course Mth173 Please remember to use the following email address. ctosborne@earthlink.net This may have been the first assignment under the orientation, but I couldn't remember. These are all the assignments that were saved under the vhmthphy file. I did try to get into the other assignments and it gave me error messages. I tried them at home and at work.
......!!!!!!!!...................................
12:15:35 `q001. You should copy and paste these instructions to a word processor for reference. However you can always view them, as well as everything else that has appeared in this box, by clicking the 'Display Everything' button. 1. For the next question or answer, you click on 'Next Question / Answer' button above the box at top left until a question has been posed. Once a question has been posed you are to answer before you click again on this button. 2. Before clicking for an answer, type your best answer to the current question into the box to the right, then clip on the 'Enter Answer' button. 3. After entering your answer you will click on 'Next Question / Answer' to view the answer to the question. Do not tamper with the information displayed in the left-hand box. 4. If your answer was incorrect, incomplete or would otherwise require revision, you will enter a self-critique. If you learned something from the answer, you need to restate it in your own words in order to reinforce your learning. If there is something you feel you should note for future reference, you should make a note in your own words. Go to the response box (the right-hand box) and type in a self-critique and/or notes, as appropriate. Do not copy and paste anything from the left-hand box, since that information will be saved in any case. 5. If you wish to save your response to your Notes file you may choose to click on the 'Save As Notes' button rather than the 'Enter Answer' button. Doing so will save your work for your future reference. Your work will be saved in a Notes file in the c:\vhmthphy folder. The title of the Notes file will also include the name you gave when you started the program. 6. After clicking either the 'Enter Response' or the 'Save as Notes' button, click on 'Next Question / Answer' and proceed in a similar manner. In the right-hand box briefly describe your understanding of these instructions, then click 'Enter Answer'.
......!!!!!!!!...................................
RESPONSE --> After reading the problem question, typed the answer in the right hand box and enter response. After this process has been done click next question to see what the correct answer is. If the answer that I typed in was wrong put in my own words what I learned and what response I have to the correct answer that was displayed confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:15:53 Your answer has been noted. Enter 'ok' in the Response Box and click on Enter Response, then click on Next Question/Answer for the first real question.
......!!!!!!!!...................................
RESPONSE --> ok confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:17:40 `q002. Note that there are 10 questions in this assignment. The questions are of increasing difficulty--the first questions are fairly easy but later questions are very tricky. The main purposes of these exercises are to refine your thinking about rates, and to see how you process challenging information. Continue as far as you can until you are completely lost. Students who are prepared for the highest-level math courses might not ever get lost. If you make $50 in 5 hr, then at what rate are you earning money?
......!!!!!!!!...................................
RESPONSE --> If you make $50 in 5 hours then you would be making $10 per hour. This could derived by dividing $50 by 5 hours. confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:18:43 The rate at which you are earning money is the number of dollars per hour you are earning. You are earning money at the rate of 50 dollars / (5 hours) = 10 dollars / hour. It is very likely that you immediately came up with the $10 / hour because almosteveryone is familiar with the concept of the pay rate, the number of dollars per hour. Note carefully that the pay rate is found by dividing the quantity earned by the time required to earn it. Time rates in general are found by dividing an accumulated quantity by the time required to accumulate it. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
......!!!!!!!!...................................
RESPONSE --> All was understood confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:20:24 `q003.If you make $60,000 per year then how much do you make per month?
......!!!!!!!!...................................
RESPONSE --> If you make $60,000 per year, then you would make $5,000 per month. This can derived by dividing $60,000 by 12 months(equal to one year). confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:21:06 Most people will very quickly see that we need to divide $60,000 by 12 months, giving us 60,000 dollars / (12 months) = 5000 dollars / month. Note that again we have found a time rate, dividing the accumulated quantity by the time required to accumulate it. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
......!!!!!!!!...................................
RESPONSE --> All was understood confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:22:51 `q004. Suppose that the $60,000 is made in a year by a small business. Would be more appropriate to say that the business makes $5000 per month, or that the business makes an average of $5000 per month?
......!!!!!!!!...................................
RESPONSE --> If a small company makes $60,000 in one year then it would be more appropriate to say that the business makes an average of $5,000 per month than it would be to say it made $5,000 per month. Most companies do not make the same amount of money each month. confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:23:34 Small businesses do not usually make the same amount of money every month. The amount made depends on the demand for the services or commodities provided by the business, and there are often seasonal fluctuations in addition to other market fluctuations. It is almost certain that a small business making $60,000 per year will make more than $5000 in some months and less than $5000 in others. Therefore it is much more appropriate to say that the business makes and average of $5000 per month. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
......!!!!!!!!...................................
RESPONSE --> All is understood. confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:25:13 `q005. If you travel 300 miles in 6 hours, at what average rate are you covering distance, and why do we say average rate instead of just plain rate?
......!!!!!!!!...................................
RESPONSE --> When you have traveled 300 miles in 6 hours you are traveling an average of 50 miles per hour. It is an average rate, because you don't always travel the same speed and you may have stops or delalys because of traffic. confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:26:18 The average rate is 50 miles per hour, or 50 miles / hour. This is obtained by dividing the accumulated quantity, the 300 miles, by the time required to accumulate it, obtaining ave rate = 300 miles / ( 6 hours) = 50 miles / hour. Note that the rate at which distance is covered is called speed. The car has an average speed of 50 miles/hour. We say 'average rate' in this case because it is almost certain that slight changes in pressure on the accelerator, traffic conditions and other factors ensure that the speed will sometimes be greater than 50 miles/hour and sometimes less than 50 miles/hour; the 50 miles/hour we obtain from the given information is clearly and overall average of the velocities. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
......!!!!!!!!...................................
RESPONSE --> All is understood. I did however forget to put the formula for how I got the answer. confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:29:07 `q006. If you use 60 gallons of gasoline on a 1200 mile trip, then at what average rate are you using gasoline, with respect to miles traveled?
......!!!!!!!!...................................
RESPONSE --> If you traveled 1200 miles and used only 60 gallons of gas, then you only used .05 or 1/20th gallon of gas per mile. This is derived by 60/1200. confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:31:04 The rate of change of one quantity with respect to another is the change in the first quantity, divided by the change in the second. As in previous examples, we found the rate at which money was made with respect to time by dividing the amount of money made by the time required to make it. By analogy, the rate at which we use fuel with respect to miles traveled is the change in the amount of fuel divided by the number of miles traveled. In this case we use 60 gallons of fuel in 1200 miles, so the average rate it 60 gal / (1200 miles) = .05 gallons / mile. Note that this question didn't ask for miles per gallon. Miles per gallon is an appropriate and common calculation, but it measures the rate at which miles are covered with respect to the amount of fuel used. Be sure you see the difference. Note that in this problem we again have here an example of a rate, but unlike previous instances this rate is not calculated with respect to time. This rate is calculated with respect to the amount of fuel used. We divide the accumulated quantity, in this case miles, by the amount of fuel required to cover those miles. Note that again we call the result of this problem an average rate because there are always at least subtle differences in driving conditions that require the use of more fuel on some miles than on others. It's very important to understand the phrase 'with respect to'. Whether the calculation makes sense or not, it is defined by the order of the terms. In this case gallons / mile tells you how many gallons you are burning, on the average, per mile. This concept is not as familiar as miles / gallon, but except for familiarity it's technically no more difficult. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
......!!!!!!!!...................................
RESPONSE --> I did respond in miles per gallon on this question, but do understand that the question did not ask for miles per gallon. confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:32:46 `q007. The word 'average' generally connotes something like adding two quantities and dividing by 2, or adding several quantities and dividing by the number of quantities we added. Why is it that we are calculating average rates but we aren't adding anything?
......!!!!!!!!...................................
RESPONSE --> Each one of our problems gave us totals of each product that we were asked to calculate, therfore we only have to divide the to different figures in the question. confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:33:34 The word 'average' in the context of the dollars / month, miles / gallon types of questions we have been answering was used because we expect that in different months different amounts were earned, or that over different parts of the trip the gas mileage might have varied, but that if we knew all the individual quantities (e.g., the dollars earned each month, the number of gallons used with each mile) and averaged them in the usual manner, we would get the .05 gallons / mile, or the $5000 / month. In a sense we have already added up all the dollars earned in each month, or the miles traveled on each gallon, and we have obtained the total $60,000 or 1200 miles. Thus when we divide by the number of months or the number of gallons, we are in fact calculating an average rate. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
......!!!!!!!!...................................
RESPONSE --> All was understood. confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:45:50 `q008. In a study of how lifting strength is influenced by various ways of training, a study group was divided into 2 subgroups of equally matched individuals. The first group did 10 pushups per day for a year and the second group did 50 pushups per day for year. At the end of the year to lifting strength of the first group averaged 147 pounds, while that of the second group averaged 162 pounds. At what average rate did lifting strength increase per daily pushup?
......!!!!!!!!...................................
RESPONSE --> When 2 groups of individuals do pushups each day for a year then we can say that the average rate of lifting strength increased by 1.41% each day. The first group did a total of 3,650 pushups in a years time (10 pushups x 365 days) and the second group did 18,200 pushups in a years time (50 pushups x 365 days). Both groups did a total of 60 pushups per day (10 + 50) and had a total of 21,900 pushups (3,650 + 18,250) in a years time. To derive at the 1.41% average rate of lifting strength daily you would take the totaal increase in lbs lifted which would be 309(147 + 162) and divide this by the total of pushups done in a year 21,900 (3,650 by the first group + 18,250 by the second group). confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:48:14 The second group had 15 pounds more lifting strength as a result of doing 40 more daily pushups than the first. The desired rate is therefore 15 pounds / 40 pushups = .375 pounds / pushup. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
......!!!!!!!!...................................
RESPONSE --> When reading this problem I had the understanding that both groups would be added together. However, you only need to take the difference of the lbs that are being lifted and the difference of the pushups and divided the lbs by the pushups to get the average rate. confidence assessment: 1
.................................................
......!!!!!!!!...................................
12:56:50 `q009. In another part of the study, participants all did 30 pushups per day, but one group did pushups with a 10-pound weight on their shoulders while the other used a 30-pound weight. At the end of the study, the first group had an average lifting strength of 171 pounds, while the second had an average lifting strength of 188 pounds. At what average rate did lifting strength increase with respect to the added shoulder weight?
......!!!!!!!!...................................
RESPONSE --> The average lifting strength would have an average increase in rate of .85. The difference of the pounds that was put on the shoulders of the participants was 20lbs and the difference of lifting strength was 17lbs. By dividing the average lifting strength (17lbs) by the difference of the lbs (20lbs) then you could get and average increase in lifting weight of .85lbs. confidence assessment: 2
.................................................
......!!!!!!!!...................................
12:57:24 The difference in lifting strength was 17 pounds, as a result of a 20 pound difference in added weight. The average rate at which strength increases with respect added weight would therefore be 17 lifting pounds / (20 added pounds) = .85 lifting pounds / added pound. The strength advantage was .85 lifting pounds per pound of added weight, on the average. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
......!!!!!!!!...................................
RESPONSE --> All was understood. confidence assessment: 3
.................................................
......!!!!!!!!...................................
13:00:23 `q010. During a race, a runner passes the 100-meter mark 12 seconds after the start and the 200-meter mark 22 seconds after the start. At what average rate was the runner covering distance between those two positions?
......!!!!!!!!...................................
RESPONSE --> The difference of the meters marked was 100 meters (200m - 100m) and the difference of the seconds was 10 seconds (22s - 12s). The average rate that was covered was 10 meters (100m/10sec). confidence assessment: 3
.................................................
......!!!!!!!!...................................
13:00:49 The runner traveled 100 meters between the two positions, and required 10 seconds to do so. The average rate at which the runner was covering distance was therefore 100 meters / (10 seconds) = 10 meters / second. Again this is an average rate; at different positions in his stride the runner would clearly be traveling at slightly different speeds. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
......!!!!!!!!...................................
RESPONSE --> All was understood. confidence assessment: 3
.................................................
......!!!!!!!!...................................
13:07:28 `q011. During a race, a runner passes the 100-meter mark moving at 10 meters / second, and the 200-meter mark moving at 9 meters / second. What is your best estimate of how long it takes the runner to cover the 100 meter distance?
......!!!!!!!!...................................
RESPONSE --> The first runner would take 10 seconds to get to the 100-meter mark 100m/10m per sec). The second runner would take 11 seconds to get to the 100-meter mark (200m/9m per second/2). The average rate at which a runner can cover the 100 meter distance is 10.5 seconds. confidence assessment: 3
.................................................
......!!!!!!!!...................................
13:09:28 At 10 meters/sec, the runner would require 10 seconds to travel 100 meters. However the runner seems to be slowing, and will therefore require more than 10 seconds to travel the 100 meters. We don't know what the runner's average speed is, we only know that it goes from 10 m/s to 9 m/s. The simplest estimate we could make would be that the average speed is the average of 10 m/s and 9 m/s, or (10 m/s + 9 m/s ) / 2 = 9.5 m/s. Taking this approximation as the average rate, the time required to travel 100 meters will be (100 meters) / (9.5 m/s) = 10.5 sec, approx.. Note that simply averaging the 10 m/s and the 9 m/s might not be the best way to approximate the average rate--for example we if we knew enough about the situation we might expect that this runner would maintain the 10 m/s for most of the remaining 100 meters, and simply tire during the last few seconds. However we were not given this information, and we don't add extraneous assumptions without good cause. So the approximation we used here is pretty close to the best we can do with the given information. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
......!!!!!!!!...................................
RESPONSE --> The one mistake that I made was misunderstanding that there was only one person in the problem, however I still derived that the same answer. confidence assessment: 3
.................................................
......!!!!!!!!...................................
13:12:37 `q012. We just averaged two quantities, adding them in dividing by 2, to find an average rate. We didn't do that before. Why we do it now?
......!!!!!!!!...................................
RESPONSE --> In this particular problem the totals were not given in seconds. When you are given the amounts for each second, lbs, ... then you have to get the total of what is asked o of you in the problem in order to get the average. confidence assessment: 3
.................................................
......!!!!!!!!...................................
13:13:09 In previous examples the quantities weren't rates. We were given the amount of change of some accumulating quantity, and the change in time or in some other quantity on which the first was dependent (e.g., dollars and months, miles and gallons). Here we are given 2 rates, 10 m/s and 9 m/s, in a situation where we need an average rate in order to answer a question. Within this context, averaging the 2 rates was an appropriate tactic. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
......!!!!!!!!...................................
RESPONSE --> All was understood. confidence assessment: 3
.................................................
e{???????????? assignment #002 002. Describing Graphs qa initial problems 01-18-2007 ?????????????assignment #001 001. Areas qa areas volumes misc 01-18-2007
......!!!!!!!!...................................
19:04:35 `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.
......!!!!!!!!...................................
RESPONSE --> To find the area of a rectangle you use the formula length x width, therefore the area of a rectangle whose dimensions are 4m by 3 m would be 4 x 3=12m confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:05:14 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.
......!!!!!!!!...................................
RESPONSE --> I forgot to put my power of two in. self critique assessment: 3
.................................................
......!!!!!!!!...................................
19:09:56 `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?
......!!!!!!!!...................................
RESPONSE --> The area of a triangle is 1/2 length x width. Therefore the area of a triangle whose measurement is 4m and 3 meters would be 1/2(4)(3)=6m. confidence assessment: 2
.................................................
......!!!!!!!!...................................
19:11:01 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.
......!!!!!!!!...................................
RESPONSE --> Again I forgot to put the power of the meters in the problem. self critique assessment: 3
.................................................
......!!!!!!!!...................................
19:13:16 `q003. What is the area of a parallelogram whose base is 5.0 meters and whose altitude is 2.0 meters?
......!!!!!!!!...................................
RESPONSE --> The area of a parallelogram is the length x the altitude. Therefore the measurements for a parallelogram whose base is 5m and whose altitude is 2m would be 5m x 2m, which would equal 10m ^2. confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:13:56 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.
......!!!!!!!!...................................
RESPONSE --> All is understood self critique assessment: 3
.................................................
......!!!!!!!!...................................
19:16:11 `q004. What is the area of a triangle whose base is 5.0 cm and whose altitude is 2.0 cm?
......!!!!!!!!...................................
RESPONSE --> To find the area of a triangle you would multiply length times width. Therefore it would be (1/2)5cm x 2cm = 5cm^2 confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:16:45 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.
......!!!!!!!!...................................
RESPONSE --> All is understood self critique assessment: 3
.................................................
......!!!!!!!!...................................
19:22:05 `q005. What is the area of a trapezoid with a width of 4.0 km and average altitude of 5.0 km?
......!!!!!!!!...................................
RESPONSE --> The area of a trapezoid with a width of 4km and an average altitude of 5km would be (4km x 5km) + 1/2(4km x 5km)=30km^2 confidence assessment: 1
.................................................
......!!!!!!!!...................................
19:24:16 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.
......!!!!!!!!...................................
RESPONSE --> In remembering the shape of a trapezoid I did not see that it could be turned into a rectangle. I saw the square and the triangle, therefore, I missed the viewing of the rectangle. self critique assessment: 2
.................................................
......!!!!!!!!...................................
19:27:12 `q006. What is the area of a trapezoid whose width is 4 cm in whose altitudes are 3.0 cm and 8.0 cm?
......!!!!!!!!...................................
RESPONSE --> In order to find the area of this trapezoid I would have to first add the two altitudes together and divide by two in order to find the average altitude of the trapezoid. I would add (3+8)/2=5.5 to get the average altitude. Then 4cm x 5.5cm=22cm^2 confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:27:47 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.
......!!!!!!!!...................................
RESPONSE --> The previous question helped me to remember and understand the concept of the area of a traqezoid. self critique assessment: 3
.................................................
......!!!!!!!!...................................
19:29:48 `q007. What is the area of a circle whose radius is 3.00 cm?
......!!!!!!!!...................................
RESPONSE --> The area of a circle is pi(3.14) times radius ^2. Therefore, 3.14 x 3^2= 226.26. confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:31:27 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.
......!!!!!!!!...................................
RESPONSE --> When figuring the answer for area, I put my decimal in the wrong place. This gave me an answer that I should have seen was wrong. self critique assessment: 3
.................................................
......!!!!!!!!...................................
19:33:39 `q008. What is the circumference of a circle whose radius is exactly 3 cm?
......!!!!!!!!...................................
RESPONSE --> To find the circumference of a circle you would use the formula 2 x pi x r. This would give you 2 x 3.14 x 3cm = 18.8cm confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:35:02 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.
......!!!!!!!!...................................
RESPONSE --> In this problem I rounded like the last problem was, so I left the four off of my answer. However, I did the problem correctly and understand the problem. self critique assessment: 3
.................................................
......!!!!!!!!...................................
19:37:45 `q009. What is the area of a circle whose diameter is exactly 12 meters?
......!!!!!!!!...................................
RESPONSE --> Since the formula for a circle is 3.14 x r^2. You would have to divided 12 by 2 to find the radius, since this problem gives you the measure of diameter. Therefore, the answer would be 3.14 x 6^2cm=113.04 cm^2 confidence assessment:
.................................................
......!!!!!!!!...................................
19:38:47 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.
......!!!!!!!!...................................
RESPONSE --> I understand self critique assessment: 3
.................................................
......!!!!!!!!...................................
19:47:57 `q010. What is the area of a circle whose circumference is 14 `pi meters?
......!!!!!!!!...................................
RESPONSE --> In doing the problem you would have to use the formula for the area of a circle and work backwards. You would use 14 = 3.14r^2 and solve for r. You would first move the 3.14 to the opposite side of the equals and get 14/3.14=4.459 rounded. Then you would find the squar root of 4.459, which would give you 2.11 rounded. Therefore the answer would be 2.11cm. confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:52:13 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.
......!!!!!!!!...................................
RESPONSE --> I read the problem as the 14cm to be the area of the circle, therefore, I found only the radius using the formula for area and not circumference. self critique assessment: 3
.................................................
......!!!!!!!!...................................
19:57:06 `q011. What is the radius of circle whose area is 78 square meters?
......!!!!!!!!...................................
RESPONSE --> To find the radius of a circle with an area of 78 you would work backwards for the area formula. Therefore, it would be 78=3.14r^2. You would first have 78/3.14 which would give you 24.84 and then find the square root of 24/84 which would equal 4.98cm for the radius. confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:58:52 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.
......!!!!!!!!...................................
RESPONSE --> The only thing with my answer to this problem was rounding. I am not sure when to round and when not to. Also, when rounding, how many places would I round. Does this even matter? self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:00:20 `q012. Summary Question 1: How do we visualize the area of a rectangle?
......!!!!!!!!...................................
RESPONSE --> I visualize the area of a rectangle by seeing the inside of a four sided figure of which opposite sides are equal. confidence assessment: 1
.................................................
......!!!!!!!!...................................
20:01:03 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.
......!!!!!!!!...................................
RESPONSE --> I was not sure what the question was asking. I understand now. self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:03:05 `q013. Summary Question 2: How do we visualize the area of a right triangle?
......!!!!!!!!...................................
RESPONSE --> I would visualize a right triangle by splitting the triangle in half and both parts would make a square. Therefore, I would use take 1/2 of the formula for a square. 1/2(L)(W) confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:03:30 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.
......!!!!!!!!...................................
RESPONSE --> Understand self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:06:44 `q014. Summary Question 3: How do we calculate the area of a parallelogram?
......!!!!!!!!...................................
RESPONSE --> The area of a parallelogram would be calculated by finding the measurement of the base multiplied by the altitude, because you could take the extra part of the shape and make a rectangle. Therefore, you would use the formula base times the altitude. confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:07:08 The area of a parallelogram is equal to the product of its base and its altitude. The altitude is measured perpendicular to the base.
......!!!!!!!!...................................
RESPONSE --> My explanation was much longer. self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:08:35 `q015. Summary Question 4: How do we calculate the area of a trapezoid?
......!!!!!!!!...................................
RESPONSE --> We can use the formula l x w to find the are of a trapezoid. confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:09:18 We think of the trapezoid being oriented so that its two parallel sides are vertical, and we multiply the average altitude by the width.
......!!!!!!!!...................................
RESPONSE --> I need to remember that it is the average altitude and not just the altitude for the area of a trapezoid. self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:10:07 `q016. Summary Question 5: How do we calculate the area of a circle?
......!!!!!!!!...................................
RESPONSE --> We use the formula area equals pi (3.14) multiplied by the radius ^2 confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:10:25 We use the formula A = pi r^2, where r is the radius of the circle.
......!!!!!!!!...................................
RESPONSE --> Understood self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:14:30 `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?
......!!!!!!!!...................................
RESPONSE --> To find the circumference of a circle we would use the formula A=pi(r^2). To avoid confusing this formula with the formula for circumference you would have to remember that in the formula for circumference C=2(pi)(r) you multiply 2 by 3.14 and by the radius, but in the formula for area you have to multiply pi and the radius after you raised it to the second power. confidence assessment: 1
.................................................
......!!!!!!!!...................................
20:15:32 01-18-2007 20:15:32 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.
......!!!!!!!!...................................
NOTES -------> Circumference is not measured in squared units. This will help me to remember how to separate the two formulas.
.......................................................!!!!!!!!...................................
20:16:49 `q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
......!!!!!!!!...................................
RESPONSE --> I took notes on paper and noted how the information for the problems were worded in order to better understand the process of getting the correct answer. confidence assessment: 3
.................................................
???f~??v??????assignment #001
.................................................
qa prelim Here is assignment 2 ??a??C??}?????assignment #002 002. Describing Graphs qa initial problems 02-14-2007
......!!!!!!!!...................................
20:19:02 `q001. You will frequently need to describe the graphs you have constructed in this course. This exercise is designed to get you used to some of the terminology we use to describe graphs. Please complete this exercise and email your work to the instructor. Note that you should do these graphs on paper without using a calculator. None of the arithmetic involved here should require a calculator, and you should not require the graphing capabilities of your calculator to answer these questions. Problem 1. We make a table for y = 2x + 7 as follows: We construct two columns, and label the first column 'x' and the second 'y'. Put the numbers -3, -2, -1, -, 1, 2, 3 in the 'x' column. We substitute -3 into the expression and get y = 2(-3) + 7 = 1. We substitute -2 and get y = 2(-2) + 7 = 3. Substituting the remaining numbers we get y values 5, 7, 9, 11 and 13. These numbers go into the second column, each next to the x value from which it was obtained. We then graph these points on a set of x-y coordinate axes. Noting that these points lie on a straight line, we then construct the line through the points. Now make a table for and graph the function y = 3x - 4. Identify the intercepts of the graph, i.e., the points where the graph goes through the x and the y axes.
......!!!!!!!!...................................
RESPONSE --> For the problems y=3x-4 I used the numbers for my x column -3, -2, -1, 0, 1, 2, 3 and the corresponding y column for each of these numbers is -13, -10, -7, -4, -1, 2 and 5. When putting these points onto the graph it shows an increase for the slope of 3. The point that is crossed over on the x-axis is (4,0) and the point where it is crossed over on' the y-axis is (1 1/3, 0). confidence assessment: 2
.................................................
......!!!!!!!!...................................
20:47:38 The graph goes through the x axis when y = 0 and through the y axis when x = 0. The x-intercept is therefore when 0 = 3x - 4, so 4 = 3x and x = 4/3. The y-intercept is when y = 3 * 0 - 4 = -4. Thus the x intercept is at (4/3, 0) and the y intercept is at (0, -4). Your graph should confirm this.
......!!!!!!!!...................................
RESPONSE --> self critique assessment:
.................................................
......!!!!!!!!...................................
20:49:34 `q002. Does the steepness of the graph in the preceding exercise (of the function y = 3x - 4) change? If so describe how it changes.
......!!!!!!!!...................................
RESPONSE --> The steepness of the graph for the problem y = 3x - 4 is the same throughout the problem. The slope is increasing by 3 each time the interval increases by 1. confidence assessment: 3
.................................................
I am not sure what assignment this is because it says assignment 1 and assignment 2 on it. ??????????R???d? assignment #001 001. Rates qa rates 02-21-2007 ???????????????assignment #002 002. Volumes qa areas volumes misc 02-21-2007
......!!!!!!!!...................................
19:31:52 `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?
......!!!!!!!!...................................
RESPONSE --> To find the volume of a rectangular solid you would use the formula v=lwh. Therefore you would have v=3x5x7, volume would be 105cm^2 confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:35:16 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.
......!!!!!!!!...................................
RESPONSE --> My answer was correct accept for the correct power notation. This is where I forget and am reviewing the information as I go forward in the problems. self critique assessment: 3
.................................................
......!!!!!!!!...................................
19:38:08 `q002. What is the volume of a rectangular solid whose base area is 48 square meters and whose altitude is 2 meters?
......!!!!!!!!...................................
RESPONSE --> You would use the formula for volume which is v=A*h. Therefore you would get v=48*2m, which would give you 96cm^3 confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:38:32 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.
......!!!!!!!!...................................
RESPONSE --> Review is good. self critique assessment: 3
.................................................
......!!!!!!!!...................................
19:40:50 `q003. What is the volume of a uniform cylinder whose base area is 20 square meters and whose altitude is 40 meters?
......!!!!!!!!...................................
RESPONSE --> In the previous problem it was noted that the formula for volume was used with other solid shapes, therefore, you would still use the same formula. In using that formula you would have v=20m^2*40m= 800m^3 for the volume of a uniform cylinder. confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:41:18 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.
......!!!!!!!!...................................
RESPONSE --> Understand self critique assessment: 3
.................................................
......!!!!!!!!...................................
19:45:11 `q004. What is the volume of a uniform cylinder whose base has radius 5 cm and whose altitude is 30 cm?
......!!!!!!!!...................................
RESPONSE --> In this problem it does not state that it is a solid therefore you would use the formula for the volumer of a cylinder. V=pi*r^2*h V=(3.14)(5cm^2)(30cm) V=2355cm^2 confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:49:42 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.
......!!!!!!!!...................................
RESPONSE --> Why would you not multiply the pi to the rest of the problem in the answer 750 pi cm^3 self critique assessment: 0
.................................................
......!!!!!!!!...................................
19:58:00 `q005. Estimate the dimensions of a metal can containing food. What is its volume, as indicated by your estimates?
......!!!!!!!!...................................
RESPONSE --> In estimating a 16oz can of food to get the volume, I got radius of 1.5 inches and height of 3.25 inches. Therefore, I would use the formula V=A*h and have V=pi 2.25 in * 3.25 in V=7.3125 pi inches ^3 confidence assessment: 1
.................................................
......!!!!!!!!...................................
20:03:38 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.
......!!!!!!!!...................................
RESPONSE --> I am still unclear as to why the answer is noted with the pi in it instead of using 3.14 to solve the rest of the problem. self critique assessment: 0
.................................................
......!!!!!!!!...................................
20:04:56 `q006. What is the volume of a pyramid whose base area is 50 square cm and whose altitude is 60 cm?
......!!!!!!!!...................................
RESPONSE --> By using the fomula V=A*h you would get V=50cm^2 * 60cm V=3000cm ^3 confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:06:47 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.
......!!!!!!!!...................................
RESPONSE --> On this problem I confused a pyramid with the other shapes that we had been using in the problems. Therefore, I used the wrong formula to answer the problem. self critique assessment: 1
.................................................
......!!!!!!!!...................................
20:10:07 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.
......!!!!!!!!...................................
RESPONSE --> self critique assessment:
.................................................
......!!!!!!!!...................................
20:16:58 `q008. What is a volume of a sphere whose radius is 4 meters?
......!!!!!!!!...................................
RESPONSE --> We would use the formula for finding the volume of a sphere, which is V=4/3 * pi * r^3 V= 4/3 * pi * 4m^3 V=85 cm pi ^3 confidence assessment: 1
.................................................
......!!!!!!!!...................................
20:22:09 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.
......!!!!!!!!...................................
RESPONSE --> When I put my answer of 85 pi m^3 I rounded my answer after dividing 256/3. Would this be proper notation? self critique assessment: 1
.................................................
......!!!!!!!!...................................
20:27:52 `q009. What is the volume of a planet whose diameter is 14,000 km?
......!!!!!!!!...................................
RESPONSE --> For this problem I would use the formula for a sphere V=4/3 * pi * r^3 V=4/3 * pi * 14000/2km ^3 V = 1,372,000,000,000 pi km^3 confidence assessment: 1
.................................................
......!!!!!!!!...................................
20:28:37 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.
......!!!!!!!!...................................
RESPONSE --> I understand how you want the answer noted, but again don't understand why. self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:30:45 `q010. Summary Question 1: What basic principle do we apply to find the volume of a uniform cylinder of known dimensions?
......!!!!!!!!...................................
RESPONSE --> The basic principle that we can apply to a known dimension of not just a cylinder but many shapes is the area. confidence assessment: 1
.................................................
......!!!!!!!!...................................
20:31:20 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.
......!!!!!!!!...................................
RESPONSE --> I understood the concept, but did not respond in this way. self critique assessment: 1
.................................................
......!!!!!!!!...................................
20:33:22 `q011. Summary Question 2: What basic principle do we apply to find the volume of a pyramid or a cone?
......!!!!!!!!...................................
RESPONSE --> We have to keep in mind that these two shapes get smaller and smaller toward the point of the shape. Therefore we would use the formula V=1/3 * pi *r^2 *h confidence assessment: 1
.................................................
......!!!!!!!!...................................
20:33:45 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.
......!!!!!!!!...................................
RESPONSE --> I understand self critique assessment:
.................................................
......!!!!!!!!...................................
20:34:30 `q012. Summary Question 3: What is the formula for the volume of a sphere?
......!!!!!!!!...................................
RESPONSE --> The volume of a sphere is as follows: V=4/3 *pi *r^3 confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:34:38 The volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere.
......!!!!!!!!...................................
RESPONSE --> self critique assessment:
.................................................
......!!!!!!!!...................................
20:35:28 `q013. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
......!!!!!!!!...................................
RESPONSE --> I have saved numerous notations in the answers to review. I also worked problems out on paper and double checked answers. confidence assessment: "