course I just went ahead and combined the rest of the assignments into one page so wouldn't have to flip back and forth.Thanks,
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query Logarithms, Logarithmic Functions, Logarithmic Equations 1. For what value of x will the function y = log{base 2}(x) first reach 4
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RESPONSE --> x would be e^4 and that would reach 4 near 53 confidence assessment: 2
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** ln(x) = 4 translates to x = e^4, which occurs at x = 55 approx. ln(x) = 2 translates to x = e^2, which occurs at x = 7.4 approx. ln(x) = 3 translates to x = e^3, which occurs at x = 20 approx. **
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RESPONSE --> I see, I didn't take it out far enough. self critique assessment: 2
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for what value of x will the function y = ln(x) first reach y = 4?
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RESPONSE --> that would be e^y=x so it would be around the same answer as the last, somewhere around 53. confidence assessment: 2
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y = ln(x) means that e^y = x. The function y = ln(x) will first reach y = 4 when x = e^4 = 54.6 approx. **
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RESPONSE --> I was close!! self critique assessment: 2
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3. Explain why the negative y axis is an asymptote for a log{base b}(x) function explain why this is so only if b > 1
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RESPONSE --> it is the opposite of y=b^x. b/c it gives it a minus axis. confidence assessment: 2
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** the log{base b} function is the inverse of the y = b^x function. Assuming b > 1, large negative values of x lead to positive b^x values near zero, resulting in a horizontal asymptotes along the negative x axis. When the columns are reversed we end up with small positive values of x associated with large negative y, resulting in a vertical asymptote along the negative y axis. You can take a negative power of any positive b, greater than 1 or not. For b > 1, larger negative powers make the result smaller, and the negative x axis is an asymptote. For b < 1, larger positive powers make the result smaller, and the positive x axis becomes the asymptote. **
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RESPONSE --> ok self critique assessment: 2
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5. What are your estimates for the values of b for the two exponential functions on the given graph?
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RESPONSE --> My estimates would b approx. 3 and 8 confidence assessment: 2
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** The exponential function y = A b^x has y intercept (0, A) and basic point (1, A * b). Both graphs pass thru (0, 1) so A = 1. The x = 1 points are (1, 3.5) and (1, 7.3) approx.. Thus for the first A * b = 3.5; since A = 1 we have b = 3.5. For the second we similarly conclude that b = 7.3. So the functions are y = 3.5^x and y = 7.3^x, approx.. **
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RESPONSE --> self critique assessment: 2
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At what points will each of the logarithmic functions reach the values y = 2, y = 3 and y = 4?
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RESPONSE --> Approx .50 and 1. confidence assessment: 2
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** Using trial and error we find that for y = 3.5^x, the x values .55, .88 and 1.11 give us y = 2, 3 and 4. We find that for y = 7.3^x, the x values .35, .55 and .70 give us y = 2, 3 and 4. **
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RESPONSE --> I forgot to do the 3 but I kinda see. self critique assessment: 2
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7. What is the decibel level of a sound which is 10,000 times as loud as hearing threshold intensity?
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RESPONSE --> I think it would be 40. confidence assessment: 2
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dB = 10 log(I / I0). If the sound is 10,000 times as loud as hearing threshold intensity then I / I0 = 10,000. log(10,000) = 4, since 10^4 = 10,000. So dB = 10 log(I / I0) = 10 * 4 = 40.
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RESPONSE --> I got it. self critique assessment: 2
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What are the decibel levels of sounds which are 100, 10,000,000 and 1,000,000,000 times louder than threshold intensity?
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RESPONSE --> So if the last was 40 you just multiply them by their amount of zeros by ten to get 20,70, and 90 confidence assessment: 2
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10 log(100) = 10 * 2 = 20, so a sound which is 100 times hearing threshold intensity is a 20 dB sound. 10 log(10,000,000) = 10 * 7 = 70, so a sound which is 100 times hearing threshold intensity is a 70 dB sound. 10 log(1,000,000,000) = 10 * 9 = 90, so a sound which is 100 times hearing threshold intensity is a 90 dB sound.
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RESPONSE --> I get this. self critique assessment: 2
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how can you easily find these decibel levels without using a calculator?
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RESPONSE --> Like I said before just multiply by the number of zeros. confidence assessment: 2
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08:10:20 Since 10^1, 10^2, 10^3, ... are 10, 100, 1000, ..., the power of 10 is the number of zeros in the result. Since the log of a number is the power to which 10 must be raised to get this number, the log of one of these numbers is equal to the number of zeros.
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RESPONSE --> self critique assessment: 3
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08:11:15 What are the decibel levels of sounds which are 500, 30,000,000 and 7,000,000,000 times louder than threshold intensity?
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RESPONSE --> I would say half would be 25 75 and 100 confidence assessment: 2
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10 log(500) = 10 * 2.699 = 26.99 so this is a 26.99 dB sound. 10 log(30,000,000) = 10 * 7.477 = 74.77 so this is a 74.77 dB sound. 10 log(7,000,000,000) = 10 * 9.845 = 98.45 so this is a 98.45 dB sound.
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RESPONSE --> I got them wrong but I think I understand. self critique assessment: 2
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8. If a sound measures 40 decibels, then what is the intensity of the sound, as a multiple of the hearing threshold intensity?
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RESPONSE --> Just divide by 10 to get 4. confidence assessment: 2
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** Let x be the ratio I / I0. Then we solve the equation 40= 10*log(x). You solve this by dividing by 10 to get log(x) = 4 then translating this to exponential form x = 10^4 = 10,000. The sound is 10,000 times the hearing threshold intensity, so I = 10,000 I0. **
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RESPONSE --> self critique assessment: 2
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Answer the same question for sounds measuring 20, 50, 80 and 100 decibels.
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RESPONSE --> 2,5,8, and 100 confidence assessment: 2
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** since dB = 10 log(I / I0) we have log(I/I0) = dB / 10. Translating to exponential form this tells us that I / I0 = 10^(dB/10) wo that I = I0 * 10^(dB/10). For a 20 dB sound this gives us I = I0 * 10^(20/10) = I0 * 10^2 = 100 I0. The sound is 100 times the intensity of the hearing threshold sound. For a 50 dB sound this gives us I = I0 * 10^(50/10) = I0 * 10^5 = 100,000 I0. The sound is 100,000 times the intensity of the hearing threshold sound. For an 80 dB sound this gives us }I = I0 * 10^(80/10) = I0 * 10^8 = 100,000,000 I0. The sound is 100,000,000 times the intensity of the hearing threshold sound. For a 100 dB sound this gives us I = I0 * 10^(100/10) = I0 * 10^10 = 10,000,000,000 I0. The sound is 10,000,000,000 times the intensity of the hearing threshold sound. **
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RESPONSE --> self critique assessment: 2
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assignment #019 019. `query 19 Precalculus I 12-09-2008
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`questionNumber 190000 explain the steps in fitting an exponential function to data
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RESPONSE --> you can solve equations with 2 points, you have to readjust bigger data Then you put it into the correct format. confidence assessment: 2
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08:21:46 `questionNumber 190000 ** If you have two points you can solve the simultaneous equations. If you have a more extensive data set you use transformations. For exponential data you plot log(y) vs. x. If the graph is a straight line then you have a good fit. If the slope is m and the vertical intercept is b then your graph gives you log(y) = m x + b. You invert the transformation using the inverse function 10^x, obtaining 10^log(y) = 10^(mx+b) so that y = 10^(mx) * 10^b, and then rearrange this into the desired form. Note that if a power function fits the data then log y vs. log x will give a straight line so that log y = m log x + b. In this case our solution will be y = 10^b * x^m, a power function rather than an exponential function. **
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RESPONSE --> I can see how the answer was achieved. self critique assessment: 2
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ۅԝz˶J assignment #020 020. `query 20 Precalculus I 12-09-2008
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08:24:34 What are the zeros of f(x) = 2x - 6 and g(x) = x + 2?
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RESPONSE --> Just solve the equations to get 3 and 2. confidence assessment: 2
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** f(x) = 2x - 6 is zero when 2x - 6 = 0. This equation is easily solved to yield x = 3. g(x) = x + 2 is zero when x + 2 = 0. This equation is easily solved to yield x = -2. **
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RESPONSE --> I forgot to carry the - in the two. self critique assessment: 2
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What does the quadratic formula give you for the zeros of the quadratic polynomial q(x)?
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RESPONSE --> The quadratic formula gives you qx=fx*gx confidence assessment: 2
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** We get q(x) = f(x) * g(x) = (2x - 6) ( x + 2) = 2x ( x + 6) - 6 ( x + 2) = 2 x^2 + 4 x - 6 x - 12 = 2 x^2 - 2 x - 12. This polynomial is zero, by the quadratic formula, when and only when x = [ -(-2) +- sqrt( (-2)^2 - 4(2)(-12) ] / (2 * 2) = [ 2 +- sqrt( 100) ] / 4 = [ 2 +- 10 ] / 4. Simplifying we get x = (2+10) / 4 = 3 or x = (2 - 10) / 4 = -2. This agrees with the fact that f(x) = 0 when and only when x = -3, and g(x) = 0 when and only when x = 2. The only was f(x) * g(x) can be zero is for either f(x) or g(x) to be zero. **
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RESPONSE --> I see, I didn't finish far enough self critique assessment: 2
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08:26:56 2. If z1 and z1 are the zeros of x^2 - x + 6, then what is the evidence that x^2-x + 6=(x - z1) * (x - z2)?
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RESPONSE --> confidence assessment:
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2. If z1 and z1 are the zeros of x^2 - x + 6, then what is the evidence that x^2-x + 6=(x - z1) * (x - z2)?
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RESPONSE --> just plug it in to get zero for both zs. confidence assessment: 2
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08:27:38 ** z1 and z2 both give zero when plugged into x^2 - x + 6 and also into (x-z1)(x-z2). (x-z1)(x-z2) gives an x^2 term, matching the x^2 term of x^2 - x + 6. Since the zeros and the highest-power term match both functions are obtained from the basic y = x^2 function by the same vertical stretch, both have parabolic graphs and both have the same zeros. They must therefore be identical. **
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RESPONSE --> i see self critique assessment: 2
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3. Explain why, if the quadratic polynomial f(x) = a x^2 + bx + c has no zeros, that polynomial cannot be the product of two linear polynomials.
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RESPONSE --> linear has zeros. confidence assessment: 2
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** If f(x) has linear factors, then if any of these linear factors is zero, multiplying it by the other factors will yield zero. Any linear factor can be set equal to zero and solved for x. Thus if f(x) has linear factors, it has zeros. So if f(x) has no zeros, it cannot have linear factors. **
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RESPONSE --> i got it. self critique assessment: 2
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assignment #021 021. `query 21 Precalculus I 12-09-2008
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What are the possible number of linear and irreducible quadratic factors for a polynomnial of degree 6?
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RESPONSE --> You could have the full 6 or half of that -3 confidence assessment: 2
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** You can have as many as 6 linear and 3 irreducible quadratic factors for a polynomial of degree 6. For a polynomial of degree 6: If you have no irreducible quadratic factors then to have degree 6 you will need 6 linear factors. If you have exactly one irreducible quadratic factor then this factor is of degree 2 and you will need 4 linear factors. If you have exactly two irreducible quadratic factors then the product of these factors is of degree 4 and you will need 2 linear factors. If you have three irreducible quadratic factors then the product of these factors is of degree 6 and you can have no linear factors. For a polynomial of degree 7: If you have no irreducible quadratic factors then to have degree 7 you will need 7 linear factors. If you have exactly one irreducible quadratic factor then this factor is of degree 2 and you will need 5 linear factors to give you degree 7. If you have exactly two irreducible quadratic factors then the product of these factors is of degree 4 and you will need 3 linear factors to give you degree 7. If you have three irreducible quadratic factors then the product of these factors is of degree 6 and you will need 1 linear factor to give you degree 7. **
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RESPONSE --> I see self critique assessment: 2
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For a degree 6 polynomial with one irreducible quadratic factor and four linear factors list the possible numbers of repeated and distinct zeros.
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RESPONSE --> you could have 4 distinct, 2 distinct, or 1 distinct and I think it would be the same with repeated. confidence assessment: 2
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......!!!!!!!!...................................** there could be 1 root repeated 4 times, 2 roots with 1 repeated 3 times and the other distinct from it, 2 distinct roots each repeated twice, three distinct roots with one of them repeated twice, or four distinct roots. Explanation:
You have one zero for every linear factor, so there will be four zeros. Since the degree is even the far-left and far-right behaviors will be the same, either both increasing very rapidly toward +infinity or both decreasing very rapidly toward -infinity. You can have 4 distinct zeros, which will result in a graph passing straight thru the x axis at each zero, passing one way (up or down) through one zero and the opposite way (down or up) through the next. You can have 2 repeated and 2 distinct zeros. At the repeated zero the graph will just touch the x axis as does a parabola at its vertex. The graph will pass straight through the x axis at the two distinct zeros. You can have 3 repeated and 1 distinct zero. At the 3 repeated zeros the graph will level off at the instant it passes thru the x axis, in the same way the y = x^3 graph levels off at x = 0. The graph will pass through the x axis at the one distinct zero. You can have two pairs of 2 repeated zeros. At each repeated zero the graph will just touch the x axis as does a parabola at its vertex. Since there are no single zeros (or any other zeros repeated an odd number of times) the graph will not pass through the x axis, so will remain either entirely above or below the x axis except at these two points. You can have four repeated zeros. At the repeated zero the graph will just touch the x axis, much as does a parabola at its vertex except that just as the y = x^4 function is somewhat flatter near its 'vertex' than the y = x^2 function, the graph will be flatter near this zero than would be a parabolic graph. **
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RESPONSE --> I think I understand where that is coming from self critique assessment: 2
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Describe a typical graph for each of these possibilities. Describe by specifying the shape of the graph at each of its zeros, and describe the far-left and far-right belavior of the graph.
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RESPONSE --> For the 1 distinct it is a curve in the positive quadrant, for 2 distinct it would also be above the x axis for a while, and for 4 it would go through the x axis i think 4 times, thus the four distincts. confidence assessment: 2
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ERRONEOUS STUDENT SOLUTION AND INSTRUCTOR CORRESION for 4 distincts, the graph curves up above the x axis then it crosses it 4 timeswithe the line retreating from the direction it came from for say 2 distincts and 2 repeats, line curves above x axis then it kisses the axis then it crosses it twice, retreating to the side it came from for 1 distinct and 3 it curves above x axis then it crosses once and kisses, finnaly heading off to the opposite side it came from INSTRUCTOR COMMENTS: {The graph can't go off in th opposite direction. Since it is a product of four linear factors and any number of quadratic factors its degree is even so its large-x behavior is the same for large positive as for large negative x. It doesn't kiss at a degree-3 root, it acts like the y = x^3 polynomial, leveling off just for an instant as it passes through the zero and on to the other side of the axis. **
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RESPONSE --> I t hink I understand. self critique assessment: 2
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assignment #022 022. `query 22 Precalculus I 12-10-2008
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Explain why the function y = x^-p has a vertical asymptote at x = 0.
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RESPONSE --> The possibilities of how x and h can be near each other are endless thus the graph has a vertical asymptote at x=0 confidence assessment: 2
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** x^-p = 1 / x^p. As x gets closer to 0, x^p gets closer to 0. Dividing 1 by a number which gets closer and closer to 0 gives us a result with larger and larger magnitude. There is no limit to how close x can get to 0, so there is no limit to how many times x^p can divide into 1. This results in y = x^p values that approach infinite distance from the x axis. The graph therefore approaches a vertical limit. **
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RESPONSE --> I am pretty sure I understand the basic concept. self critique assessment: 2
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Explain why the function y = (x-h)^-p has a vertical asymptote at x = h.
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RESPONSE --> Like I said before the possibilities of how close they can get are endless thus making a vertical asymptote as x=h confidence assessment: 2
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** (x-h)^-p = 1 / (x-h)^p. As x gets closer to h, (x-h)^p gets closer to 0. Dividing 1 by a number which gets closer and closer to 0 gives us a result with larger and larger magnitude. There is no limit to how close x can get to h, so there is no limit to how many times (x-h)^p can divide into 1. This results in y = (x-h)^p values that approach infinite distance from the x axis as x approaches h. The graph therefore approaches a vertical limit. This can also be seen as a horizontal shift of the y = x^-p function. Replacing x by x - h shifts the graph h units in the x direction, so the asymptote at x = 0 shifts to x = h. **
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RESPONSE --> I think I get this. self critique assessment: 2
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Explain why the function y = (x-h)^-p is identical to that of x^-p except for the shift of h units in the x direction.
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RESPONSE --> It keeps the shape but changes the x amount confidence assessment: 2
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STUDENT ANSWER: You end up with the exact same y values but at the different position of x changed by the h value. INSTRUCTOR COMMENT: Good start. More specifically the x value at which a given y value occurs is shifted h units, so that for example y = x^p is zero when x = 0, but y = (x - h)^p is zero when x = h.
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RESPONSE --> I didn't get this one but by reading the explanation, I think I comprehend it. self critique assessment: 2
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assignment #022 022. `query 22 Precalculus I 12-10-2008 assignment #023 023. `query 23 Precalculus I 12-11-2008
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Query problem 2. Describe the sum of the two graphs.
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RESPONSE --> There are blue and black grapsh The black graph is at 8,3,0,-1,0,3, and 8 if x=-3,-2,-1,0,1,2,and 3 and the blue graph has them at 1.5,.6,.1,-.1,-.6 and -.8 using the same x amounts. confidence assessment: 2
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** The 'black' graph takes values 8, 3, 0, -1, 0, 3, 8 at x = -3, -2, -1, 0, 1, 2, 3. The 'blue' graph takes approximate values 1.7, .8, .2, -.1, -.4, -.6, -.8 at the same x values. The 'blue' graph takes value zero at approximately x = -.4. The sum of the two graphs will coincide with the 'blue' graph where the 'black' graph is zero, which occurs at x = -1 and x = 1. The sum will coincide with the 'black' graph where the 'blue' graph is zero, which occurs at about x = -.4. **
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RESPONSE --> I see self critique assessment: 2
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Where it is the sum graph higher than the 'black' graph, and where is it lower? Answer by giving specific intervals.
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RESPONSE --> It is higher when their is a + blue and lower when the blue is - confidence assessment: 2
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** The sum of the graphs is higher than the 'black' graph where the 'blue' graph is positive, lower where the 'blue' graph is negative. The 'blue' graph is positive on the interval from x = -3 to x = -.4, approx.. This interval can be written [-3, -.4), or -3 <= x < -.4. **
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RESPONSE --> I forgot to write the intervals but I understand. self critique assessment: 2
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Where it is the sum graph higher than the 'blue' graph, and where is it lower? Answer by giving specific intervals.
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RESPONSE --> It is the same as with the black graph. Where the black graph is + the blue is higher, where the black graph is - the blue is lower. confidence assessment: 2
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** The sum of the graphs is higher than the 'blue' graph where the 'black' graph is positive, lower where the 'black' graph is negative. The 'black' graph is positive on the interval from x = -1 to x = 1, not including the endpoints of the interval. This interval can be written (-1, 1) or -1 < x < 1. **
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RESPONSE --> I think I get this self critique assessment: 2
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Where does thus sum graph coincide with the 'black' graph, and why? Give your estimate of the specific coordinates of the point or points where this occurs.
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RESPONSE --> I think they meet at the zero at approx x=-1 and 1. confidence assessment: 2
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** The sum coincides with the 'black' graph where the 'blue' graph is zero, which occurs at about x = -.4. The coordinates would be about (-.4, -.7), on the 'black' graph. **
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RESPONSE --> I think I get it. self critique assessment: 2
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Where does thus sum graph coincide with the 'blue' graph, and why? Give your estimate of the specific coordinates of the point or points where this occurs.
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RESPONSE --> It would be through x=-1 confidence assessment: 2
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** The sum coincides with the 'blue' graph where the 'black' graph is zero, which occurs at x = -1 and x = 1. The coordinates would be about (-1, .2) and (1, -.4), on the 'blue' graph. **
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RESPONSE --> I didn't finish it out on the last problem but i see how the answer was achieved. self critique assessment: 2
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Query problem 3 Describe the quotient graph obtained by dividing the 'black' graph by the 'blue' graph. You should answer the following questions: Where it is the quotient graph further from the x axis than the 'black' graph, and where is it closer? Answer by giving specific intervals, and explaining why you believe these to be the correct intervals. Where it is the quotient graph on the same side of the x axis as the 'black' graph, and where is it on the opposite side, and why? Answer by giving specific intervals. Where does thus quotient graph coincide with the 'black' graph, and why? Give your estimate of the specific coordinates of the point or points where this occurs. Where does the quotient graph have vertical asymptote(s), and why? Describe the graph at each vertical asymptote.
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RESPONSE --> the black is not a parabola but the blue appears to be. The ""quotient would be more away from the x line than the black graph when the blue is 1 away from the origin. When the black is zero the blue won't be which means the ""quotient"" will be. confidence assessment: 2
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** The 'black' graph is periodic, passing through 0 at approximately x = -3.1, 0, 3.1, 6.3. This graph has peaks with y = 1.5, approx., at x = 1.6 and 7.8, approx., and valleys with y = -1.5 at x = -1.6 and x = 4.7 approx. The 'blue' graph appears to be parabolic, passing thru the y axis at x = -1 and reaching a minimum value around y = -1.1 somewhere near x = 1. This graph passes thru the x axis at x = 5.5, approx., and first exceeds y = 1 around x = 7.5. The quotient will be further from the x axis than the 'black' graph wherever the 'blue' graph is within 1 unit of the origin, since division by a number whose magnitude is less than 1 gives a result whose magnitude is greater than the number being divided. This will occur to the left of x = 1, and between about x = 2 and x = 7.5. Between about x = 0 and x = 1 the 'blue' graph is more than 1 unit from the x axis and the quotient graph will be closer to the x axis than the 'black' graph. The same is true for x > 7.5, approx.. The 'black' graph is zero at or near x = -3.1, 0, 3.1, 6.3. At both of these points the 'blue' graph is nonzero so the quotient will be zero. The 'blue' graph is negative for x < 5.5, approx.. Since division by a negative number gives us the opposite sign as the number being divided, on this interval the quotient graph will be on the opposite side of the x axis from the 'black' graph. The 'blue' graph is positive for x > 5.5, approx.. Since division by a positive number gives us the same sign as the number being divided, on this interval the quotient graph will be on the same side of the x axis as the 'black' graph. The quotient graph will therefore start at the left with positive y values, about 3 times as far from the x axis as the 'black' graph (this since the value of the 'blue' graph is about -1/3, and division by -1/3 reverses the sign and gives us a result with 3 times the magnitude of the divisor). The quotient graph will have y value about 2.5 at x = -1.6, where the 'black' graph 'peaks', but the quotient graph will 'peak' slightly to the left of this point due to the increasing magnitude of the 'blue' graph. The quotient graph will then reach y = 0 / (-1) = 0 at x = 0 and, since the 'black' graph then becomes positive while the 'blue' graph remains negative, the quotient graph will become negative. Between x = 0 and x = 2 the magnitude of the 'blue' graph is a little greater than 1, so the quotient graph will be a little closer to the x axis than the 'black' graph (while remaining on the other side of the x axis). At x = 3.1 approx. the 'black graph is again zero, so the quotient graph will meet the x axis at this point. Past x = 3.1 the quotient graph will become positive, since the signs of both graphs are negative. As we approach x = 5.5, where the value of the 'blue' graph is zero, the quotient will increase more and more rapidly in magnitude (this since the result of dividing a negative number by a negative number near zero is a large positive number, larger the closer the divisor is to zero). The result will be a vertical asymptote at x = 5.5, with the y value approaching +infinity as x approaches 5.5 from the left. Just past x = 5.5 the 'blue' values become positive. Dividing a negative number by a positive number near zero results in a very large negative value, so that on this side of x = 5.5 the asymptote will rise up from -infinity. The quotient graph passes through the x axis near x = 6.3, where the 'black' graph is again zero. To the right of this point both graphs have positive values and the quotient graph will be positive. Around x = 7.5, where the 'blue' value is 1, the graph will coincide with the 'black' graph, giving us a point near (7.5, 1.3). Past this point the 'blue' value is greater than 1 so that the quotient graph will become nearer the x axis than the 'black' graph, increasingly so as x (and hence the 'blue' value) increases. This will result in a 'peak' of the quotient graph somewhere around x = 7.5, a bit to the left of the peak of the 'black' graph. **
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RESPONSE --> I see self critique assessment: 2 assignment #023 023. `query 23 Precalculus I 12-11-2008 assignment #024 024. `query 24 Precalculus I 12-12-2008
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09:05:41 Explain why, when either f(x) or g(x) is 0, then the product function also has a 0 for that value of x.
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RESPONSE --> When zero is multiplied with any number the answer is zero. confidence assessment: 2
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09:05:50 STUDENT RESPONSE: If you multiply any number by zero and you get zero.
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RESPONSE --> I got it. self critique assessment: 3
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09:06:33 Explain why, when the magnitude | f(x) | of f(x) is greater than 1 (i.e., when the graph of f(x) is more than one unit from the x axis), then the product function will be further from the x axis than the g(x) function.
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RESPONSE --> If you multiply by one you get the same number, thus any number greater than one will yield a higher number than the original one. confidence assessment: 2
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09:06:49 STUDENT RESPONSE: If you multiply a number by another number greater than 1, the result is greater than the original number. If you multiply a number by another number whose magnitude is greater than 1, the result will have greater magnitude that the original number. If | f(x) | > 1 then the magnitude of f(x) * g(x) will be greater than the magnitude of g(x). The magnitude of g(x) at a given value of x is its distance from the x axis, so when the magnitude increases so does the distance from the x axis.
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RESPONSE --> I understand this. self critique assessment: 2
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09:07:28 Explain why, when the magnitude | f(x) | of f(x) is less than 1 (i.e., when the graph of f(x) is less than one unit from the x axis), then the product function will be closer to the x axis than the g(x) function.
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RESPONSE --> If a number is multiplied by something less than one, then the number will be reduced. confidence assessment: 2
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09:07:39 STUDENT RESPONSE: If you multiply a number by another number less than 1, the result is less than the original number. If you multiply a number by another number whose magnitude is less than 1, the result will have a lesser magnitude that the original number. If | f(x) | < 1 then the magnitude of f(x) * g(x) will be less than the magnitude of g(x). The magnitude of g(x) at a given value of x is its distance from the x axis, so when the magnitude decreases so does the distance from the x axis.
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RESPONSE --> I get this. self critique assessment: 2
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09:10:40 Explain why, when f(x) and g(x) are either both positive or both negative, the product function is positive. When f(x) and g(x) have opposite signs the product function is negative.
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RESPONSE --> A negative times a negative is a positive and a positive times a positive is positive. THen two different signs would be negative. confidence assessment: 2
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09:10:51 STUDENT RESPONSE: This is basic multiplication: + * + = +, - * - = -, + * - = -. The product of like signs is positive, the product of unlike signs is negative. Since the product function results from multiplication of the two functions, these rules apply.
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RESPONSE --> I got it. self critique assessment: 2
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09:14:06 Explain why, when f(x) = 1, the graph of the product function coincides with the graph of g(x).
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RESPONSE --> because it is the same g(x)*1 is the same as g(x) confidence assessment: 2
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09:14:23 STUDENT RESPONSE: g(x) * 1 = g(x)
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RESPONSE --> I think I understand this stuff pretty good. self critique assessment: 2
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09:16:05 problem 4 Sketch graphs for y = f(x) = 2^x and y = g(x) = .5 x, for -2 < x < 2. Use your graphs to predict the shape of the y = g(x) * f(x) graph. Describe the graphs of the two functions, and explain how you used these graphs predict the shape of the graph of the product function.
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RESPONSE --> If g(x) is negative the graph is negaitve Where g is positive the graph will be positive. If the graph is positive it will rise steeper. confidence assessment: 2
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09:16:17 STUDENT RESPONSE: Where g(x) is - and f(x) is + graph will be -. where g(x) =0 graph will be at 0where both are + graph will be positive and rise more steeply. y=2^x asymptote negative x axis y intercept (0,1) y=.5x linear graph passing through (0,0) rising 1 unit for run of 2 units INSTRUCTOR COMMENT: It follows that since one function is negative for x < 0 while the other is always positive the product will be negative for x < 0, and since both functions are positive for x > 0 the product will be positive for x > 0. Since one function is zero at x=0 the product will be 0 at x = 0. For x > 0 the exponential rise of the one graph and the continuing rise of the other imply that the graph will rise more and more rapidly, without bound, for large positive x. For x < 0 one function is positive and the other is negative so the graph will be below the x axis. For large negative x, one graph approaches 0 while the other keeps increasing in magnitude; it's not immediatly clear which function 'wins'. However the exponential always 'beats' a fixed power so the graph will be asymptotic to the negative x axis. It will reach a minimum somewhere to the left of the x axis, before curving back toward the x axis and becoming asymptotic. **
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RESPONSE --> I think I get this. self critique assessment: 2
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assignment #025 025. `query 25 Precalculus I 12-12-2008
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09:18:50 Have you completed Test 2? If so give comments, insights, etc.. This is the only question on this Query.
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RESPONSE --> I have yet to complete test 2. confidence assessment: 2
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assignment #026 026. `query 26 Precalculus I 12-12-2008
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09:23:40 Give the rabbit populations for the first 12 months. Explain how each new population is obtained, and what your method for obtaining the new population has to do with the assumed nature of rabbit reproduction.
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RESPONSE --> 2 newborns after the first month. so they have to mature to add two more after another month and so you would have another pair of babies then it would go 3,5,and then 8, then 13,21,34 and so on... confidence assessment: 2
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09:23:59 ** Our assumptions are that rabbits require a month after birth to mature, mature rabbbits produce a pair of newborns every month starting the first month after they reach maturity, and rabbits never die. Every month the number of newborn pairs is equal to the number of mature pairs in the preceding month, which is equal to the total number of pairs from the month before that. Since all the rabbits from the preceding month are still present, the total number in the new month will be equal to the total number from the preceding month plus the number of newborns, which is equal to the total number from the preceding month plus the month before that. If we start with 1 pair of newborns then 1 month later we have a pair of mature rabbits, so after another month we have 2 pairs of rabbits. Our first three numbers are therefore 1, 1 and 2. In the next month we will have the 2 pairs we had in the preceding month plus 1 pair of newborns from the pair we had the month before that for a total of 3 pairs. In the next month we will have the 3 pairs we had in the preceding month plus 2 pairs of newborns from the pair we had the month before that for a total of 5 pairs. In the next month we will have the 5 pairs we had in the preceding month plus 3 pairs of newborns from the pair we had the month before that for a total of 8 pairs. The pattern continues 8 + 5 = 13 13 + 8 = 21 21 + 13 = 34 etc. . **
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RESPONSE --> I like these kind of questions better than earlier formulas. self critique assessment: 2
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09:24:24 Explain how the rule a(n) = a(n-1) + a(n-2) is related to the enumeration of the rabbit population.
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RESPONSE --> Just plug the numbers into the equations. confidence assessment: 2
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09:24:37 STUDENT ANSWER: a(n) is total number of rabbits for month(n) a(n-1) is total for previous month a(n-2) is total for 2 months previous which is total mature for 1 month previous INSTRUCTOR COMMENT: Good. In terms of the solution given for the preceding exercise: In the nth we will have the a(n-1) pairs we had in the preceding month plus a(n-2) pairs of newborns from the a(n-2) pairs we had the month before that for a total of a(n) = a(n-1) + a(n-2) pairs.
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RESPONSE --> That is what I was trying to say. confidence assessment: 2
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09:25:51 What are your ratios a(n) / a(n-1) for n = 1, 2, 3, 4, ..., 10, and what does your graph of ratio vs. n look like?
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RESPONSE --> My ratios are approx. 1.1,2.1,1.7,1.5,1.7,1.6,1.69,and so on. My graph has less activity as I go on. confidence assessment: 2
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09:26:01 ** The ratios are 1, 2, 1.5, 1.66, 1.6, 1.625, 1.615, 1.619, 1.6176 1.6181 1.61797 etc. The graph is jagged, up one time, down the next, but jumping less and less each time. If you were to make a horizontal line through two successive graph points, all subsequent points would be 'squeezed' between these lines. COMMON ERROR: The ratios are 1, 2, 1.5, 1.7, 1.6, 1.6, 1.6, ... . You rounded off before you could see the difference in the last 3 results. If you don't see a difference in the ratios you need to use more significant figures--carry your calculation out to enough decimal places that the differences show. **
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RESPONSE --> I see. self critique assessment: 2
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09:28:19 problem 8 Use the Fibonacci sequence data points for n = 5 and n = 10 to obtain an exponential model in the form y = a b^x.
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RESPONSE --> If you use the points 5,8 and 10,89 you plug it into a b^x to get the equations 8 = a b^5 89 = a b^10 then divide and substitute to get 8 = a times1.619^5 a = 8 /1.619^5 = aprox. .711 confidence assessment: 2
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09:28:31 ** You would use the points (5, 8) and (10,89) for your y = A b^x model. You would get the equations 8 = A b^5 89 = A b^10 Dividing the second equation by the first you would get 11.125 = b^5, so b = (11.125)^1/5 = 1.619. Substituting into the first equation we get 8 = A * (1.619)^5 A = 8 /(1.619)^5 = .711. So the model is y = .719 * 1.619^x. **
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RESPONSE --> I think I understand. self critique assessment: 2
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assignment #026 026. `query 26 Precalculus I 12-12-2008 ended
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