course For whatever reason the send file copied assignments to the same file. This is the second time I've done Assignment 3 (volumes) and I don't know what I could have done to have both assignments end up on the same send file.Any comments or suggestions? }Toů]cassignment #001
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10:35:55 `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 --> 4*3=12m^2 confidence assessment: 3
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10:36:18 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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10:37:22 `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?
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RESPONSE --> .5*3*4=6m^ confidence assessment: 3
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10:37:37 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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10:38:22 `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 --> a= 5 * 2 a = 10m^2 confidence assessment: 3
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10:38:30 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 --> self critique assessment: ok
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10:39:35 `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 --> .5 *5 *2=5cm^2 confidence assessment: 3
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10:39:51 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: 3
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10:40:41 `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 --> 5 *4 = 20km^2 confidence assessment: 3
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10:40:59 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: 3
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10:41:51 `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 --> 4 * 5.5 = 22 cm^2 confidence assessment: 3
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10:42:08 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: 3
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10:43:13 `q007. What is the area of a circle whose radius is 3.00 cm?
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RESPONSE --> 3.14*[3^2]=28.26 cm^2 confidence assessment: 3
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10:43:22 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 --> self critique assessment: 3
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10:44:19 `q008. What is the circumference of a circle whose radius is exactly 3 cm?
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RESPONSE --> 3.14*6=18.84 cm confidence assessment: 3
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10:44:39 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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10:45:43 `q009. What is the area of a circle whose diameter is exactly 12 meters?
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RESPONSE --> 3.14*6^2=11.04 m^2 confidence assessment: 3
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10:46:08 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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10:47:28 `q010. What is the area of a circle whose circumference is 14 `pi meters?
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RESPONSE --> 3.14*7^2 =153.9 pi meters^2 confidence assessment: 3
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10:47:41 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: 3
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10:51:01 `q011. What is the radius of circle whose area is 78 square meters?
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RESPONSE --> 78 =3.14[ *r^2] 4.98= r confidence assessment: 3
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10:51:17 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: 3
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10:51:46 `q012. Summary Question 1: How do we visualize the area of a rectangle?
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RESPONSE --> length * width confidence assessment: 3
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10:51:58 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: 6
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10:52:34 `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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10:52:46 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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10:53:06 `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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10:53:23 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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10:54:07 `q015. Summary Question 4: How do we calculate the area of a trapezoid?
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RESPONSE --> altitude * width confidence assessment: 3
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10:54:23 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: 3
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10:55:06 `q016. Summary Question 5: How do we calculate the area of a circle?
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RESPONSE --> pi *r^2 confidence assessment: 3
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10:55:22 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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10:56:45 `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 = pi *diameter know the difference between a diameter and a radius confidence assessment: 3
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10:57:03 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: 2
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10:58:09 `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 the formulas copied in my notes that I keep at home. confidence assessment: 3
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֦ڼvyiȀvߩл assignment #002 002. Volumes qa areas volumes misc 06-07-2008
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11:57:41 `q001. There are 9 questions and 4 summary questions in this assignment. What is the volume of a rectangular solid whose dimensions are exactly 3 cm by 5 cm by 7 cm?
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RESPONSE --> 3cm*5cm*7cm=105cm^3 confidence assessment: 3
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11:58:14 If we orient this object so that its 3 cm dimension is its 'height', then it will be 'resting' on a rectangular base whose dimension are 5 cm by 7 cm. This base can be divided into 5 rows each consisting of 7 squares, each 1 meter by 1 meter. There will therefore be 5 * 7 = 35 such squares, showing us that the area of the base is 35 m^2. Above each of these base squares the object rises to a distance of 3 meters, forming a small rectangular tower. Each such tower can be divided into 3 cubical blocks, each having dimension 1 meter by 1 meter by 1 meter. The volume of each 1-meter cube is 1 m * 1 m * 1 m = 1 m^3, also expressed as 1 cubic meter. So each small 'tower' has volume 3 m^3. The object can be divided into 35 such 'towers'. So the total volume is 35 * 3 m^3 = 105 m^3. This construction shows us why the volume of a rectangular solid is equal to the area of the base (in this example the 35 m^2 of the base) and the altitude (in this case 3 meters). The volume of any rectangular solid is therefore V = A * h, where A is the area of the base and h the altitude. This is sometimes expressed as V = L * W * h, where L and W are the length and width of the base. However the relationship V = A * h applies to a much broader class of objects than just rectangular solids, and V = A * h is a more powerful idea than V = L * W * h. Remember both, but remember also that V = A * h is the more important.
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RESPONSE --> ok self critique assessment: 3
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11:59:38 `q002. What is the volume of a rectangular solid whose base area is 48 square meters and whose altitude is 2 meters?
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RESPONSE --> 48m^2 *2m = 96m^3 confidence assessment: 3
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11:59:56 Using the idea that V = A * h we find that the volume of this solid is V = A * h = 48 m^2 * 2 m = 96 m^3. Note that m * m^2 means m * (m * m) = m * m * m = m^2.
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RESPONSE --> ok self critique assessment: 3
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12:05:19 `q003. What is the volume of a uniform cylinder whose base area is 20 square meters and whose altitude is 40 meters?
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RESPONSE --> v= 3.14( 2.5^2 *40) v= 785 m^3 confidence assessment: 2
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12:07:42 V = A * h applies to uniform cylinders as well as to rectangular solids. We are given the altitude h and the base area A so we conclude that V = A * h = 20 m^2 * 40 m = 800 m^3. The relationship V = A * h applies to any solid object whose cross-sectional area A is constant. This is the case for uniform cylinders and uniform prisms.
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RESPONSE --> I calculated the radius to use in the formula: A= pi *r^2 *height. self critique assessment: 2
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12:12:11 `q004. What is the volume of a uniform cylinder whose base has radius 5 cm and whose altitude is 30 cm?
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RESPONSE --> area= 3.14 *7^2 a = 153.86 volume = 30 *153.86 volume = 4615.8 cm^3 confidence assessment: 2
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12:16:51 The cylinder is uniform, which means that its cross-sectional area is constant. So the relationship V = A * h applies. The cross-sectional area A is the area of a circle of radius 5 cm, so we see that A = pi r^2 = pi ( 5 cm)^2 = 25 pi cm^2. Since the altitude is 30 cm the volume is therefore V = A * h = 25 pi cm^2 * 30 cm = 750 pi cm^3. Note that the common formula for the volume of a uniform cylinder is V = pi r^2 h. However this is just an instance of the formula V = A * h, since the cross-sectional area A of the uniform cylinder is pi r^2. Rather than having to carry around the formula V = pi r^2 h, it's more efficient to remember V = A * h and to apply the well-known formula A = pi r^2 for the area of a circle.
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RESPONSE --> If the radius of the circle is 7 how can the area be calculated using a 5? self critique assessment: 2
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12:19:23 `q005. Estimate the dimensions of a metal can containing food. What is its volume, as indicated by your estimates?
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RESPONSE --> 500 in^3 confidence assessment: 3
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12:21:04 People will commonly estimate the dimensions of a can of food in centimeters or in inches, though other units of measure are possible (e.g., millimeters, feet, meters, miles, km). Different cans have different dimensions, and your estimate will depend a lot on what can you are using. A typical can might have a circular cross-section with diameter 3 inches and altitude 5 inches. This can would have volume V = A * h, where A is the area of the cross-section. The diameter of the cross-section is 3 inches so its radius will be 3/2 in.. The cross-sectional area is therefore A = pi r^2 = pi * (3/2 in)^2 = 9 pi / 4 in^2 and its volume is V = A * h = (9 pi / 4) in^2 * 5 in = 45 pi / 4 in^3. Approximating, this comes out to around 35 in^3. Another can around the same size might have diameter 8 cm and height 14 cm, giving it cross-sectional area A = pi ( 4 cm)^2 = 16 pi cm^2 and volume V = A * h = 16 pi cm^2 * 14 cm = 224 pi cm^2.
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RESPONSE --> My mistake was using the diameter and not the radius. self critique assessment: 2
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12:23:31 `q006. What is the volume of a pyramid whose base area is 50 square cm and whose altitude is 60 cm?
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RESPONSE --> confidence assessment:
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12:23:55 We can't use the V = A * h idea for a pyramid because the thing doesn't have a constant cross-sectional area--from base to apex the cross-sections get smaller and smaller. It turns out that there is a way to cut up and reassemble a pyramid to show that its volume is exactly 1/3 that of a rectangular solid with base area A and altitude h. Think of putting the pyramid in a box having the same altitude as the pyramid, with the base of the pyramid just covering the bottom of the box. The apex (the point) of the pyramid will just touch the top of the box. The pyramid occupies exactly 1/3 the volume of that box. So the volume of the pyramid is V = 1/3 * A * h. The base area A is 30 cm^2 and the altitude is 60 cm so we have V = 1/3 * 50 cm^2 * 60 cm = 1000 cm^3.
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RESPONSE --> ok self critique assessment: 1
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12:26:36 `q007. What is the volume of a cone whose base area is 20 square meters and whose altitude is 9 meters?
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RESPONSE --> 1/3 * 20 * 9 = 60m^3 confidence assessment: 2
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12:26:50 Just as the volume of a pyramid is 1/3 the volume of the 'box' that contains it, the volume of a cone is 1/3 the volume of the cylinder that contains it. Specifically, the cylinder that contains the cone has the base of the cone as its base and matches the altitude of the cone. So the volume of the cone is 1/3 A * h, where A is the area of the base and h is the altitude of the cone. In this case the base area and altitude are given, so the volume of the cone is V = 1/3 A * h = 1/3 * 20 m^2 * 9 m = 60 m^3.
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RESPONSE --> ok self critique assessment: 3
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12:32:59 `q008. What is a volume of a sphere whose radius is 4 meters?
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RESPONSE --> v = 4/3 3.14 *4^3 v = 268 m^3 confidence assessment: 2
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12:45:44 `q009. What is the volume of a planet whose diameter is 14,000 km?
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RESPONSE --> v= 4/3 * 3.14 *700^3 v = 1436026667km^3 confidence assessment: 3
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12:46:08 The planet is presumably a sphere, so to the extent that this is so the volume of this planet is V = 4/3 pi r^3, where r is the radius of the planet. The diameter of the planet is 14,000 km so the radius is half this, or 7,000 km. It follows that the volume of the planet is V = 4/3 pi r^3 = 4/3 pi * (7,000 km)^3 = 4/3 pi * 343,000,000,000 km^3 = 1,372,000,000,000 / 3 * pi km^3. This result can be approximated to an appropriate number of significant figures.
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RESPONSE --> ok self critique assessment: 3
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12:47:24 `q010. Summary Question 1: What basic principle do we apply to find the volume of a uniform cylinder of known dimensions?
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RESPONSE --> pi *r^2 * h confidence assessment: 3
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12:47:55 The principle is that when the cross-section of an object is constant, its volume is V = A * h, where A is the cross-sectional area and h the altitude. Altitude is measure perpendicular to the cross-section.
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RESPONSE --> ok self critique assessment: 3
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12:48:54 `q011. Summary Question 2: What basic principle do we apply to find the volume of a pyramid or a cone?
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RESPONSE --> 1/3 *area of the base * height confidence assessment: 3
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12:49:09 The volumes of these solids are each 1/3 the volume of the enclosing figure. Each volume can be expressed as V = 1/3 A * h, where A is the area of the base and h the altitude as measured perpendicular to the base.
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RESPONSE --> ok self critique assessment: 3
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12:50:37 `q012. Summary Question 3: What is the formula for the volume of a sphere?
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RESPONSE --> V =4/3 * pi * r^3 confidence assessment: 3
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^]{ϤO봼ڨ assignment #002 002. Volumes qa areas volumes misc 06-07-2008
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13:18:14 `q001. There are 9 questions and 4 summary questions in this assignment. What is the volume of a rectangular solid whose dimensions are exactly 3 cm by 5 cm by 7 cm?
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RESPONSE --> 3*5*7=105cm^3 confidence assessment: 3
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13:18:33 If we orient this object so that its 3 cm dimension is its 'height', then it will be 'resting' on a rectangular base whose dimension are 5 cm by 7 cm. This base can be divided into 5 rows each consisting of 7 squares, each 1 meter by 1 meter. There will therefore be 5 * 7 = 35 such squares, showing us that the area of the base is 35 m^2. Above each of these base squares the object rises to a distance of 3 meters, forming a small rectangular tower. Each such tower can be divided into 3 cubical blocks, each having dimension 1 meter by 1 meter by 1 meter. The volume of each 1-meter cube is 1 m * 1 m * 1 m = 1 m^3, also expressed as 1 cubic meter. So each small 'tower' has volume 3 m^3. The object can be divided into 35 such 'towers'. So the total volume is 35 * 3 m^3 = 105 m^3. This construction shows us why the volume of a rectangular solid is equal to the area of the base (in this example the 35 m^2 of the base) and the altitude (in this case 3 meters). The volume of any rectangular solid is therefore V = A * h, where A is the area of the base and h the altitude. This is sometimes expressed as V = L * W * h, where L and W are the length and width of the base. However the relationship V = A * h applies to a much broader class of objects than just rectangular solids, and V = A * h is a more powerful idea than V = L * W * h. Remember both, but remember also that V = A * h is the more important.
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RESPONSE --> ok self critique assessment: 3
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13:19:19 `q002. What is the volume of a rectangular solid whose base area is 48 square meters and whose altitude is 2 meters?
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RESPONSE --> 2*48=96cm^3 confidence assessment: 3
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13:19:33 Using the idea that V = A * h we find that the volume of this solid is V = A * h = 48 m^2 * 2 m = 96 m^3. Note that m * m^2 means m * (m * m) = m * m * m = m^2.
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RESPONSE --> ok self critique assessment: 3
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13:20:38 `q003. What is the volume of a uniform cylinder whose base area is 20 square meters and whose altitude is 40 meters?
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RESPONSE --> 20*40=800m^3 confidence assessment: 3
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13:20:51 V = A * h applies to uniform cylinders as well as to rectangular solids. We are given the altitude h and the base area A so we conclude that V = A * h = 20 m^2 * 40 m = 800 m^3. The relationship V = A * h applies to any solid object whose cross-sectional area A is constant. This is the case for uniform cylinders and uniform prisms.
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RESPONSE --> ok self critique assessment: 3
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13:23:01 `q004. What is the volume of a uniform cylinder whose base has radius 5 cm and whose altitude is 30 cm?
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RESPONSE --> 3.14*25=78.5cm^3 confidence assessment: 3
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13:23:44 The cylinder is uniform, which means that its cross-sectional area is constant. So the relationship V = A * h applies. The cross-sectional area A is the area of a circle of radius 5 cm, so we see that A = pi r^2 = pi ( 5 cm)^2 = 25 pi cm^2. Since the altitude is 30 cm the volume is therefore V = A * h = 25 pi cm^2 * 30 cm = 750 pi cm^3. Note that the common formula for the volume of a uniform cylinder is V = pi r^2 h. However this is just an instance of the formula V = A * h, since the cross-sectional area A of the uniform cylinder is pi r^2. Rather than having to carry around the formula V = pi r^2 h, it's more efficient to remember V = A * h and to apply the well-known formula A = pi r^2 for the area of a circle.
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RESPONSE --> I did not mtiply the heght. self critique assessment: 2
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13:25:20 `q005. Estimate the dimensions of a metal can containing food. What is its volume, as indicated by your estimates?
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RESPONSE --> 130 in^3 confidence assessment: 3
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13:25:44 People will commonly estimate the dimensions of a can of food in centimeters or in inches, though other units of measure are possible (e.g., millimeters, feet, meters, miles, km). Different cans have different dimensions, and your estimate will depend a lot on what can you are using. A typical can might have a circular cross-section with diameter 3 inches and altitude 5 inches. This can would have volume V = A * h, where A is the area of the cross-section. The diameter of the cross-section is 3 inches so its radius will be 3/2 in.. The cross-sectional area is therefore A = pi r^2 = pi * (3/2 in)^2 = 9 pi / 4 in^2 and its volume is V = A * h = (9 pi / 4) in^2 * 5 in = 45 pi / 4 in^3. Approximating, this comes out to around 35 in^3. Another can around the same size might have diameter 8 cm and height 14 cm, giving it cross-sectional area A = pi ( 4 cm)^2 = 16 pi cm^2 and volume V = A * h = 16 pi cm^2 * 14 cm = 224 pi cm^2.
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RESPONSE --> ok self critique assessment: 3
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13:26:45 `q006. What is the volume of a pyramid whose base area is 50 square cm and whose altitude is 60 cm?
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RESPONSE --> 1/3 *50 *60 = 1000cm^3 confidence assessment: 3
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13:26:59 We can't use the V = A * h idea for a pyramid because the thing doesn't have a constant cross-sectional area--from base to apex the cross-sections get smaller and smaller. It turns out that there is a way to cut up and reassemble a pyramid to show that its volume is exactly 1/3 that of a rectangular solid with base area A and altitude h. Think of putting the pyramid in a box having the same altitude as the pyramid, with the base of the pyramid just covering the bottom of the box. The apex (the point) of the pyramid will just touch the top of the box. The pyramid occupies exactly 1/3 the volume of that box. So the volume of the pyramid is V = 1/3 * A * h. The base area A is 30 cm^2 and the altitude is 60 cm so we have V = 1/3 * 50 cm^2 * 60 cm = 1000 cm^3.
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RESPONSE --> ok self critique assessment: 3
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13:27:49 `q007. What is the volume of a cone whose base area is 20 square meters and whose altitude is 9 meters?
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RESPONSE --> 1/3*20*9=60m^3 confidence assessment: 3
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13:28:01 Just as the volume of a pyramid is 1/3 the volume of the 'box' that contains it, the volume of a cone is 1/3 the volume of the cylinder that contains it. Specifically, the cylinder that contains the cone has the base of the cone as its base and matches the altitude of the cone. So the volume of the cone is 1/3 A * h, where A is the area of the base and h is the altitude of the cone. In this case the base area and altitude are given, so the volume of the cone is V = 1/3 A * h = 1/3 * 20 m^2 * 9 m = 60 m^3.
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RESPONSE --> ok self critique assessment: 3
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13:29:56 `q008. What is a volume of a sphere whose radius is 4 meters?
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RESPONSE --> 4/3 *[p *4^3 = 268m^3 confidence assessment: 3
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13:30:11 The volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere. In this case r = 4 m so V = 4/3 pi * (4 m)^3 = 4/3 pi * 4^3 m^3 = 256/3 pi m^3.
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RESPONSE --> ok self critique assessment: 3
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13:31:22 `q009. What is the volume of a planet whose diameter is 14,000 km?
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RESPONSE --> 4/3 *3.14 *700^3= 1436026667 km^3 confidence assessment: 3
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13:31:52 The planet is presumably a sphere, so to the extent that this is so the volume of this planet is V = 4/3 pi r^3, where r is the radius of the planet. The diameter of the planet is 14,000 km so the radius is half this, or 7,000 km. It follows that the volume of the planet is V = 4/3 pi r^3 = 4/3 pi * (7,000 km)^3 = 4/3 pi * 343,000,000,000 km^3 = 1,372,000,000,000 / 3 * pi km^3. This result can be approximated to an appropriate number of significant figures.
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RESPONSE --> ok self critique assessment: 3
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13:32:36 `q010. Summary Question 1: What basic principle do we apply to find the volume of a uniform cylinder of known dimensions?
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RESPONSE --> V =A *h confidence assessment: 3
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13:32:49 The principle is that when the cross-section of an object is constant, its volume is V = A * h, where A is the cross-sectional area and h the altitude. Altitude is measure perpendicular to the cross-section.
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RESPONSE --> ok self critique assessment:
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13:33:32 `q011. Summary Question 2: What basic principle do we apply to find the volume of a pyramid or a cone?
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RESPONSE --> 1/3 * B * h confidence assessment: 3
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13:33:43 The volumes of these solids are each 1/3 the volume of the enclosing figure. Each volume can be expressed as V = 1/3 A * h, where A is the area of the base and h the altitude as measured perpendicular to the base.
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RESPONSE --> ok self critique assessment: 3
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13:34:27 `q012. Summary Question 3: What is the formula for the volume of a sphere?
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RESPONSE --> 4/3 *3014 * r^3 confidence assessment: 3
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13:34:43 The volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere.
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RESPONSE --> ok self critique assessment: 3
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13:35:34 `q013. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> The appropiate formulas are in my notebook. confidence assessment: 3
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