^-.5)^x =12 * .61^x
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18:06:38 ** 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 I see now
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18:08:06 what is b for the function y = .007 ( e^(.71 x) )?
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RESPONSE --> 12 (e^.71)^x = 12 * 2.04^x so b=2.041
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18:08:13 ** 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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18:09:28 what is b for the function y = -13 ( e^(3.9 x) )?
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RESPONSE --> = 12 (e^3.9)^x = 12 so b = 49.4
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18:09:37 ** 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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18:11:03 List these functions, each in the form y = A b^x.
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RESPONSE --> y=12(.606^x) , y=.007(2.03^x) , y=13(49.40^x)
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18:11:10 ** 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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18:12:47 ** A proportionality to the cube would be E = k v^3. **
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RESPONSE --> Didnt understand the question
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18:13:56 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 --> vertical is the clock time and horizontal is temperature
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18:14:03 ** 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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18:14:40 what is the meaning of the equation H(30) = 10?
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RESPONSE --> when the time is 30 the temperature is 10
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18:14:46 ** This means that when clock time t is 30, the temperature H is 10. **
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RESPONSE --> got it
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18:15:11 What is the meaning of the vertical intercept?
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RESPONSE --> Temperature
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18:15:33 ** This is the value of H when t = 0--i.e., the temperature at clock time 0. **
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RESPONSE --> didnt understand question
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18:16:47 What is the meaning of the horizontal intercept?
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RESPONSE --> value of time at temperature of 0
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18:16:51 ** 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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18:20:38 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 --> the slope is 1.8 by taking the difference in freezing and boiling in F and same in C then dividing. F = 1.8 * 20 + 32= 68 and C =32 / (-.8) =-40. meaning both F and C are the same at this Temp.
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18:20:49 ** 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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