course Mth 163 This is the second time sending the work from 08/29 no reply was ever posted on my access page. {WxyStudent Name:
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21:16:12 `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 --> Area of a rectangle = length x width A=4m x 3m A=12m^2
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21:16:38 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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21:19:09 `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?
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RESPONSE --> Area of a triangle = (1/2)(base * height) A=1/2 (3m * 4m) A=1/2 (12m) A=6m^2
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21:19:26 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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21:22:34 `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 parallelogram is basically a rectangle that has been shifted. The altitude makes a right triangle which could be taken off and attatched to the opposide side, now making the figure a rectangle so the area can be solved with the formula A=bh. A= 5.0m * 2.0m A= 10.0m^2
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21:22:48 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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21:25:55 `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 --> A=1/2bh A= 1/2 (5.0cm * 2.0cm) A= 1/2 (10cm^2) A= 5cm^2
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21:26:11 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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21:31:16 `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 --> Finding the average altitude essentially makes the trapezoid a rectangle, so the area should be found my multiplying the average altitude by the width. A = avg. alt * width A = 5.0km * 4.0km A = 20km^2
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21:31:30 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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21:34:40 `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 --> Find the average altitude: 3.0 cm + 8.0 cm = 11.0 cm / 2 = 5.5 cm A = avg. altitude * width A = 5.5 cm * 4 cm A = 22.0 cm^2
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21:34:50 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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21:37:57 `q007. What is the area of a circle whose radius is 3.00 cm?
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RESPONSE --> The area of a circle = pi * r^2 A = pi * 3.00^2 A = pi * 9cm^2 A = 28.27cm^2
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21:39:09 09-05-2006 21:39:09 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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NOTES -------> round to significant figures.
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21:42:24 `q008. What is the circumference of a circle whose radius is exactly 3 cm?
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RESPONSE --> Circumference = 2 pi r r being radius C= 2 pi 3cm C= 6 pi cm C= 18.9 cm
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21:42:47 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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21:46:22 `q009. What is the area of a circle whose diameter is exactly 12 meters?
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RESPONSE --> You need the radius to find the area of a circle, the problem gives the diameter, 12 m, so the diameter is 2 * r, to find the radius, divide the diameter by 2. r = 12m /2 r = 6m Area = pi r^2 A = pi 6^2 A = 36 cm^2 pi A = approx 113.10m^2
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21:46:39 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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21:49:56 `q010. What is the area of a circle whose circumference is 14 `pi meters?
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RESPONSE --> Circumference = 2 pi r therefore, when given a circumference of 14 pi m, you can divide 14 by 2 to find the radius (7). Area = pi r^2 A= pi 7^2 A= 49 m^2 pi A=153.9 m^2
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21:51:18 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 --> 49 pi m^2 does equal approx. 153.9m^2 Should the answer be left in terms of pi because that is the form in which the information was given?
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21:55:42 `q011. What is the radius of circle whose area is 78 square meters?
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RESPONSE --> A = pi r^2 So 78 m^2 needs to be divided by pi and then take the square root of the number. 78 / pi = approx 24.8 m^2 sqrt 24.8 m^2 = approx 4.98 m
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21:56:26 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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21:57:32 `q012. Summary Question 1: How do we visualize the area of a rectangle?
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RESPONSE --> A rectangle can be divided up into square units, the number of square units inside of a rectangle tells the area.
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21:57:56 09-05-2006 21:57:56 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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NOTES -------> rectangle a = lw
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21:59:52 `q013. Summary Question 2: How do we visualize the area of a right triangle?
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RESPONSE --> Any triangle can be put with another triangle of the same size to equal a square. A square's area is found by bh. The area of a triangle is one-half the size of a square with the same base and height measurements so A = 1/2bh
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22:00:00 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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22:01:01 `q014. Summary Question 3: How do we calculate the area of a parallelogram?
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RESPONSE --> The part of a parallelogram that juts out can be moved over to the opposite side and flipped to make a rectangle, therefore, the area of a parallelogram is found my multiplying the base times the height. A = bh
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22:01:22 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 --> height = altitude
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22:02:25 `q015. Summary Question 4: How do we calculate the area of a trapezoid?
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RESPONSE --> To find the area of a trapezoid, you must first find the average altitude by adding the two altitudes then dividing by two, then multiply this number by the width. A = avg. alt. * width
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22:02:37 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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22:03:08 `q016. Summary Question 5: How do we calculate the area of a circle?
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RESPONSE --> The area of a circle is found by multiplying pi times the squared radius. A = pi r^2
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22:03:13 We use the formula A = pi r^2, where r is the radius of the circle.
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22:04:24 `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 --> Circumference = 2 pi r This formula is very similar to the formula for finding the area of a circle (A=pi r^2), however, circumference does not give an answer in square units.
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22:04:31 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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22:07:36 `q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> Areas are square units within the form given. Squares, rectangles, and parallelograms use the formula A = bh to find the area. A triangle is half of a square, therefore , to find the area use A = 1/2 b h. A Trapezoid area is found by multiplying the average altitude by the width. A circle's area is found by multiplying pi r^2. A circle's circumference ( C = 2 pi r ) is not in square units.
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22:07:48 This ends the first assignment.
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22:08:03 002. Volumes
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22:10:58 `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 --> Volume of a rectangle = lwh V = 3cm * 5cm * 7cm V = 15 cm^2 * 7cm V = 105 cm ^ 3
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22:12:58 09-05-2006 22:12:58 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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NOTES -------> V = A * h A being the area of the base
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22:13:51 `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 --> V=Ah V=48m^2 * 2 m V= 96 m^3
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22:13:58 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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22:16:01 `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 --> The volume would still be found by multiplying the area of the base by the height, so 20 m^2 times 40 m = 800m^3
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22:17:08 09-05-2006 22:17:08 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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NOTES -------> V= Ah can be applied to any object whose base area is constant
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22:21:30 `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 --> Find the area of the base by using A = pi r^2. A = pi 5^2 A = 25 cm^2 pi This information goes into V = Ah V = (25 cm^2 pi) * 30 cm V = 2356.2 cm^3
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22:23:09 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 --> 750 pi cm^3 = approx 2356.2cm^3 Do you leave in terms of pi to avoid rounding?
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22:26:22 `q005. Estimate the dimensions of a metal can containing food. What is its volume, as indicated by your estimates?
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RESPONSE --> I estimate the radius of the metal can to be 2 in and the height to be 5 in. A = pi r^2 A = pi 2^2 A = 4 in^2 pi V = 4 in^2 pi * 5 in V = 20 in^3 pi
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22:26:54 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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22:31:30 `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 --> The volume of a pyramid = 1/3 * A * h V = 1/3 ( 50cm^3) (60 cm) V = 1/3 ( 3000cm^3) V = 1000cm^3
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22:32:02 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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22:34:18 `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 --> V = 1/3 *A * h V = 1/3 ( 20 m^2)(9m) V = 1/3 (180 m^3) V = 60 m^3
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22:34:36 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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22:38:41 `q008. What is a volume of a sphere whose radius is 4 meters?
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RESPONSE --> Volume of a sphere = 4/3 pi r^3 V = 4/3 pi 4^3 V = 4/3 pi 64m^3 V = 256/3 m^3 pi V = 85 1/3 m^3 pi
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22:39:06 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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22:46:56 `q009. What is the volume of a planet whose diameter is 14,000 km?
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RESPONSE --> V = 4/3 pi r^3 The diameter is 14,000 km so divided by 2 = 7000 km, the radius. V = 4/3 pi (7000km)^3 V = 4/3 pi 3.43*10^11 V = (1.372 * 10^12 / 3) km^3 pi
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22:47:34 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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22:48:17 `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 --> The area of the base ( found through A = pi r^2) multiplied by the height gives the volume of a uniform cylinder.
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22:48:30 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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22:49:09 `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 --> The volume of a pyramid or cone is found by multiplying 1/3 times the area of the base times the height. V = 1/3Ah
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22:49:20 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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22:49:38 `q012. Summary Question 3: What is the formula for the volume of a sphere?
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RESPONSE --> Volume of a sphere = 4/3 pi r^3
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22:49:43 The volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere.
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22:51:44 `q013. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> The volume of a rectangle, cube, cylinder, any object with a uniform altitude can be found by multiplying the area of the base by the height. The Volume of a pyramid and cone can be found by mulitplying 1/3 times the area of the base times the height of the object. The volume of a sphere is found by multiplying 4/3 times pi times the cubed radius.
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22:51:46 This ends the second assignment.
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22:52:24 003. Misc: Surface Area, Pythagorean Theorem, Density
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22:57:49 `q001. There are 10 questions and 5 summary questions in this assignment. What is surface area of a rectangular solid whose dimensions are 3 meters by 4 meters by 6 meters?
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RESPONSE --> The total surface area of a rectangular solid would be found by adding all of the individual areas of the six faces of a rectangular solid. SA = 2lw + 2lh + 2wh l=6m h=3m w=4m SA = (2*6*4) + (2*6*3) + (2*4*3) SA = 48 + 36 + 24 SA = 108 m^2
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22:58:06 A rectangular solid has six faces (top, bottom, front, back, left side, right side if you're facing it). The pairs top and bottom, right and left sides, and front-back have identical areas. This solid therefore has two faces with each of the following dimensions: 3 m by 4 m, 3 m by 6 m and 4 m by 6 m, areas 12 m^2, 18 m^2 and 24 m^2. Total area is 2 * 12 m^2 + 2 * 18 m^2 + 2 * 24 m^2 = 108 m^2.
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23:06:09 `q002. What is the surface area of the curved sides of a cylinder whose radius is five meters and whose altitude is 12 meters? If the cylinder is closed what is its total surface area?
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RESPONSE --> The SA of the curved sides would be found by finding the circumference of the circle (C = 2 pi r = 2 pi 5 m = 10 pi m) then multiplying it by the height of the cylinder ( 10 pi m * 12 m = 120 pi m^2). If the cylinder were closed, take the area of the curved sides ( 120 pi m^2 ) and add it to the area of the circles on each end ( A = pi r^2 = pi 5^2 = 25 m^2 pi * 2 = 50 pi m^2) 120 pi m^2 + 50 pi m^2 = 170 pi m^2
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23:06:29 The circumference of this cylinder is 2 pi r = 2 pi * 5 m = 10 pi m. If the cylinder was cut by a straight line running up its curved face then unrolled it would form a rectangle whose length and width would be the altitude and the circumference. The area of the curved side is therefore A = circumference * altitude = 10 pi m * 12 m = 120 pi m^2. If the cylinder is closed then it has a top and a bottom, each a circle of radius 5 m with resulting area A = pi r^2 = pi * (5 m)^2 = 25 pi m^2. The total area would then be total area = area of sides + 2 * area of base = 120 pi m^2 + 2 * 25 pi m^2 = 170 pi m^2.
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23:09:06 `q003. What is surface area of a sphere of diameter three cm?
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RESPONSE --> Surface area of a sphere is found by 4 pi r^2. SA = 4 pi 1.5^2 cm SA = 4 pi 2.25 cm^2 SA = 9 pi cm^2
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23:09:19 The surface area of a sphere of radius r is A = 4 pi r^2. This sphere has radius 3 cm / 2, and therefore has surface area A = 4 pi r^2 = 4 pi * (3/2 cm)^2 = 9 pi cm^2.
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23:12:52 `q004. What is hypotenuse of a right triangle whose legs are 5 meters and 9 meters?
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RESPONSE --> In the pythagorean theorem ( c^2 = a^2 + b^2 ) c represents the hypotenuse where a and b represent the legs. c^2 = 5^2 m + 9^2 m c^2 = 25 m + 81 m c^2 = 106 m c = sqrt 106 m
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23:13:15 The Pythagorean Theorem says that the hypotenuse c of a right triangle with legs a and b satisfies the equation c^2 = a^2 + b^2. So, since all lengths are positive, we know that c = sqrt(a^2 + b^2) = sqrt( (5 m)^2 + (9 m)^2 ) = sqrt( 25 m^2 + 81 m^2) = sqrt( 106 m^2 ) = 10.3 m, approx.. Note that this is not what we would get if we made the common error of assuming that sqrt(a^2 + b^2) = a + b; this would tell us that the hypotenuse is 14 m, which is emphatically not so. There is no justification whatsoever for applying a distributive law (like x * ( y + z) = x * y + x * z ) to the square root operator.
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23:15:23 `q005. If the hypotenuse of a right triangle has length 6 meters and one of its legs has length 4 meters what is the length of the other leg?
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RESPONSE --> c^2 = a^2 + b^2 6^2 m = 4^2 m + b^2 36 m = 16 m + b^2 20 m = b^2 sqrt 20 m = b b = approx. 4.5 m
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23:15:44 If c is the hypotenuse and a and b the legs, we know by the Pythagorean Theorem that c^2 = a^2 + b^2, so that a^2 = c^2 - b^2. Knowing the hypotenuse c = 6 m and the side b = 4 m we therefore find the unknown leg: a = sqrt( c^2 - b^2) = sqrt( (6 m)^2 - (4 m)^2 ) = sqrt(36 m^2 - 16 m^2) = sqrt(20 m^2) = sqrt(20) * sqrt(m^2) = 2 sqrt(5) m, or approximately 4.4 m.
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23:20:34 `q006. If a rectangular solid made of a uniform, homogeneous material has dimensions 4 cm by 7 cm by 12 cm and if its mass is 700 grams then what is its density in grams per cubic cm?
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RESPONSE --> V = lwh V = 4*7*12 V = 336 cm^3 700 g / 336 cm^3 = approx 2.08 g/cm^3
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23:21:14 The volume of this solid is 4 cm * 7 cm * 12 cm = 336 cm^3. Its density in grams per cm^3 is the number of grams in each cm^3. We find this quantity by dividing the number of grams by the number of cm^3. We find that density = 700 grams / (336 cm^3) = 2.06 grams / cm^3. Note that the solid was said to be uniform and homogeneous, meaning that it's all made of the same material, which is uniformly distributed. So each cm^3 does indeed have a mass of 2.06 grams. Had we not known that the material was uniform and homogeneous we could have said that the average density is 2.06 grams / cm^3, but not that the density is 2.06 grams / cm^3 (the object could be made of two separate substances, one with density less than 2.06 grams / cm^3 and the other with density greater than 2.06 g / cm^3, in appropriate proportions; neither substance would have density 2.06 g / cm^3, but the average density could be 2.06 g / cm^3).
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23:25:38 `q007. What is the mass of a sphere of radius 4 meters if its average density is 3,000 kg/cubic meter?
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RESPONSE --> V = 4/3 pi r^3 V = 4/3 pi 4^3 m V = 4/3 pi 64 m^3 V = 256/3m^3 pi V = 85.33 pi m^3 V = 268.08 m^3 * 3000 kg = 804247.72 kg
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23:26:49 A average density of 3000 kg / cubic meter implies that, at least on the average, every cubic meter has a mass of 3000 kg. So to find the mass of the sphere we multiply the number of cubic meters by 3000 kg. The volume of a sphere of radius 4 meters is 4/3 pi r^3 = 4/3 * pi (4m)^3 = 256/3 * pi m^3. So the mass of this sphere is mass = density * volume = 256 / 3 * pi m^3 * 3000 kg / m^3 = 256,000 * pi kg. This result can be approximated to an appropriate number of significant figures.
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RESPONSE --> 256000 pi kg = approx. 804.247.72 kg
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23:37:44 `q008. If we build a an object out of two pieces of material, one having a volume of 6 cm^3 at a density of 4 grams per cm^3 and another with a volume of 10 cm^3 at a density of 2 grams per cm^3 then what is the average density of this object?
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RESPONSE --> If both of the volumes were the same, the average density would be ( 4g/cm^3 + 2 g/cm^3 = 6 g/cm^3 / 2 ) 3 g/cm^3, however the volumes are different with the first material being 3/5 the volume of the second one. So multiply 4 g/cm^3 by 3/5 to get 12/5 g/cm^3 which is added to the volume of the second material 2 g/cm^3 equalling 22/5 g/cm^3 then divide by 2, giving 22/10 g/cm^3 or 2 1/5 g/cm^3 as the average density.
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23:39:07 09-05-2006 23:39:07 The first piece has a mass of 4 grams / cm^3 * 6 cm^3 = 24 grams. The second has a mass of 2 grams / cm^3 * 10 cm^3 = 20 grams. So the total mass is 24 grams + 20 grams = 44 grams. The average density of this object is average density = total mass / total volume = (24 grams + 20 grams) / (6 cm^3 + 10 cm^3) = 44 grams / (16 cm^3) = 2.75 grams / cm^3.
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NOTES -------> average density = total mass / total volume
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23:44:56 `q009. In a large box of dimension 2 meters by 3 meters by 5 meters we place 27 cubic meters of sand whose density is 2100 kg/cubic meter, surrounding a total of three cubic meters of cannon balls whose density is 8,000 kg per cubic meter. What is the average density of the material in the box?
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RESPONSE --> avg. density = total mass/ total volume sand: 2100 kg/m^3 * 27 m^3 = 56700 kg cannon: 8000 kg/ m^3 * 3 m^3 = 24000 kg avg density = (56700 kg + 24000 kg ) / (27 m^3 + 3 m^3 ) = 80700 kg / 30 m^3 = 2690 kg/ m^3
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23:45:09 We find the average density from the total mass and the total volume. The mass of the sand is 27 m^3 * 2100 kg / m^3 = 56,700 kg. The mass of the cannonballs is 3 m^3 * 8,000 kg / m^3 = 24,000 kg. The average density is therefore average density = total mass / total volume = (56,700 kg + 24,000 kg) / (27 m^3 + 3 m^3) = 80,700 kg / (30 m^3) = 2,700 kg / m^3, approx..
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23:53:26 `q010. How many cubic meters of oil are there in an oil slick which covers 1,700,000 square meters (between 1/2 and 1 square mile) to an average depth of .015 meters? If the density of the oil is 860 kg/cubic meter the what is the mass of the oil slick?
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RESPONSE --> 1700000m^2 * .015 m = 25500 m^3 Mass = density * volume 860 kg/m^3 * 25500 m^3 = 21930000 kg
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23:53:57 The volume of the slick is V = A * h, where A is the area of the slick and h the thickness. This is the same principle used to find the volume of a cylinder or a rectangular solid. We see that the volume is V = A * h = 1,700,000 m^2 * .015 m = 25,500 m^3. The mass of the slick is therefore mass = density * volume = 860 kg / m^3 * 24,400 m^3 = 2,193,000 kg. This result should be rounded according to the number of significant figures in the given information.
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23:55:40 `q011. Summary Question 1: How do we find the surface area of a cylinder?
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RESPONSE --> The surface area of a cylinder is found by adding the area of the curved surface which is made up of an extended circumference ( 2 pi r h) and the two circle areas ( pi r^2) SA = 2 pi r h + 2 pi r^2
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23:55:51 The curved surface of the cylinder can be 'unrolled' to form a rectangle whose dimensions are equal to the circumference and the altitude of the cylinder, so the curved surface has volume Acurved = circumference * altitude = 2 pi r * h, where r is the radius and h the altitude. The top and bottom of the cylinder are both circles of radius r, each with resulting area pi r^2. {]The total surface area is therefore Acylinder = 2 pi r h + 2 pi r^2.
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23:56:11 `q012. Summary Question 2: What is the formula for the surface area of a sphere?
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RESPONSE --> SA = 4 pi r^2
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23:56:14 The surface area of a sphere is A = 4 pi r^2, where r is the radius of the sphere.
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23:56:55 `q013. Summary Question 3: What is the meaning of the term 'density'.
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RESPONSE --> Density is the mass/ volume. How many grams are in a cubic meter for example.
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23:57:12 The average density of an object is its mass per unit of volume, calculated by dividing its total mass by its total volume. If the object is uniform and homogeneous then its density is constant and we can speak of its 'density' as opposed to its 'average density'
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23:58:09 `q014. Summary Question 4: If we know average density and mass, how can we find volume?
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RESPONSE --> Dividing the mass by the average density will give the volume.
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23:58:17 Since mass = ave density * volume, it follows by simple algebra that volume = mass / ave density.
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23:59:35 09-05-2006 23:59:35 `q015. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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NOTES -------> SA of a rectangle = 2lw + 2 lh + 2 wh SA of a cylinder = 2 pi r h + 2 pi r^2 SA of a sphere = 4 pi r^2 Pythagorean Theorem c^2 = a^2 + b^2 Density = mass/volume
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23:59:37 This ends the third assignment.
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00:00:08 004. Units of volume measure
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