course Mth271 \E횰fբ`assignment #005
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11:07:57 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 --> Slope represents rise/run, which in this trapezoid, represents the velocity change/change in time. Because here we have the change in velocity and time, we can accurately say that the slope represents average rate of change. confidence assessment: 2
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11:08:08 the specific idea is that ave rate of depth change is [change in depth / change in time] ; rise represents change in depth and run represents change in time so slope = rise/run represents ave rate of depth change. **
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RESPONSE --> self critique assessment: 3
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11:09:19 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 of a trapezoid is the average height times the width. In this case area represents the average altitude and multiplied by the width in seconds. confidence assessment: 2
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11:10:00 The average altitude represents the avg. velocity. The area of a trapezoid involves the altitude, which represents the avg. velocity, and the width, which represents the change in clock time. When you multiply ave altitude by width you are representing ave vel * change in clock time, which gives change in position. This reasoning isn't confined to velocities. For any rate vs. clock time graph, average altitude represents approximate average rate, which multiplied by the change in time (not by the time itself) gives you the change in quantity **
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RESPONSE --> I forgot to mention the end result of area is the average velocity. self critique assessment: 2
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11:11:37 text problem 0.5 #8 add x/(2-x) + 2/(x-2)
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RESPONSE --> x/(2-x) + 2/(x-2) = -x/(x-2) + 2/ (x-2) = 2-x/(x-2) confidence assessment: 2
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11:12:51 common denominator could be [ (2-x)(x-2) ]. In this case we have x / (2-x) + 2 / (x-2) = [ (x-2) / (x-2) ] * [ x / (2-x) ] + [ (2-x) / (2-x) ] * [ 2 / (x-2) ] = x(x-2) / [ (2-x)(x-2) ] + 2 (2-x) / [ (2-x)(x-2) ] = [x(x-2) + 2(2-x) ] / [ (2-x)(x-2) ] = [ x^2 - 2x + 4 - 2x ] / [ (2-x)(x-2) ] = (x^2-4x+4) / [ -x^2+4x-4 ] = (x-2)^2 / [-(x-2)^2] = -1. NOTE however that there is a SIMPLER SOLUTION: We can note that x-2 = -(2-x) so that the original problem is -x/(x-2) + 2 /(x-2) = (-x + 2) / (x-2) = -(x-2)/(x-2) = -1. **
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RESPONSE --> I solved according to the simpler solution, but failed to simplify further to get -1. I just missed the step but understand it now. self critique assessment: 2
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11:18:00 text problem 0.5 #48 cost = 6 x + 900,000 / x, write as single fraction and determine cost to store 240 units
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RESPONSE --> Cost= ((6x^2) + 900,000) / x Cost = ((6(240^2) + 900,000) /240 Cost= ((6 * 57600) + 900,000)/240 Cost= (345,600 + 900,000)/240 Cost = 1,245,600 / 240 Cost= 5190 confidence assessment: 2
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11:18:11 express with common denominator x: [x / x] * 6x + 900,000 / x = 6x^2 / x + 900,000 / x = (6x^2 + 900,000) / x so cost = (6x^2+900,000)/x Evaluating at x = 240 we get cost = (6 * 240^2 + 900000) / 240 = 5190. **
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RESPONSE --> self critique assessment: 3
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{o곻oҲ^Ϋ assignment #004 004. `query 4 Applied Calculus I 02-14-2009
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11:20:20 #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 --> Q(t) = 550mg(.89)^t confidence assessment: 3
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11:20:26 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 --> self critique assessment: 3
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11:21:34 How much antibiotic is present at 3:00 p.m.?
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RESPONSE --> Because the time interval from 10:00 am to 3:00pm is 5 hours, we use 5 as the t value Q(5) = 550mg(.89)^5 Q(5)= 307.12mg confidence assessment: 3
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11:21:42 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 --> self critique assessment: 3
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11:24:08 Describe your graph and explain how it was used to estimate half-life.
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RESPONSE --> My graph starts at 500mg, slowly decreasing to 171.50 at Q(11). The graph is shaped in a downward slope. From the graph, I could estimate that the doubling time was between Q(5) and Q(6) confidence assessment: 3
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11:25:08 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 --> I didn't explain how I actually set up the graph, but that is how I did it. I just didn't put in how to set up. self critique assessment: 2
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11:25:44 What is the equation to find the half-life? What is the most simplified form of this equation?
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RESPONSE --> Q(dt) = 550mg (.89)^dt = 275mg confidence assessment: 3
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11:26:55 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, I didn't simplify but I did get that answer. self critique assessment: 2
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11:33:16 #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 --> Through trial and error, I found that the t values between -25 (which yields .092) and -32 (which yields .047) lie betwee 0.05 and 0.1. Between values 73 (which yields .00095) and 56 ( which yields .0048) lies between .005 and .001. confidence assessment: 2
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11:36:10 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 --> I went through the same process, but I arrived at different intervals, but didn't go as far as this process did. I can see now that I should have tried the numbers in between the whole numbers, going to decimals. self critique assessment: 2
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11:36:49 explain why the negative t axis is a horizontal asymptote for this function.
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RESPONSE --> The negative t axis aprroaches the y axis closer and closer, approaching zero at a much smaller number each interval. confidence assessment: 2
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11:37:29 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 --> I also see that as t approaches infinity, so does 1.1^t. self critique assessment: 2
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11:40:33 #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 --> Using the equation y = Ae^kx, we see that b would equal e^k. y = 12 (e^-.5x) y= 12 (e^-.5) ^x y = 12 (.607)^x This shows that .607 is the b value of this equation. confidence assessment: 2
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11:40:41 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 --> self critique assessment: 3
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11:41:56 what is b for the function y = .007 ( e^(.71 x) )?
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RESPONSE --> y = .007 (e^.71x) y = .007 (e^.71) ^ x y = .007 (2.034) ^ x We see from this equation that the b value is 2.034 ( or 2.03). confidence assessment: 3
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11:42:09 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 --> self critique assessment: 3
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11:44:12 what is b for the function y = -13 ( e^(3.9 x) )?
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RESPONSE --> y = -13 (e^3.9x) y = -13 (e^3.9)^x y = -13 ( 49.402)^x b = 49.402 confidence assessment: 2
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11:44:19 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 --> self critique assessment: 3
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11:45:57 List these functions, each in the form y = A b^x.
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RESPONSE --> y = 12 (e^-.5)^x y = 12 (.607)^x y = .007(e^.71)^x y = .007 (2.034)^x y = -13 (e^3.9)^x y= -13 (49.402)^x confidence assessment: 3
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11:46:19 The functions are y=12(.6065^x) y=.007(2.03399^x) and y=-13(49.40244^x) **
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RESPONSE --> My decimals are rounded, is this okay?
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11:47:59 0.4.44 (was 0.4.40 find all real zeros of x^2+5x+6
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RESPONSE --> x^2 + 5x + 6 = (x + 3) (x + 2) = Zeros are -3 and -2 confidence assessment: 2
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11:48:08 We can factor this equation to get (x+3)(x+2)=0. (x+3)(x+2) is zero if x+3 = 0 or if x + 2 = 0. We solve these equations to get x = -3 and x = -2, which are our two solutions to the equation. COMMON ERROR AND COMMENT: Solutions are x = 2 and x = 3. INSTRUCTOR COMMENT: This error generally comes after factoring the equation into (x+2)(x+3) = 0, which is satisfied for x = -2 and for x = -3. The error could easily be spotted by substituting x = 2 or x = 3 into this equation; we can see quickly that neither gives us the correct solution. It is very important to get into the habit of checking solutions. **
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RESPONSE --> self critique assessment: 3
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11:48:52 Explain how these zeros would appear on the graph of this function.
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RESPONSE --> The graph would be closed brackets between -3 and -2. confidence assessment: 1
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11:49:30 We've found the x values where y = 0. The graph will therefore go through the x axis at x = -2 and x = -3. **
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RESPONSE --> Oh ok, I don't really know what I was thinking, but since -3 and -2 are zeros, I can clearly see that the graph will cross the x-axis at these two point.s self critique assessment: 2.
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11:50:22 0.4.50 (was 0.4.46 x^4-625=0
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RESPONSE --> x^4 - 625 = 0 x^4 = 625 x +-(5) The zeros are positive or negative 5. confidence assessment: 2
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11:51:03 Common solution: x^4 - 625 = 0. Add 625 to both sides: x^4 = 625. Take fourth root of both sides, recalling that the fourth power is `blind' to the sign of the number: x = +-625^(1/4) = +-[ (625)^(1/2) ] ^(1/2) = +- 25^(1/2) = +-5. This is a good and appropriate solution. It's also important to understand how to use factoring, which applies to a broader range of equations, so be sure you understand the following: We factor the equation to get (x^2 + 25) ( x^2 - 25) = 0, then factor the second factor to get (x^2 + 25)(x - 5)(x + 5) = 0. Our solutions are therefore x^2 + 25 = 0, x - 5 = 0 and x + 5 = 0. The first has no solution and the solution to the second two are x = 5 and x = -5. The solutions to the equation are x = 5 and x = -5. **
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RESPONSE --> self critique assessment: 3
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11:53:18 0.4.70 (was 0.4.66 P = -200x^2 + 2000x -3800. Find the x interval for which profit is >1000
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RESPONSE --> P= -200x^2 + 2000x - 3800 > 1000 P= -200x^2 + 2000x - 4800 > 0 P= x^2 - 10x + 24 > 0 P= (x-6) (x-4)>0 confidence assessment: 0
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11:54:44 You have to solve the inequality 1000<-200x^2+2000x-3800, which rearranges to 0<-200x^2+2000x-4800. This could be solved using the quadratic formula, but it's easier if we simplify it first, and it turns out that it's pretty easy to factor when we do: -200x^2+2000x-4800 = 0 divided on both sides by -200 gives x^2 - 10 x + 24 = 0, which factors into (x-6)(x-4)=0 and has solutions x=4 and x=6. So -200x^2+2000x-4800 changes signs at x = 4 and x = 6, and only at these values (gotta go thru 0 to change signs). At x = 0 the expression is negative. Therefore from x = -infinity to 4 the expression is negative, from x=4 to x=6 it is positive and from x=6 to infinity it's negative. Thus your answer would be the interval (4,6). COMMON ERROR: x = 4 and x = 6. INSTRUCTOR COMMENT: You need to use these values, which are the zeros of the function, to split the domain of the function into the intervals 4 < x < 6, (-infinity, 4) and (6, infinity). Each interval is tested to see whether the inequality holds over that interval. COMMON ERROR: Solution 4 > x > 6. INSTRUCTOR COMMENT: Look carefully at your inequalities. 4 > x > 6 means that 4 > x AND x > 6. There is no value of x that is both less than 4 and greater than 6. There are of course a lot of x values that are greater than 4 and less than 6. The inequality should have been written 4 < x < 6, which is equivalent to the interval (4, 6). **
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RESPONSE --> I didn't finish the problem, I got to the simplifying to (x-6) (x-4). I can see now that the answer must be 4>x>6.
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