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01:27:27 Where f(x) = x^3, what are f(-2), f(-a), f(x-4) and f(x) - 4?
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RESPONSE --> If f(x) = x^3 the we can substitute to get the following
f(-2) = -2^2 = -2*-2*-2 = -8 f(-a) = -a^2 = -a*-a*-a f(x-4) = (x-4)^3 = (x-4)(x-4)(x-4) = (x^2-4x-4x+16)(x-4) f(x) - 4 = (x^3) - 4.................................................
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01:29:32 ** COMMON ERROR WITH COMMENT: Where f(x) = x^3, f(-2) = -2^3
INSTRUCTOR CRITIQUE: Write that f(-2) = (-2)^3. By order of operations -2^3 = -(2^3) (you exponentiate first, before you apply the negative, which is effectively a multiplication by -1). For an odd power like 3 it makes no difference in the end---in both cases the result is -8 COMMON ERROR WITH COMMENT: f(-a) = -a^3. INSTRUCTOR COMMENT:f(-a) = (-a)^3, which because of the odd power is the same as -a^3. If g(x) = x^2, g(-a) would be (-a)^2 = a^2. ANSWERS TO THE REMAINING TWO QUESTIONS: f(x-4) = (x-4)^3. If you expand that you get (x^2 - 8 x + 16) ( x - 4) = x^3 - 12 x^4 + 48 x - 64. In more detail the expansion is as follows: (x-4)^3 = (x-4)(x-4)(x-4) = [ (x-4)(x-4) ] * (x-4) = [ x ( x - 4) - 4 ( x - 4) ] ( x - 4) = (x^2 - 4 x - 4 x + 16) ( x - 4) = (x^2 - 8x + 16) ( x - 4) = (x^2 - 8x + 16) * x - (x^2 - 8x + 16) * 4 = x^2 * x - 8 x * x + 16 * x - x^2 * 4 - (-8x) * 4 - 16 * 4 = x^3 - 8 x^2 + 16 x - 4 x^2 + 32 x - 64 = x^3 - 12 x^2 + 48 x - 64. f(x) - 4 = x^3 - 4. **......!!!!!!!!...................................
RESPONSE --> Ok, I have made notes as to how to write out the negative coefficients with an exponent.
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01:33:09 Where f(x) = 2^x, find f(2), f(-a), f(x+3) and f(x) + 3?
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RESPONSE --> Where f(x) = 2^x
f(2) = 2^2 f(2) = 4 f(-a) = (2)^(-a) f(x+3) = 2^(x+3) f(x) + 3 = 2^x + 3.................................................
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01:33:22 ** Where f(x) = 2^x we have:
f(2)= 2^2 or 4; f(a) = 2^a; f(x+3) = 2^(x+3); and f(x) + 3 = 2^x + 3. **......!!!!!!!!...................................
RESPONSE --> ok
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01:33:29 13:06:29
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RESPONSE --> ok
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01:35:34 query functions given by meaningful names. What are some of the advantages of using meaninful names for functions?
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RESPONSE --> It is advantageous to use meaningful names in describing functions so that the reader knows what type of equation the author is using in order to arrive at the function conclusion.
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01:35:45 ** TWO STUDENT RESPONSES:
Using the function notation gives more meaning to our equations. Example.........!!!!!!!!...................................
RESPONSE --> ok
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01:36:59 'depth(t) = ' is a lot more understandable than ' y = '
I read of the pros and cons in using names. I think using names helps to give meaning to the equation, especially if it's one you haven't looked at for a couple days, you know right away what you were doing by reading the 'meaningful' notation.**......!!!!!!!!...................................
RESPONSE --> This is very true. By understanding what it is that you are looking for and looking back on work performed it can be very advantageous so that you know what type of function it was.
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01:38:44 What were your results for value(0), value(2), value(t+3) and value(t+3)/value(t)?
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RESPONSE --> Value(0) = $1000(1.07)^0 = $1000(1) = $1000
Value(1) = $1000(1.07)^1 = $1000(1.07) = $1070 Value(2) = $1000(1.07)^2 = $1000(1.1449) = $1144.9 Value(t+3) = $1000(1.07)^(t+3) Value(t+3)/Value(t) = $1000(1.07)^(t+3) / $1000(1.07)^t.................................................
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01:39:56 ** Substitute very carefully and show your steps:
value(0) = $1000(1.07)^0 = $ 1000 value(2) = $1000(1.07)^2 = $1144.90 value(t + 3) = $1000(1.07)^(t + 3) value(t + 3) / value (t) = [$1000(1.07)^(t + 3)] / [ $1000(1.07)^t] , which we simplify. The $1000 in the numerator can be divided by the $1000 in the denominator to give us value(t+3) / value(t) = 1.07^(t+3) / [ 1.07^t]. By the laws of exponents 1.07^(t+3) = 1.07^t * 1.07^3 so we get value(t+3) / value(t) = 1.07^t * 1.07^3 / [ 1.07^t]. The 1.07^t divides out and we end up with value(t+3) / value(t) = 1.07^3. **......!!!!!!!!...................................
RESPONSE --> Ok. I have made notes in my notebook as to the simplification using the exponents for the final problem.
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01:40:02 13:14:30
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RESPONSE --> ok
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01:46:01 What did you get for illumination(distance)/illumination(2*distance)? Show your work on this one.
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RESPONSE --> Illumination(1) = 50 / 1^2 = 50 / 1 = 50
Illumination(2) = 50 / 2^2 = 50 / 4 = 12.5 Illumination(3) = 50 / 3^2 = 50 / 9 = 5.56 Illumination(distance) / illumination(2*distance) = (50 / distance^2) / (50 / 2*distance^2).................................................
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01:48:32 ** We substitute carefully and literally to get
illumination (distance) / illumination (2*distance) = [50 / distance^2] / [50 / (2*distance)^2] which is a complex fraction and needs to be simplified. You invert the denominator and multiply to get [ 50 / distance^2 ] * [ (2 * distance)^2 / 50 ] = (2 * distance)^2 / distance^2 = 4 * distance^2 / distance^2 = 4. **......!!!!!!!!...................................
RESPONSE --> Ok, I have made notes on this subject.
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01:49:50 query #3. Sketch a reasonable graph of y = f(x), if it is known that f(2) = 80, f(5) = 40 and f(10) = 25. Explain how you constructed your graph.
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RESPONSE --> To construct the graph I used the initail points of (2,80), (5,40) and (10, 25). Then I sketched a general estimated graph for the remaining points.
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01:50:06 ** STUDENT ANSWER WITH INSTRUCTOR COMMENT: I used graph paper counted each small square as 2 units in each direction, I put a dot for the points (2,80), (5,40), & (10,25) and connected the dots with lines.
INSTRUCTOR COMMENT: The points could be connected with straight lines, but you might also have used a smooth curve. **......!!!!!!!!...................................
RESPONSE --> Ok.
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01:50:43 what is your estimate of value of x for which f(x) = 60?
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RESPONSE --> The estimated value of x when f(x) = 60 is 3.5.
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01:50:55 **If your graph was linear from the point (2, 80) to the point (5, 40) you would estimate x = 3.5, since 3.5 is halfway from 2 to 5 and 60 is halfway from 80 to 40.
However with f(10) = 25 a straightforward smooth curve would be decreasing at a decreasing rate and the y = 60 value would probably occur just a bit earlier, perhaps somewhere around x = 3.3**......!!!!!!!!...................................
RESPONSE --> Ok.
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01:51:25 what is your estimate of the value f(7)?
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RESPONSE --> The estimated value for f(7) is 34.
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01:51:42 ** STUDENT RESPONSE WITH INSTRUCTOR COMMENT: My estimation for the value of f(7) was f(7) = 34.
A linear graph between (5, 40) and (10, 25) would have slope -3 and would therefore decline by 6 units between x = 5 and x = 7, giving your estimate of 34. However the graph is probably decreasing at a decreasing rate, implying a somewhat greater decline between 5 and 7 that would be implied by a straight-line approximation. A better estimate might be f(7) = 32 or 33. **......!!!!!!!!...................................
RESPONSE --> ok
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01:53:03 what is your estimate of the difference between f(7) and f(9)?
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RESPONSE --> My estimate of the difference between f(7) and f(9) is 3 units. Though obviously with a varying decrease in the graph it may be more than that.
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01:53:15 ** If f(7) = 32 and f(9) = 27, not unreasonable estimates, then the difference is 5. **
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RESPONSE --> ok
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01:53:20 13:29:11
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RESPONSE --> ok
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01:54:55 what is your estimate of the difference in x values between the points where f(x) = 70 and where f(x) = 30?
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RESPONSE --> I mistakenly put the answer to this question in the answer box for the last question. The answer I got for the previous question was 6. So for this question I estimated the difference between f(x) = 70 and f(x) = 30 to be 3 units.
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01:55:20 ** f(x) will be 70 somewhere around x = 3 and f(x) will be 30 somewhere around x = 8 or 9. The difference in these x values is about 5 or 6.
On your graph you could draw horizontal lines at y = 70 and at y = 30. You could then project these lines down to the x axis. The distance between these projection lines, which again should probably be around 5 or 6, can be estimated by the scale of the graph. **......!!!!!!!!...................................
RESPONSE --> ok
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01:55:29 query #4. temperature vs. clock time function y = temperature = T(t), what is the symbolic expression for ...
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RESPONSE --> ok
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01:55:48 ** STUDENT SOLUTION WITH INSTRUCTOR COMMENT:
The temperature at time t = 3; T(3)The temperature at time t = 5; T(5) The change in temperature between t = 3 and t = 5; T(3) - T(5) The order of the expressions is important. For example the change between 30 degrees and 80 degrees is 80 deg - 30 deg = 50 degrees; the change between 90 degrees and 20 degrees is 20 deg - 90 deg = -70 deg. The change between T(3) and T(5) is T(5) - T(3). When we specify the change in a quantity we subtract the earlier value from the later. INSTRUCTOR COMMENT: To average two numbers you add them and divide by 2. The average of the temperatures at t = 3 and t = 5 is therefore [T(3) + T(5)] /2 **......!!!!!!!!...................................
RESPONSE --> ok
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02:03:03 What equation would we solve to find the clock time when the model predicts a temperature of 150? How would we find the length of time required for the temperature to fall from 80 to 30?
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RESPONSE --> We would use the equation y = 150 - t to solve the equation to find clock time when the model predicts a temperature of 150.
Y = (80 - t) + (30 - t) / 2 is the equation we would use to find the length of time needed for the temperature from 80 to 30..................................................
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02:03:58 ** GOOD STUDENT SOLUTION:
To find the clock time when the model predicts a temperature of 150 we would solve the equation T(t) = 150. To find the length of time required for the temperature to fall from 80 to 30 we would first solve the equation T(t) = 80 to get the clock time at 80 degrees then find the t value or clok time when T(t) = 30. Then we could subtract the smaller t value from the larger t value to get the answer. [ value of t at T(t) = 30] - [ value of t at T(t) = 80)] **......!!!!!!!!...................................
RESPONSE --> I am saving this as notes as I am apparently confused on this issue.
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*&*& i}` assignment #004 hQrQԐzՇ Physics I Class Notes 09-18-2005
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03:22:54 How do we reason out the process of determining acceleration from rest given displacement and time duration?
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RESPONSE --> We can find the acceleration from rest given the displacement and time for displacement by dividing the displacement by the time it takes to travel the length of the displacement.
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03:23:35 ** COMMON ERROR: By first finding the average velocity. Also we must understand this gives us the change in velocity which gives us the acceleration.
INSTRUCTOR COMMENT: Average acceleration is not related to average velocity. It is related to the change in velocity, but change in velocity is not the same thing as acceleration. Acceleration is the rate of change of velocity. ANSWER TO QUESTION: Average acceleration is defined to be the average rate at which velocity changes, which is change in velocity/change in clock time, or equivalently change in velocity/time duration. However we do not know the change in velocity, nor from the given information can we directly determine the change in velocity. So we have to look at what we can reason out from what we know. Given displacement from rest and time duration we calculate average velocity: vAve = `ds / `dt = displacement/time duration. Since the initial velocity is 0, the final velocity is double the average so we find the final velocity by doubling the average velocity. Now that we know the initial velocity 0 and the final velocity we can find change in velocity by subtracting initial vel from final vel. Dividing change in velocity by a time duration we finally obtain the average acceleration. **......!!!!!!!!...................................
RESPONSE --> ok
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ةы`xѹG assignment #002 hQrQԐzՇ Physics I 09-18-2005
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12:20:37 ** Questions about velocity, average velocity, acceleration, etc. are very confusing because so many of the concepts have similar definitions. People have trouble distinguishing things like average velocity, which for uniform acceleration can be obtained in a process that adds two velocities, from average acceleration, which involves subtracting two velocities; one of these processes involves dividing by 2 and the other dividing by the time interval `dt.
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RESPONSE --> ok
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12:21:55 It is essential to keep the definitions and the meanings of the terms very clear and to work everything from definitions. It is equally important to have a good common-sense understanding of every definition so you can develop the intuition to make sense of everything you do.
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RESPONSE --> ok
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12:22:03 That inevitably takes people a little time. But in the process you develop the habits you will need to succeed in the course. **
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RESPONSE --> ok
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ת֏HwV assignment # hQrQԐzՇ Liberal Arts Mathematics I 09-18-2005 "