course Mth 158 ۰\ЄǔXDStudent Name:
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16:57:47 `q001. It will be very important in this course for your instructor to see and understand the process of visualization and reasoning you use when you solve problems. This exercise is designed to give you a first experience with these ideas, and your instructor a first look at your work. Answer the following questions and explain in commonsense terms why your answer makes sense.
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RESPONSE --> ok
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16:58:32 For each question draw a picture to make sense out of the situation, and include a description of the picture. Samples Sample question and response Question: If a bundle of shingles covers 30 square feet, how many bundles are required to cover a 600 square foot roof? Response: We might draw a picture of a rectangle representing the area, dividing the rectangle into a number of smaller rectangles each representing the area covered by a single bundle. This makes it clear that we are dividing the roof area into 1-bundle areas, and makes it clear why we are going to have to divide. Reasoning this problem out in words, we can say that a single bundle would cover 30 square feet. Two bundles would cover 60 square feet. Three bundles would cover 90 square feet. We could continue in this manner until we reach 600 square feet. However, this would be cumbersome. It is more efficient to use the ideas of multiplication and division. We imagine grouping the 600 square feet into 30 square foot patches. There will be 600 / 30 patches and each will require exactly one bundle. We therefore require 600 / 30 bundles = 20 bundles. {}Your responses might not be as clear as the above, though they might be even more clear. I won't be looking for perfection, though I wouldn't object to it, but for a first effort at visualizing a situation and communicating a reasoning process. This is not something you are used to doing and it might take a few attempts before you can achieve good results, but you will get better every time you try. {}You might be unsure of what to do on a specific question. In such a case specific questions and expressions of confusion are also acceptable responses. Such a response must include your attempts to come up with a picture and reason out an explanation. For example your response might be Sample expression of confusion: I've drawn a picture of a pile of bundles and a roof but I'm not sure how to connect the two. I tried multiplying the number of bundles by the square feet of the roof but I got 18,000, and I know it won't take 18,000 bundles to cover the roof. How do you put the area covered by a bundle together with the roof area to get the number of bundles required? A poor response would be something like 'I don't know how to do #17'. This response reveals nothing of your attempt to understand the question and the situation. Nor does it ask a specific question. Incidentally, you might be tempted to quote rules or formulas about rates and velocities in answering these questions. Don't. This exercise isn't about being able to memorize rules and quote them. It is about expanding your ability to visualize, reason and communicate.
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RESPONSE --> ok
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16:59:21 In your own words briefly summarize the instructions and the intent of this exercise.
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RESPONSE --> Read the question and draw an illustration to answer the question and give a response on what you would draw to answer the question.
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17:00:07 `q001. If you earn 50 dollars in 5 hours, at what average rate are you earning money, in dollars per hour?
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RESPONSE --> 50/5 = 10 dollars per hour.
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17:00:36 If you travel 300 miles in 6 hours, at what average rate are you traveling, in miles per hour?
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RESPONSE --> 300/6 = 50 miles per hour
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17:01:15 `q002. If a ball rolling down a grooved track travels 40 centimeters in 5 seconds, at what average rate is the ball moving, in centimeters per second?
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RESPONSE --> 40/5 = 8 cm per s
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17:02:09 The preceding three questions illustrate the concept of a rate. In each case, to find the rate we divided the change in some quantity (the number of dollars or the distance, in these examples) by the time required for the change (the number of hours or seconds, in these examples). Explain in your own words what is meant by the idea of a rate.
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RESPONSE --> Rate is when you divide the change in some quantity by the time required for the change.
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17:02:44 `q003. If you are earning money at the average rate of 15 dollars per hour, how much do you earn in 6 hours?
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RESPONSE --> 15*6 = 90 dollars
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17:03:12 If you are traveling at an average rate of 60 miles per hour, how far do you travel in 9 hours?
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RESPONSE --> 60*9 = 540 miles
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17:03:39 `q004. If a ball travels at and average rate of 13 centimeters per second, how far does it travel in 3 seconds?
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RESPONSE --> 13*3 = 39 cm
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17:04:54 In the preceding three exercises you turned the concept of a rate around. You were given the rate and the change in the clock time, and you calculated the change in the quantity. Explain in your own words how this increases your understanding of the concept of a rate.
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RESPONSE --> You multiplied the rate by the time to figure out the change in the quantity.
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17:05:27 `q005. How long does it take to earn 100 dollars at an average rate of 4 dollars per hour?
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RESPONSE --> 100/4 = 25 hours
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17:06:04 How long does it take to travel 500 miles at an average rate of 25 miles per hour?
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RESPONSE --> 500/25 = 20 hours
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17:06:30 `q006. How long does it take a rolling ball to travel 80 centimeters at an average rate of 16 centimeters per second?
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RESPONSE --> 80/16 = 5 seconds
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17:07:41 In the preceding three exercises you again expanded your concept of the idea of a rate. Explain how these problems illustrate the concept of a rate.
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RESPONSE --> You took the distance and divided it by the mph going to get the amount of time taken.
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ăZw{z Student Name: assignment #001 001. Areas
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17:19:20 `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
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17:19:27 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
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17:19:46 `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?
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RESPONSE --> 4*3 = 12m^2
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17:20:25 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
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17:20:40 `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 --> 5*2 = 10m^2
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17:20:57 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
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17:21:36 `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*2 = 10cm^2 * 1/2 = 5cm^2
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17:21:44 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
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17:22:17 `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 --> 4*5 = 20km^2
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17:22:29 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
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17:23:27 `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*(3*8)= 96 / 1/2 = 48 cm^2
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17:24:15 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
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17:24:48 `q007. What is the area of a circle whose radius is 3.00 cm?
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RESPONSE --> 3*3.14 = 9.42 cm
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17:26:09 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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RESPONSE --> ok
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17:26:27 `q008. What is the circumference of a circle whose radius is exactly 3 cm?
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RESPONSE --> 3*3.14 = 9.42
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17:26:58 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
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17:27:44 `q009. What is the area of a circle whose diameter is exactly 12 meters?
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RESPONSE --> 6^2*3.14 = 113.04m^2
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17:28:16 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
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17:28:46 `q010. What is the area of a circle whose circumference is 14 `pi meters?
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RESPONSE --> 7^2*3.14 = 153.86
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17:29:46 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
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17:30:16 `q011. What is the radius of circle whose area is 78 square meters?
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RESPONSE --> no idea
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17:30:35 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
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17:30:47 `q012. Summary Question 1: How do we visualize the area of a rectangle?
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RESPONSE --> A=L*W
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17:30:52 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
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17:31:07 `q013. Summary Question 2: How do we visualize the area of a right triangle?
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RESPONSE --> A=L*W
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17:31:16 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 --> oh ok
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17:31:28 `q014. Summary Question 3: How do we calculate the area of a parallelogram?
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RESPONSE --> A=1/2*b*h
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17:31:39 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
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17:31:55 `q015. Summary Question 4: How do we calculate the area of a trapezoid?
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RESPONSE --> A=l*w*h
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17:32:01 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
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17:32:14 `q016. Summary Question 5: How do we calculate the area of a circle?
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RESPONSE --> A=r^2*3.14
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17:32:22 We use the formula A = pi r^2, where r is the radius of the circle.
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RESPONSE --> ok
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17:32:47 `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 --> The circumference of a circle confuses me.
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17:33:01 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
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17:33:51 `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 learned that I really need to review this thoroughly because I do not remember all the formulas like I need to.
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17:33:57 This ends the first assignment.
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RESPONSE --> ok
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17:34:06 002. Volumes
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RESPONSE --> ok
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17:34:31 `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
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17:35:15 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
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17:35:44 `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 --> 48*2 = 96 m^2
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17:35:52 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
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17:36:17 `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
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17:36:23 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
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17:36:49 `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 --> 5*30 = 150cm^2
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17:37:07 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 --> ok
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17:37:59 `q005. Estimate the dimensions of a metal can containing food. What is its volume, as indicated by your estimates?
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RESPONSE --> 5*3 = 15 in
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17:38:28 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
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17:38:52 `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 --> 50*60 = 3000cm^2
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17:39:13 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
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17:39:40 `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 --> 20*9*1/3 = 60m^2
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17:39:45 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
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17:40:10 `q008. What is a volume of a sphere whose radius is 4 meters?
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RESPONSE --> 4*3.14=12.56m^2
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17:40:32 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
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17:41:00 `q009. What is the volume of a planet whose diameter is 14,000 km?
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RESPONSE --> don't know
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17:41:10 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
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17:41:34 `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=4/3 pi r^3
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17:41:41 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 --> oh ok
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17:41:55 `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 --> V=1/3 * A* h
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17:41:59 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
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17:42:12 `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
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17:42:17 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
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17:42:33 `q013. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> I really need to look over the volume formulas
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17:42:36 This ends the second assignment.
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RESPONSE --> ok
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syQ Student Name: assignment #003 003. Misc: Surface Area, Pythagorean Theorem, Density
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18:12:15 `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 --> 2*12+2*18+2*24= 108m^3
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18:12:22 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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RESPONSE --> ok
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18:14:44 `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 --> Circumfence = 2*3.14*5 = 31.4m A = 31.4m * 12m = 376.8m^2
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18:15:30 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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RESPONSE --> ok
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18:16:13 `q003. What is surface area of a sphere of diameter three cm?
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RESPONSE --> 4*3.14*3^2 = 113.04cm
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18:17:08 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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RESPONSE --> oh I forgot to divide the radius by 2
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18:18:06 `q004. What is hypotenuse of a right triangle whose legs are 5 meters and 9 meters?
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RESPONSE --> c^2 = 5^2 + 9^2 = 25 + 81 = 106m^2
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18:18:58 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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RESPONSE --> I didn't take the square root of the 106 m^2
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18:21:55 `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 --> 6^2 + 4^2 = 36 + 16 = 52 = squa (52) = 7.2m
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18:22:22 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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RESPONSE --> oh okay so you would subtract the two, I get it.
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18:24:17 `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 --> 4*7*12 = 336m^2 700 / 336 = 2.08 g/cm^3
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18:24:27 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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RESPONSE --> ok
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18:25:14 `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 --> 3000/4 = 750 mass
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18:25:25 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 --> oh ok
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18:27:02 `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 --> 6*4 = 24 g/cm^3 10*2 = 20 g/cm^3 24+20 = 44 g/cm^3
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18:27:21 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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RESPONSE --> oh ok
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18:28:27 `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 --> 2*3*5= 30m^3 I found the area but have no clue on the rest of this problem.
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18:28:48 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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RESPONSE --> oh ok
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18:30:11 `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 --> 1700000 * .015 = 25500m^3 / 860 = 29.7 mass
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18:30:26 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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RESPONSE --> ok
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18:30:47 `q011. Summary Question 1: How do we find the surface area of a cylinder?
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RESPONSE --> A = circumfrence 8 altitude
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18:30:56 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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RESPONSE --> ok
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18:31:13 `q012. Summary Question 2: What is the formula for the surface area of a sphere?
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RESPONSE --> A = 4 pi r^2
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18:31:16 The surface area of a sphere is A = 4 pi r^2, where r is the radius of the sphere.
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RESPONSE --> ok
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18:33:29 `q013. Summary Question 3: What is the meaning of the term 'density'.
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RESPONSE --> it's the ratio of the mass of a substance to its volume
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18:33:34 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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RESPONSE --> ok
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18:33:52 `q014. Summary Question 4: If we know average density and mass, how can we find volume?
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RESPONSE --> Volume = density * mass
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18:34:04 Since mass = ave density * volume, it follows by simple algebra that volume = mass / ave density.
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RESPONSE --> oh I had it confused
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18:34:27 `q015. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> I need to review all of these past 3 q_a sessions.
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18:34:30 This ends the third assignment.
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RESPONSE --> ok
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18:34:34 004. Units of volume measure
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RESPONSE --> ok
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18:35:12 `q001. There are 10 questions and 5 summary questions in this assignment. How many cubic centimeters of fluid would require to fill a cubic container 10 cm on a side?
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RESPONSE --> I don't know how to find this out.
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18:35:40 The volume of the container is 10 cm * 10 cm * 10 cm = 1000 cm^3. So it would take 1000 cubic centimeters of fluid to fill the container.
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RESPONSE --> oh ok
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18:36:32 `q002. How many cubes each 10 cm on a side would it take to build a solid cube one meter on a side?
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RESPONSE --> 10*10*10*10
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18:36:52 It takes ten 10 cm cubes laid side by side to make a row 1 meter long or a tower 1 meter high. It should therefore be clear that the large cube could be built using 10 layers, each consisting of 10 rows of 10 small cubes. This would require 10 * 10 * 10 = 1000 of the smaller cubes.
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RESPONSE --> ok
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18:37:18 `q003. How many square tiles each one meter on each side would it take to cover a square one km on the side?
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RESPONSE --> 1000
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18:37:45 It takes 1000 meters to make a kilometer (km). To cover a square 1 km on a side would take 1000 rows each with 1000 such tiles to cover 1 square km. It therefore would take 1000 * 1000 = 1,000,000 squares each 1 m on a side to cover a square one km on a side. We can also calculate this formally. Since 1 km = 1000 meters, a square km is (1 km)^2 = (1000 m)^2 = 1,000,000 m^2.
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RESPONSE --> ok
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18:38:43 `q004. How many cubic centimeters are there in a liter?
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RESPONSE --> 1000cm^3
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18:38:48 A liter is the volume of a cube 10 cm on a side. Such a cube has volume 10 cm * 10 cm * 10 cm = 1000 cm^3. There are thus 1000 cubic centimeters in a liter.
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RESPONSE --> ok
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18:39:26 `q005. How many liters are there in a cubic meter?
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RESPONSE --> 1000 liters
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18:39:30 A liter is the volume of a cube 10 cm on a side. It would take 10 layers each of 10 rows each of 10 such cubes to fill a cube 1 meter on a side. There are thus 10 * 10 * 10 = 1000 liters in a cubic meter.
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RESPONSE --> ok
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18:40:14 `q006. How many cm^3 are there in a cubic meter?
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RESPONSE --> 1000000cm^3
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18:40:18 There are 1000 cm^3 in a liter and 1000 liters in a m^3, so there are 1000 * 1000 = 1,000,000 cm^3 in a m^3. It's important to understand the 'chain' of units in the previous problem, from cm^3 to liters to m^3. However another way to get the desired result is also important: There are 100 cm in a meter, so 1 m^3 = (1 m)^3 = (100 cm)^3 = 1,000,000 cm^3.
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RESPONSE --> ok
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18:40:46 `q007. If a liter of water has a mass of 1 kg the what is the mass of a cubic meter of water?
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RESPONSE --> 1000kg
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18:40:56 Since there are 1000 liters in a cubic meter, the mass of a cubic meter of water will be 1000 kg. This is a little over a ton.
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RESPONSE --> ok
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18:41:39 `q008. What is the mass of a cubic km of water?
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RESPONSE --> 5000
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18:41:51 A cubic meter of water has a mass of 1000 kg. A cubic km is (1000 m)^3 = 1,000,000,000 m^3, so a cubic km will have a mass of 1,000,000,000 m^3 * 1000 kg / m^3 = 1,000,000,000,000 kg. In scientific notation we would say that 1 m^3 has a mass of 10^3 kg, a cubic km is (10^3 m)^3 = 10^9 m^3, so a cubic km has mass (10^9 m^3) * 1000 kg / m^3 = 10^12 kg.
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RESPONSE --> oh ok
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18:42:34 `q009. If each of 5 billion people drink two liters of water per day then how long would it take these people to drink a cubic km of water?
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RESPONSE --> I don't know how to figure this out
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18:42:43 5 billion people drinking 2 liters per day would consume 10 billion, or 10,000,000,000, or 10^10 liters per day. A cubic km is (10^3 m)^3 = 10^9 m^3 and each m^3 is 1000 liters, so a cubic km is 10^9 m^3 * 10^3 liters / m^3 = 10^12 liters, or 1,000,000,000,000 liters. At 10^10 liters per day the time required to consume a cubic km would be time to consume 1 km^3 = 10^12 liters / (10^10 liters / day) = 10^2 days, or 100 days. This calculation could also be written out: 1,000,000,000,000 liters / (10,000,000,000 liters / day) = 100 days.
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RESPONSE --> oh ok
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18:44:57 `q010. The radius of the Earth is approximately 6400 kilometers. What is the surface area of the Earth? If the surface of the Earth was covered to a depth of 2 km with water that what would be the approximate volume of all this water?
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RESPONSE --> 4/3 * 3.14 * 3201^2 = 42,898,265.52
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18:45:09 The surface area would be A = 4 pi r^2 = 4 pi ( 6400 km)^2 = 510,000,000 km^2. A flat area of 510,000,000 km^2 covered to a depth of 2 km would indicate a volume of V = A * h = 510,000,000 km^2 * 2 km = 1,020,000,000 km^3. However the Earth's surface is curved, not flat. The outside of the 2 km covering of water would have a radius 2 km greater than that of the Earth, and therefore a greater surface area. However a difference of 2 km in 6400 km will change the area by only a fraction of one percent, so the rounded result 1,020,000,000,000 km^3 would still be accurate.
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RESPONSE --> ok
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18:45:21 `q011. Summary Question 1: How can we visualize the number of cubic centimeters in a liter?
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RESPONSE --> 10*10*10
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18:45:28 Since a liter is a cube 10 cm on a side, we visualize 10 layers each of 10 rows each of 10 one-centimeter cubes, for a total of 1000 1-cm cubes. There are 1000 cubic cm in a liter.
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RESPONSE --> ok
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18:45:34 `q012. Summary Question 2: How can we visualize the number of liters in a cubic meter?
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RESPONSE --> 10*10*10
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18:45:39 Since a liter is a cube 10 cm on a side, we need 10 such cubes to span 1 meter. So we visualize 10 layers each of 10 rows each of 10 ten-centimeter cubes, for a total of 1000 10-cm cubes. Again each 10-cm cube is a liter, so there are 1000 liters in a cubic meter.
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RESPONSE --> ok
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18:46:19 `q013. Summary Question 3: How can we calculate the number of cubic centimeters in a cubic meter?
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RESPONSE --> 10*10*10*10*10*10
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18:46:26 One way is to know that there are 1000 liters in a cubic meters, and 1000 cubic centimeters in a cubic meter, giving us 1000 * 1000 = 1,000,000 cubic centimeters in a cubic meter. Another is to know that it takes 100 cm to make a meter, so that a cubic meter is (100 cm)^3 = 1,000,000 cm^3.
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RESPONSE --> ok
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18:47:04 `q014. Summary Question 4: There are 1000 meters in a kilometer. So why aren't there 1000 cubic meters in a cubic kilometer? Or are there?
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RESPONSE --> I think that there is 1000 cubic meters in a cubic kilometer
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18:47:14 A cubic kilometer is a cube 1000 meters on a side, which would require 1000 layers each of 1000 rows each of 1000 one-meter cubes to fill. So there are 1000 * 1000 * 1000 = 1,000,000,000 cubic meters in a cubic kilometer. Alternatively, (1 km)^3 = (10^3 m)^3 = 10^9 m^3, not 1000 m^3.
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RESPONSE --> ok
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18:47:59 `q015. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> I really need to look over all these things because I obviously have forgotten so much about alegbra.
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18:48:01 This ends the fourth assignment.
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RESPONSE --> ok
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