course mth 164 Complete Assignment 5, including Class Notes, text problems and Web-based problems as specified on the Assts page.
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10:01:09 Goals for this Assignment include but are not limited to the following: 1. Construct the basic triangles corresponding to angles pi/6, pi/4, pi/3, as appropriate, in order to find the sines and cosines of these angles. 2. Using exact values construct tables and graphs for the basic trigonometric functions. 3. Using exact values construct y vs. x graphs of y = A sin(theta) or y = A cos(theta), where theta is given as a function of x. 4 . Use the technique of reversing columns and restricting domain to construct tables and graphs for the arcsin and arccos functions. 5. Given an equation in which the argument of the sine or cosines function is a function of x, solve the equation for all values of x within a given interval. Click once more on Next Question/Answer for a note on Previous Assignments.
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RESPONSE --> ok self critique assessment:
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10:07:08 `q003. Each of the original triangles has hypotenuse 1 and vertical leg 1/2. By the Pythagorean Theorem, what therefore is the length of the horizontal side of either triangle?
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RESPONSE --> In the both triangles, one angle is 90 degrees, 30 degrees and 60 degrees. The larger triangle has angles 60 degrees, 60 degrees, and 60 degrees which makes it an equlateral triangle with all 3 sides and all 3 angles equal. The length of the vertical sides of the two orginial triangles is 1/2. Using the Pythagorean theorem the lenth of the orizontal side would be 1-(1/2)^2 = sqr(3/4) = sqr(3)/2 confidence assessment: 3
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10:07:16 By the Pythagorean Theorem c^2 = a^2 + b^2, where a and b are the legs and c the hypotenuse of a right triangle. Thus b^2 = c^2 - a^2 so that b = sqrt(c^2 - a^2). In the case of this triangle, if a = 1/2 and c = 1 we get b = sqrt(1^2 - (1/2)^2) = sqrt(1 - 1/4) = sqrt(3/4)= sqrt(3) / 2.
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RESPONSE --> ok self critique assessment: 3
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10:07:48 `q004. What therefore is the y coordinate of the point on the unit circle corresponding to angular position pi/6? What is the x coordinate?
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RESPONSE --> the y coordinate would be 1/2 and the x coordinate would be sqr(3)/2 confidence assessment: 3
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10:08:10 The triangle is depicted in Figure 65. The y coordinate is 1/2 and the s coordinate is sqrt(3) / 2. The unit-circle point corresponding to angular position pi/6 is (sqrt(3)/2, 1/2), as indicated in Figure &&. Note that the approximate coordinates, accurate to 2 significant figures, are the values we used previously for this angle so that the approximate coordinates are (.87, .5).
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RESPONSE --> ok self critique assessment: 3
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10:10:22 `q005. Construct the angle corresponding to the angular position pi/4. Starting at the origin draw a line out to the pi/4 position, then directly down to the x axis and finally back along the x axis to the origin; this will form a triangle. Draw another triangle, this time starting from the origin, go into the pi/4 point, then straight back to the y-axis along horizontal line, and finally down the y-axis to the origin. What are the three angles in each triangle? How do the lengths of the two legs of each triangle compare? When these two triangles are put together what shape do they form?
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RESPONSE --> The angles would be pi/4, pi/4 and pi/2 or 45, 45, 90 degrees. The lenghts of the two legs would be equal since they are both opposite equal angles. Together these two triangle form a rectangle. confidence assessment: 3
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10:10:31 The triangles are shown in Figure 46. Each triangle has a horizontal and a vertical side, so one of the angles formed is a right angle pi/2. The central angle in each is pi/4, so it follows that the remaining angle, in order that the three angles add up to pi, is also pi/4. These triangles, having two equal angles, are therefore both isosceles, with equal legs. The two triangles form a square, since the sides of the combined figure are all horizontal or vertical and since the equal legs form the sides. If the entire figure is flipped about the theta = pi/4 line there will be no change except that the x and y sides will be interchanged. The interchanged triangles will coincide, but the x and y sides of the triangles will be interchanged, showing that the x and y coordinates of the two triangles are the same.
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RESPONSE --> ok self critique assessment: 3
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10:11:24 `q006. If the circle has radius 1 it follows that the hypotenuse of either triangle is 1. Let s stand for the lengths of the (equal) legs of either triangle. By the Pythagorean Theorem what must therefore be the length of the side s?
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RESPONSE --> 1^2 = s^2 + s^2 1 = 2s^2 1/2 = s^2 sqr(1/2) = s srq(2)/2 = s confidence assessment: 3
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10:11:29 Figure 23 shows the first triangle with the legs labeled s and the hypotenuse 1. The Pythagorean Theorem tells us that s^2 + s^2 = 1, so that 2 s^2 = 1 and s^2 = 1/2. From this we get s = 1 / sqrt(2) = sqrt(2) / 2.
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RESPONSE --> ok self critique assessment: 3
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10:11:56 `q007. What therefore is the y coordinate of the point corresponding to angular position pi/4? What is the x coordinate?
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RESPONSE --> the y coordinate would be sqr(2)/2 and the x coordinate would also be sqr(2)/2 confidence assessment: 3
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10:12:01 The y coordinate is the same as the x coordinate, sqrt(2) / 2. The unit-circle point corresponding to angular position pi/4 is therefore (sqrt(2) / 2, sqrt(2) / 2), as indicated in Figure 83. Note that the approximate coordinates, accurate to 2 significant figures, are the values we used previously for this angle so that the approximate coordinates are (.71, .71).
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RESPONSE --> ok self critique assessment: 3
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10:13:46 `q008. Sketch the triangle corresponding to the angular position pi/3. Explain why this triangle is similar to the triangle you constructed for angular position pi/6. What therefore are the x and y coordinates of the theta = pi/3 point on the unit circle?
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RESPONSE --> The triangles are similar because the interior angles are the same, ie. pi/3, pi/6 and pi/2 The x coordinate in this case would be 1/2 and the ycoordinate would be sqr(3)/2 which is reversed of the previous triangle. confidence assessment: 3
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10:13:50 The triangle is shown in Figure 88. The horizontal and vertical sides are reversed from the pi/6 triangle. So the coordinates of the points are also reversed to give us (1/2, sqrt(3)/2).
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RESPONSE --> ok self critique assessment: 3
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10:17:52 `q009. Give the x and y coordinates of the points on the unit circle corresponding to angles which are multiples of pi/6. Sketch the unit circle and all angles which are multiples of pi/6, in give their coordinates.
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RESPONSE --> angle x coordinate y coordinate pi/6 sqr(3)/2 1/2 pi/3 1/2 sqr(3)/2 pi/2 0 1 2pi/3 -1/2 sqr(3)/2 5pi/6 -sqr(3)/2 1/2 pi -1 0 7pi/6 -sqr(3)/2 -1/2 4pi/3 -1/2 -sqr(3)/2 3pi/2 0 -1 5pi/3 1/2 -sqr(3)/2 11pi/6 sqr(3)/2 -1/2 2pi 1 0 confidence assessment: 3
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10:18:00 Figure 50 shows the angles 0, pi/6, pi/3, pi/2, 3 pi/2, 5 pi/6, ..., 2 pi with accurate values for the coordinates of the points.
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RESPONSE --> ok self critique assessment: 3
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10:20:45 `q010. Give the x and y coordinates of the points on the unit circle corresponding to angles which are multiples of pi/4. Sketch the unit circle and all angles which are multiples of pi/6, in give their coordinates.
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RESPONSE --> angle x-coordinate y-coordinate 0 1 0 pi/4 sqr(2)/2 sqr(2)/2 pi/2 0 1 3pi/4 -sqr(2)/2 sqr(2)/2 pi -1 0 5pi/4 -sqr(2)/2 -sqr(2)/2 3pi/2 0 -1 7pi/4 sqr(2)/2 -sqr(2)/2 2pi 1 0 confidence assessment: 3
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10:20:54 Figure 28 shows the angles 0, pi/4, pi/2, 3 pi/4, 2 pi, 5 pi/4, ..., 2 pi with accurate values for the coordinates of the points.
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RESPONSE --> ok self critique assessment: 3
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10:26:04 `q011. Make the table for sin(theta) vs. theta with 0 <= theta < 2 pi, using an increment of pi/4. Sketch the corresponding graph. Now reverse the columns of the table and sketch the associated graph. Does your first graph represent a function? Does your second graph represent a function?
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RESPONSE --> The first graph represents a function and looks like the sine curve. The second graph is not a function because it does not pass the vertical line test confidence assessment: 2
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10:26:23 Using .71 to stand for sqrt(2) / 2 (your table should use sqrt(2)/2 where this one uses .71) we have the following table: theta sin(theta) 0 0.0 pi/4 0.71 pi/2 1.0 3 pi/4 0.71 pi 0.0 5 pi/4 -0.71 3 pi/2 -1.0 7 pi/4 -0.71 2 pi 0.0. The reversed table is sin(theta) theta 0.0 0 0.71 pi/4 1.0 pi/2 0.71 3 pi/4 0.0 pi -0.71 5 pi/4 -1.0 3 pi/2 -0.71 7 pi/4 0.0 2 pi. The corresponding graphs are indicated in figure 68. The red dots indicate points on the table; the rest of the function is filled in from what we know about the shape of the graph of y = sin(x). The graph of the sine function is the usual graph. The graph of the reversed column doesn't represent a function because for example the horizontal coordinate zero is associated with two different values, 0 and pi (and also 2 pi). Another way of saying this is that the function fails the vertical line test, which says that a graph can represent a function only if any vertical line intersecting the graph of the function intersects the graph in only 1 point. Still another way of saying the same thing is that any function must be single-valued, with any number you put into the function giving you exactly 1 value. For most numbers the first column of the inverted table we get 2 values. The numbers in the first column usually appear twice. For example sqrt(2)/2 in the first column is listed next to pi/4 and also next to 3 pi/4, 0 is listed with 0 and also with pi, etc..
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RESPONSE --> ok self critique assessment: 3
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10:27:49 `q012. Now restrict the original table of sin(theta) vs. theta to values of theta for which pi/2 <= theta <= 3 pi/2, then reverse columns of the table. Does the resulting table give you a function? Sketch the graph corresponding to the table.
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RESPONSE --> Yes it does, it passes the vertical line test confidence assessment: 3
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10:28:02 The table is as follows theta sin(theta) pi/2 1.0 3 pi/4 0.71 pi 0.0 5 pi/4 -0.71 3 pi/2 -1.0 The inverted table is as follows: sin(theta) theta 1.0 pi/2 0.71 3 pi/4 1.0 pi -0.71 5 pi/4 -1.0 3 pi/2 Each number in the first column appears exactly once in the table for the inverted function. The corresponding graphs are shown in figure 14. We see that the sine-function graph passes a horizontal line test, so that the inverted function passes a vertical-line test. Any vertical line passing through the inverted graph passes through at exactly point, showing that the graph does define a function.
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RESPONSE --> ok self critique assessment: 3
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10:28:47 `q013. If we make a table of sin(theta) vs. theta for -pi/2 <= theta <= pi/2, can this table be inverted to give us a function?
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RESPONSE --> Yes, again it passes the vertical line test confidence assessment: 3
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10:28:57 The inverted table is as follows: The table for the sine function is as follows theta sin(theta) -pi/2 0 -pi/4 -0.71 0 0 pi/4 0.71 pi/2 1.0 The inverted table is as follows: sin(theta) theta -1.0 -pi/2 -.71 -pi/4 0.0 0 0.71 pi/4 1.0 pi/2 Each number in the first column appears exactly once. The corresponding graph is shown in figure 34. Any vertical line passing through the graph passes through at exactly point, showing that the graph does define a function.
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RESPONSE --> ok self critique assessment: 3
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10:30:35 `q014. Either of the functions graphed in the previous two exercises constitute inverses of portions of the sine function. Re-label the last table you made (the inverted table), labeling the first column x and the second column arcsin(x). This function is the generally accepted inverse of the sine function. What are the values of arcsine (.71), arcsine (0) and arcsin(-1)?
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RESPONSE --> arcsin(.71) = pi/4 arcsin(0) = 0 arcsin(-1) = -pi/2 confidence assessment: 3
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10:30:42 The table is as follows: sin(theta) theta -1.0 -pi/2 -.71 -pi/4 0.0 0 0.71 pi/4 1.0 pi/2 Arcsine(.71) is the number in the second column of the table across from .71 in the first column. This number is pi/4. Arcsine(0) is the number in the second column of the table across from 0 in the first column. This number is 0. Arcsine(-1) is the number in the second column of the table across from 1 in the first column. This number is -pi/2.
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RESPONSE --> ok self critique assessment: 3
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10:32:52 `q015. What are the solutions to the following equations: arcsin(x) = pi/4; arcsin(x) = -pi/3; sin(x) = 1.
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RESPONSE --> arcsin(x) = pi/4 x = sqr(2)/2 arcsin(x) = -pi/3 x = -sqr(3)/2 sin(x) = 1 x = pi/2 confidence assessment: 3
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10:32:59 If arcsin(x) = pi/4 then x appears next to pi/4 on the table of sin(x) vs. x. In other words sin(pi/4) = x. Since sin(pi/4) = sqrt(2)/2 we have x = sqrt(2)/2. }{Another way of putting this is to say arcsin(sqrt(2)/2) = pi/4. If arcsin(x) = -pi/3, similarly, this means that x = sin(-pi/3) so that x = -1/2.
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RESPONSE --> ok self critique assessment: 3
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10:37:03 `q016. Find solutions to the following equations: sin(2x) = .87; sin(3x/4 + pi) = -.5?
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RESPONSE --> 2x = arcsin(.87) 2x = pi/3 or 2x = 2pi/3 x = pi/6 or x = pi/3 3x/4 + pi = arcsin(-.5) 3x/4 + pi = 7pi/6 or 3x/4 + pi = 11pi/6 3x/4 = pi/6 3x/4 = 5pi/6 x = 2pi/9 x = 10pi/9 confidence assessment: 3
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10:37:45 Starting with sin(2x) = .87, apply the arcsin function to both sides of the equation. We get the equation arcsin ( sin(2x) ) = arcsin(.87). The arcsin and the sine function are inverse functions; one 'undoes' the other so that arcsin(sin(2x)) = 2x. So we obtain 2x = arcsin(.87). We know from the table of the sine function that arcsin(.87) = pi/3 so we have 2x = pi/3. We divide both sides by 2 to obtain x = pi/6. Using a similar strategy for the second equation we write arcsin( sin(3x/4) ) = arcsin(-.5), which gives us 3x/4 = -pi/6 so that x = 4/3(-pi/6) = -2 pi / 9.
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RESPONSE --> ok must restrict the domain so that arcsin is a function. I was using x between 0 and 2pi self critique assessment: 2
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10:40:08 `q017. Find solutions to the following equations: sin(3x) = .38; sin(2x+pi/4) = .93.
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RESPONSE --> 3x = arcsin(.38) 3x = .39 x = .13 2x+pi/4 = arcsin(.93) 2x + pi/4 = 1.19 2x = .405 x = .2025 confidence assessment: 3
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10:40:20 If sin(3x) = .38 then we have arcsin( sin(3x) ) = arcsin(.38). .38 doesn't appear on any of our tables so we have to find arcsin(.38) using a calculator. The arcsin key is sometimes labelled arcsin, sometimes sin^-1, and is often the 2d function on the sine key. We find that arcsin(.38) is about .39. So we have 3x = .39, so that []x = .39 / 3 = .13. If sin(2x+pi/4) = .93 then arcsin(2x + pi/4) = arcsin(.93). Using the calculator to find arcsin(.93) we get 2x + pi/4 = 1.19, approx.. We easily solve for x to obtain x = (1.19 - pi/4) / 2 = .2 approx.
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RESPONSE --> ok self critique assessment: 3
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10:45:47 `q018. For what values of t do we have sin(t + pi/3) equal to .5? Give the values for which t + pi/3 lies between 0 and 2 pi. Are there values of t for which t + pi/3 lies outside the interval from 0 to 2 pi?
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RESPONSE --> t+pi/3=arcsin(.5) t+pi/3=pi/6 t=-pi/6 but since t is between 0 and 2pi this value for t is invalid t+pi/3=5pi/6 t=pi/2 t+pi/3=13pi/6 t=11pi/6 t+pi/3=17pi/6 t=5pi/2 but since t is between 0 and 2pi this value for t is invalid confidence assessment: 2
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10:46:09 sin(t+pi/3) = .5 means that sin(theta) = .5, with theta = t + pi/3. The reference-circle picture of sin(theta) = .5 is given in Figure 60. We see that this occurs at the angle arcsin(.5) = pi/6 as well as at 5 pi/6. So sin(t + pi/3) = .5 for t + pi/3 = pi/6 and for t + pi/3 = 5 pi/6. We can solve each of these equations for t: t + pi/3 = pi/6 is solved by adding -pi/3 to both sides to obtain t = - pi/6. t + pi/3 = 5 pi/6 is similarly solved to get t = pi. Any angle which is coterminal with either theta = pi/6 or with 5 pi/6 (e.g., pi/6 + 2 pi, pi /6 + 4 pi, pi/6 + 6 pi, etc. or 5 pi/6 + 2 pi, 5 pi/6 + 4 pi, etc. ) will yield the same y coordinate and therefore the same value of the sine.
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RESPONSE --> ok self critique assessment: 3
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10:49:51 `q019. What are the first four positive values of t for which sin(t + pi/3) = .5?
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RESPONSE --> Using the procedure from the previous problem t = pi/2, 11pi/6, 5pi/2, 23pi/6 confidence assessment: 2
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10:50:15 As seen in the preceding problem sin(t + pi/3) = .5 when t + pi/3 = pi/6, 5 pi/6, pi/6 + 2 pi, 5 pi/6 + 2 pi, pi/6 + 4 pi 5 pi/6 + 4 pi, etc.. We also saw that the first two solutions, corresponding to theta = pi/6 and theta = 5 pi/6, are t = -pi/6 and t = pi. Only the second of these solutions is positive. The solutions corresponding to theta = t + pi/3 = pi/6 + 2 pi, 5 pi/6 + 2 pi, pi/6 + 4 pi, 5 pi/6 + 4 pi, etc., will also be positive. Setting t + pi/3 equal to each of these angles and solving for t 2e obtain solutions t = 11 pi / 6, t = 5 pi/2, t = 25 pi/6, t = 9 pi/2. The first four positive solutions are t = pi/2, 11 pi/6, 5 pi/2, 25 pi/6.
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RESPONSE --> ok self critique assessment: 3
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