course previously submitted with old access code. ƾolڶZassignment #003
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10:14:42 `q001 A straight line connects the points (3, 5) and (7, 17), while another straight line continues on from (7, 17) to the point (10, 29). Which line is steeper and on what basis to you claim your result?
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RESPONSE --> The line with the points [3,5] and [7,17] is steeper. It has a slope of 1/3 as cpmpared to 1/4 for the second line. It has the same verticle rise over a shorter distance. confidence assessment: 3
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10:15:03 The point (3,5) has x coordinate 3 and y coordinate 5. The point (7, 17) has x coordinate 7 and y coordinate 17. To move from (3,5) to (7, 17) we must therefore move 4 units in the x direction and 12 units in the y direction. Thus between (3,5) and (7,17) the rise is 12 and the run is 4, so the rise/run ratio is 12/4 = 3. Between (7,10) and (10,29) the rise is also 12 but the run is only 3--same rise for less run, therefore more slope. The rise/run ratio here is 12/3 = 4.
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RESPONSE --> ok self critique assessment: 3
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10:19:54 `q002. The expression (x-2) * (2x+5) is zero when x = 2 and when x = -2.5. Without using a calculator verify this, and explain why these two values of x, and only these two values of x, can make the expression zero.
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RESPONSE --> When x is 2, 2-2=0 and 0 *(2x+5) = 0. when x = -2.5, 2x + 5=0 and (x-2) * 0=0 confidence assessment: 3
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10:20:13 If x = 2 then x-2 = 2 - 2 = 0, which makes the product (x -2) * (2x + 5) zero. If x = -2.5 then 2x + 5 = 2 (-2.5) + 5 = -5 + 5 = 0.which makes the product (x -2) * (2x + 5) zero. The only way to product (x-2)(2x+5) can be zero is if either (x -2) or (2x + 5) is zero. Note that (x-2)(2x+5) can be expanded using the Distributive Law to get x(2x+5) - 2(2x+5). Then again using the distributive law we get 2x^2 + 5x - 4x - 10 which simplifies to 2x^2 + x - 10. However this doesn't help us find the x values which make the expression zero. We are better off to look at the factored form.
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RESPONSE --> ok self critique assessment: 2
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10:27:17 `q003. For what x values will the expression (3x - 6) * (x + 4) * (x^2 - 4) be zero?
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RESPONSE --> x values will be [0,2,-4] 1. 3*2-6=0 and 0 times any number is 0. 2.-4+4=o and any number times 0 is 0. 3.-2^2=4 and 4-4=0; any number times 0 = 0. Since all the quantities are multiplied, any quantity with a valule of zero will make the valule of the entire expression 0. confidence assessment: 3
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10:27:43 In order for the expression to be zero we must have 3x-6 = 0 or x+4=0 or x^2-4=0. 3x-6 = 0 is rearranged to 3x = 6 then to x = 6 / 3 = 2. So when x=2, 3x-6 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero. x+4 = 0 gives us x = -4. So when x=-4, x+4 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero. x^2-4 = 0 is rearranged to x^2 = 4 which has solutions x = + - `sqrt(4) or + - 2. So when x=2 or when x = -2, x^2 - 4 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero. We therefore see that (3x - 6) * (x + 4) * (x^2 - 4) = 0 when x = 2, or -4, or -2. These are the only values of x which can yield zero.**
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RESPONSE --> ok self critique assessment: 3
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10:37:28 `q004. One straight line segment connects the points (3,5) and (7,9) while another connects the points (10,2) and (50,4). From each of the four points a line segment is drawn directly down to the x axis, forming two trapezoids. Which trapezoid has the greater area? Try to justify your answer with something more precise than, for example, 'from a sketch I can see that this one is much bigger so it must have the greater area'.
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RESPONSE --> The trapezoid formed from the points [10,2] and [50,4] should have the greater area because its base is 40 units long as compared to the first trapezoid, which has a base of 4 units of length. This when added to base 2 and multiplied by the height will gibe the second trapezoid a greater area. confidence assessment: 2
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10:38:33 Your sketch should show that while the first trapezoid averages a little more than double the altitude of the second, the second is clearly much more than twice as wide and hence has the greater area. To justify this a little more precisely, the first trapezoid, which runs from x = 3 to x = 7, is 4 units wide while the second runs from x = 10 and to x = 50 and hence has a width of 40 units. The altitudes of the first trapezoid are 5 and 9,so the average altitude of the first is 7. The average altitude of the second is the average of the altitudes 2 and 4, or 3. So the first trapezoid is over twice as high, on the average, as the first. However the second is 10 times as wide, so the second trapezoid must have the greater area. This is all the reasoning we need to answer the question. We could of course multiply average altitude by width for each trapezoid, obtaining area 7 * 4 = 28 for the first and 3 * 40 = 120 for the second. However if all we need to know is which trapezoid has a greater area, we need not bother with this step.
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RESPONSE --> self critique assessment: ok
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11:01:46 * `q005. Sketch graphs of y = x^2, y = 1/x and y = `sqrt(x) [note: `sqrt(x) means 'the square root of x'] for x > 0. We say that a graph increases if it gets higher as we move toward the right, and if a graph is increasing it has a positive slope. Explain which of the following descriptions is correct for each graph: As we move from left to right the graph increases as its slope increases. As we move from left to right the graph decreases as its slope increases. As we move from left to right the graph increases as its slope decreases. As we move from left to right the graph decreases as its slope decreases.
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RESPONSE --> As we move from left to right the graph increases as its slope increases applies to y=1/x As we move from left to right the graph decreases as its slope increases applies to y= 1/x. As we move from left to right the graph increases as its slope decreases applies to y=x^2 As we move from left to right the graph increases as its slope decreases applies to y=sqrt[x] self critique assessment: 1
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11:51:28 * `q008. At clock time t the velocity of a certain automobile is v = 3 t + 9. At velocity v its energy of motion is E = 800 v^2. What is the energy of the automobile at clock time t = 5?
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RESPONSE --> v=3[5]+9 v=24 E=800(24^2) E=460,800 self critique assessment: 2
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11:52:49 For t=5, v = 3 t + 9 = (3*5) + 9 = 24. Therefore E = 800 * 24^2 = 460800.
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RESPONSE --> ok self critique assessment: 3
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12:02:20 * `q009. Continuing the preceding problem, can you give an expression for E in terms of t?
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RESPONSE --> E= 800(3t+9)^2 self critique assessment: 2
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12:03:21 end program
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RESPONSE --> ok self critique assessment: 3
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߯Nڧ assignment #001 001. Areas qa areas volumes misc 06-06-2008
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21:35:11 `q001. There are 11 questions and 7 summary questions in this assignment. What is the area of a rectangle whose dimensions are 4 m by 3 meters.
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RESPONSE --> 12m^2 confidence assessment: 3
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21:35:54 A 4 m by 3 m rectangle can be divided into 3 rows of 4 squares, each 1 meter on a side. This makes 3 * 4 = 12 such squares. Each 1 meter square has an area of 1 square meter, or 1 m^2. The total area of the rectangle is therefore 12 square meters, or 12 m^2. The formula for the area of a rectangle is A = L * W, where L is the length and W the width of the rectangle. Applying this formula to the present problem we obtain area A = L * W = 4 m * 3 m = (4 * 3) ( m * m ) = 12 m^2. Note the use of the unit m, standing for meters, in the entire calculation. Note that m * m = m^2.
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RESPONSE --> ok self critique assessment: 3
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21:37:12 `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?
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RESPONSE --> 6m^2 confidence assessment: 3
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21:37:47 A right triangle can be joined along its hypotenuse with another identical right triangle to form a rectangle. In this case the rectangle would have dimensions 4.0 meters by 3.0 meters, and would be divided by any diagonal into two identical right triangles with legs of 4.0 meters and 3.0 meters. The rectangle will have area A = L * W = 4.0 m * 3.0 m = 12 m^2, as explained in the preceding problem. Each of the two right triangles, since they are identical, will therefore have half this area, or 1/2 * 12 m^2 = 6.0 m^2. The formula for the area of a right triangle with base b and altitude h is A = 1/2 * b * h.
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RESPONSE --> ok self critique assessment: 3
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21:40:42 `q003. What is the area of a parallelogram whose base is 5.0 meters and whose altitude is 2.0 meters?
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RESPONSE --> 10m^2 confidence assessment: 3
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21:41:02 A parallelogram is easily rearranged into a rectangle by 'cutting off' the protruding end, turning that portion upside down and joining it to the other end. Hopefully you are familiar with this construction. In any case the resulting rectangle has sides equal to the base and the altitude so its area is A = b * h. The present rectangle has area A = 5.0 m * 2.0 m = 10 m^2.
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RESPONSE --> ok self critique assessment: 3
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21:41:39 `q004. What is the area of a triangle whose base is 5.0 cm and whose altitude is 2.0 cm?
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RESPONSE --> 10m^2 confidence assessment: 3
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21:42:59 It is possible to join any triangle with an identical copy of itself to construct a parallelogram whose base and altitude are equal to the base and altitude of the triangle. The area of the parallelogram is A = b * h, so the area of each of the two identical triangles formed by 'cutting' the parallelogram about the approriate diagonal is A = 1/2 * b * h. The area of the present triangle is therefore A = 1/2 * 5.0 cm * 2.0 cm = 1/2 * 10 cm^2 = 5.0 cm^2.
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RESPONSE --> ok self critique assessment: 2
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21:52:41 `q005. What is the area of a trapezoid with a width of 4.0 km and average altitude of 5.0 km?
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RESPONSE --> 20m^2 confidence assessment: 2
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21:53:24 Any trapezoid can be reconstructed to form a rectangle whose width is equal to that of the trapezoid and whose altitude is equal to the average of the two altitudes of the trapezoid. The area of the rectangle, and therefore the trapezoid, is therefore A = base * average altitude. In the present case this area is A = 4.0 km * 5.0 km = 20 km^2.
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RESPONSE --> ok self critique assessment: 2
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21:57:57 `q006. What is the area of a trapezoid whose width is 4 cm in whose altitudes are 3.0 cm and 8.0 cm?
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RESPONSE --> A = {8+3}/2 *4 A = 22cm^2 confidence assessment: 1
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21:58:19 The area is equal to the product of the width and the average altitude. Average altitude is (3 cm + 8 cm) / 2 = 5.5 cm so the area of the trapezoid is A = 4 cm * 5.5 cm = 22 cm^2.
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RESPONSE --> ok self critique assessment: 2
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22:00:23 `q007. What is the area of a circle whose radius is 3.00 cm?
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RESPONSE --> A =3.14 *3^2 A=28.26cm^2 confidence assessment: 3
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22:00:41 The area of a circle is A = pi * r^2, where r is the radius. Thus A = pi * (3 cm)^2 = 9 pi cm^2. Note that the units are cm^2, since the cm unit is part r, which is squared. The expression 9 pi cm^2 is exact. Any decimal equivalent is an approximation. Using the 3-significant-figure approximation pi = 3.14 we find that the approximate area is A = 9 pi cm^2 = 9 * 3.14 cm^2 = 28.26 cm^2, which we round to 28.3 cm^2 to match the number of significant figures in the given radius. Be careful not to confuse the formula A = pi r^2, which gives area in square units, with the formula C = 2 pi r for the circumference. The latter gives a result which is in units of radius, rather than square units. Area is measured in square units; if you get an answer which is not in square units this tips you off to the fact that you've made an error somewhere.
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RESPONSE --> ok self critique assessment: 3
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22:02:16 `q008. What is the circumference of a circle whose radius is exactly 3 cm?
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RESPONSE --> C= 3.14 ( 3+ 3) C = 18.84 cm confidence assessment: 3
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22:02:36 The circumference of this circle is C = 2 pi r = 2 pi * 3 cm = 6 pi cm. This is the exact area. An approximation to 3 significant figures is 6 * 3.14 cm = 18.84 cm. Note that circumference is measured in the same units as radius, in this case cm, and not in cm^2. If your calculation gives you cm^2 then you know you've done something wrong.
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RESPONSE --> ok self critique assessment: 3
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22:04:34 `q009. What is the area of a circle whose diameter is exactly 12 meters?
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RESPONSE --> A= 3.14 (6^2) A = 113.04 m^2 confidence assessment: 3
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22:04:51 The area of a circle is A = pi r^2, where r is the radius. The radius of this circle is half the 12 m diameter, or 6 m. So the area is A = pi ( 6 m )^2 = 36 pi m^2. This result can be approximated to any desired accuracy by using a sufficient number of significant figures in our approximation of pi. For example using the 5-significant-figure approximation pi = 3.1416 we obtain A = 36 m^2 * 3.1416 = 113.09 m^2.
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RESPONSE --> ok self critique assessment: 3
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22:12:18 `q010. What is the area of a circle whose circumference is 14 `pi meters?
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RESPONSE --> A = 3.14 * (7^2) A = 153.9 pi meters confidence assessment: 2
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22:14:16 We know that A = pi r^2. We can find the area if we know the radius r. We therefore attempt to use the given information to find r. We know that circumference and radius are related by C = 2 pi r. Solving for r we obtain r = C / (2 pi). In this case we find that r = 14 pi m / (2 pi) = (14/2) * (pi/pi) m = 7 * 1 m = 7 m. We use this to find the area A = pi * (7 m)^2 = 49 pi m^2.
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RESPONSE --> ok self critique assessment: 2
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22:18:03 `q011. What is the radius of circle whose area is 78 square meters?
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RESPONSE --> 78 = 3.14 *d 24.8 = d d/2 = 12.4 = r confidence assessment: 3
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22:23:11 Knowing that A = pi r^2 we solve for r. We first divide both sides by pi to obtain A / pi = r^2. We then reverse the sides and take the square root of both sides, obtaining r = sqrt( A / pi ). Note that strictly speaking the solution to r^2 = A / pi is r = +-sqrt( A / pi ), meaning + sqrt( A / pi) or - sqrt(A / pi). However knowing that r and A are both positive quantities, we can reject the negative solution. Now we substitute A = 78 m^2 to obtain r = sqrt( 78 m^2 / pi) = sqrt(78 / pi) m.{} Approximating this quantity to 2 significant figures we obtain r = 5.0 m.
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RESPONSE --> ok self critique assessment: 2
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22:24:06 `q012. Summary Question 1: How do we visualize the area of a rectangle?
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RESPONSE --> length * width confidence assessment: 2
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22:24:23 We visualize the rectangle being covered by rows of 1-unit squares. We multiply the number of squares in a row by the number of rows. So the area is A = L * W.
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RESPONSE --> ok self critique assessment: 3
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22:25:44 `q013. Summary Question 2: How do we visualize the area of a right triangle?
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RESPONSE --> .5*( base * height) confidence assessment: 3
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22:26:03 We visualize two identical right triangles being joined along their common hypotenuse to form a rectangle whose length is equal to the base of the triangle and whose width is equal to the altitude of the triangle. The area of the rectangle is b * h, so the area of each triangle is 1/2 * b * h.
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RESPONSE --> ok self critique assessment: 3
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22:26:41 `q014. Summary Question 3: How do we calculate the area of a parallelogram?
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RESPONSE --> base *height confidence assessment: 3
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22:27:04 The area of a parallelogram is equal to the product of its base and its altitude. The altitude is measured perpendicular to the base.
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RESPONSE --> ok self critique assessment: 3
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22:29:12 `q015. Summary Question 4: How do we calculate the area of a trapezoid?
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RESPONSE --> altitude [ (base 1+ base 2]/ 2] confidence assessment: 3
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22:30:16 We think of the trapezoid being oriented so that its two parallel sides are vertical, and we multiply the average altitude by the width.
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RESPONSE --> ok self critique assessment: 2
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22:31:26 `q016. Summary Question 5: How do we calculate the area of a circle?
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RESPONSE --> area = 3.14 * radius^2 confidence assessment: 3
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22:31:41 We use the formula A = pi r^2, where r is the radius of the circle.
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RESPONSE --> ok self critique assessment: 3
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22:32:26 `q017. Summary Question 6: How do we calculate the circumference of a circle? How can we easily avoid confusing this formula with that for the area of the circle?
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RESPONSE --> c = 3.14 * d confidence assessment: 3
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22:32:43 We use the formula C = 2 pi r. The formula for the area involves r^2, which will give us squared units of the radius. Circumference is not measured in squared units.
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RESPONSE --> ok self critique assessment: 3
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22:33:44 `q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> I have taken notes manually in a notebook I use for this class. confidence assessment: 3
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