course Mth 271 I apologize for resubmitting the first assignment twice. I forgot to wait until I finished everything to submit all work for today.
......!!!!!!!!...................................
12:57:45 `q001. The graph of a certain function is a smooth curve passing through the points (3, 5), (7, 17) and (10, 29). Between which two points do you think the graph is steeper, on the average? Why do we say 'on the average'?
......!!!!!!!!...................................
RESPONSE --> The graph would be steeper between (10, 29) because of the distance between these two points. confidence assessment: 1
.................................................
......!!!!!!!!...................................
13:00:21 Slope = rise / run. Between points (7, 17) and (10, 29) we get rise / run = (29 - 17) / (10 - 7) =12 / 3 = 4. The slope between points (3, 5) and (7, 17) is 3 / 1. (17 - 5) / (7 -3) = 12 / 4 = 3. The segment with slope 4 is the steeper. The graph being a smooth curve, slopes may vary from point to point. The slope obtained over the interval is a specific type of average of the slopes of all points between the endpoints.
......!!!!!!!!...................................
RESPONSE --> I understand where I messed up now, I thought the question asked for one point, and not 2 points, but now, looking at the slope forms, and solving those, I can see that between points (7, 17) and (10, 29) is the steeper slope. I also understand the average concept now. It is the average because the steeper of these two points is an average of all three between the endpoints. self critique assessment: 2
.................................................
......!!!!!!!!...................................
13:02:51 2. Answer without using a calculator: As x takes the values 2.1, 2.01, 2.001 and 2.0001, what values are taken by the expression 1 / (x - 2)? 1. As the process continues, with x getting closer and closer to 2, what happens to the values of 1 / (x-2)? 2. Will the value ever exceed a billion? Will it ever exceed one trillion billions? 3. Will it ever exceed the number of particles in the known universe? 4. Is there any number it will never exceed? 5. What does the graph of y = 1 / (x-2) look like in the vicinity of x = 2?
......!!!!!!!!...................................
RESPONSE --> 1. The values of 1 / (x-2) will very slowly get larger, but at a very slow pace. 2. It will eventually, but would take a very long time. 3. Very very slowly, yes. 4. No, it will continue towards infinity 5. A straight line, on the y axis self critique assessment: 1
.................................................
......!!!!!!!!...................................
13:05:53 For x = 2.1, 2.01, 2.001, 2.0001 we see that x -2 = .1, .01, .001, .0001. Thus 1/(x -2) takes respective values 10, 100, 1000, 10,000. It is important to note that x is changing by smaller and smaller increments as it approaches 2, while the value of the function is changing by greater and greater amounts. As x gets closer in closer to 2, it will reach the values 2.00001, 2.0000001, etc.. Since we can put as many zeros as we want in .000...001 the reciprocal 100...000 can be as large as we desire. Given any number, we can exceed it. Note that the function is simply not defined for x = 2. We cannot divide 1 by 0 (try counting to 1 by 0's..You never get anywhere. It can't be done. You can count to 1 by .1's--.1, .2, .3, ..., .9, 1. You get 10. You can do similar thing for .01, .001, etc., but you just can't do it for 0). As x approaches 2 the graph approaches the vertical line x = 2; the graph itself is never vertical. That is, the graph will have a vertical asymptote at the line x = 2. As x approaches 2, therefore, 1 / (x-2) will exceed all bounds. Note that if x approaches 2 through the values 1.9, 1.99, ..., the function gives us -10, -100, etc.. So we can see that on one side of x = 2 the graph will approach +infinity, on the other it will be negative and approach -infinity.
......!!!!!!!!...................................
RESPONSE --> I see my mistake now, it was only in the fact that the numbers would getlarger slowly,they will get larger faster. I see not that we cannot divide by zero. So instead of a straight line, the function simply does not work because of this. self critique assessment: 2
.................................................
......!!!!!!!!...................................
13:24:20 `q003. One straight line segment connects the points (3,5) and (7,9) while another connects the points (10,2) and (50,4). From each of the four points a line segment is drawn directly down to the x axis, forming two trapezoids. Which trapezoid has the greater area? Try to justify your answer with something more precise than, for example, 'from a sketch I can see that this one is much bigger so it must have the greater area'.
......!!!!!!!!...................................
RESPONSE --> Area =(b1 + b2/2) * h The trapezoid containing point 10, 2 and 50, 4 has the greater area. A= (40 + 40 / 2) * 4 = 160 confidence assessment: 1
.................................................
......!!!!!!!!...................................
13:28:03 Your sketch should show that while the first trapezoid averages a little more than double the altitude of the second, the second is clearly much more than twice as wide and hence has the greater area. To justify this a little more precisely, the first trapezoid, which runs from x = 3 to x = 7, is 4 units wide while the second runs from x = 10 and to x = 50 and hence has a width of 40 units. The altitudes of the first trapezoid are 5 and 9,so the average altitude of the first is 7. The average altitude of the second is the average of the altitudes 2 and 4, or 3. So the first trapezoid is over twice as high, on the average, as the first. However the second is 10 times as wide, so the second trapezoid must have the greater area. This is all the reasoning we need to answer the question. We could of course multiply average altitude by width for each trapezoid, obtaining area 7 * 4 = 28 for the first and 3 * 40 = 120 for the second. However if all we need to know is which trapezoid has a greater area, we need not bother with this step.
......!!!!!!!!...................................
RESPONSE --> I understand this reasoning as well. I simply entered into the area formula. But now I can see how we can arrive at a better answer. The reasoning is that even though the first trapezoid is over twice as high as the second, the second is 10 times wider, therefore it must have the greater area. self critique assessment: 2
.................................................
......!!!!!!!!...................................
13:30:49 `q004. If f(x) = x^2 (meaning 'x raised to the power 2') then which is steeper, the line segment connecting the x = 2 and x = 5 points on the graph of f(x), or the line segment connecting the x = -1 and x = 7 points on the same graph? Explain the basisof your reasoning.
......!!!!!!!!...................................
RESPONSE --> The point x=-1 and x=7 would be the steeper line. When solved for f(x), the points become x=1 and x=49. The other points X=2 and x=5 solve as x=4 and x=25. The steepness between 1 and 49 is much steeper than 4 and 25. confidence assessment: 2
.................................................
......!!!!!!!!...................................
13:33:08 The line segment connecting x = 2 and the x = 5 points is steeper: Since f(x) = x^2, x = 2 gives y = 4 and x = 5 gives y = 25. The slope between the points is rise / run = (25 - 4) / (5 - 2) = 21 / 3 = 7. The line segment connecting the x = -1 point (-1,1) and the x = 7 point (7,49) has a slope of (49 - 1) / (7 - -1) = 48 / 8 = 6. The slope of the first segment is greater.
......!!!!!!!!...................................
RESPONSE --> I see my mistake now. I didn't enter y in the answer and was off by not using the slope formula. When entering these two sets of points in the equation I clearly see that the first set of points has a steeper slope than the second. self critique assessment: 2
.................................................
......!!!!!!!!...................................
13:36:13 `q005. Suppose that every week of the current millenium you go to the jewler and obtain a certain number of grams of pure gold, which you then place in an old sock and bury in your backyard. Assume that buried gold lasts a long, long time ( this is so), that the the gold remains undisturbed (maybe, maybe not so), that no other source adds gold to your backyard (probably so), and that there was no gold in your yard before.. 1. If you construct a graph of y = the number of grams of gold in your backyard vs. t = the number of weeks since Jan. 1, 2000, with the y axis pointing up and the t axis pointing to the right, will the points on your graph lie on a level straight line, a rising straight line, a falling straight line, a line which rises faster and faster, a line which rises but more and more slowly, a line which falls faster and faster, or a line which falls but more and more slowly? 2. Answer the same question assuming that every week you bury 1 more gram than you did the previous week. {}3. Answer the same question assuming that every week you bury half the amount you did the previous week.
......!!!!!!!!...................................
RESPONSE --> 1. A line that rises. This is because as the weeks pass, you are adding more and more gold. A certain amount of gold would lead to the straight line rising. 2.This would change into line that rises faster and faster because each week would lend to more and more gold being bought. 3. This would change the line into a descending line that falls faster and faster because as you decrease by half, the amount decreases drastically. confidence assessment: 2
.................................................
......!!!!!!!!...................................
13:37:15 1. If it's the same amount each week it would be a straight line. 2. Buying gold every week, the amount of gold will always increase. Since you buy more each week the rate of increase will keep increasing. So the graph will increase, and at an increasing rate. 3. Buying gold every week, the amount of gold won't ever decrease. Since you buy less each week the rate of increase will just keep falling. So the graph will increase, but at a decreasing rate. This graph will in fact approach a horizontal asymptote, since we have a geometric progression which implies an exponential function.
......!!!!!!!!...................................
RESPONSE --> 3. I see now that the line would increase, but at a decreasing rate because we are still adding gold to the yard, just smaller and smaller amounts. self critique assessment: 2
.................................................
......!!!!!!!!...................................
13:38:54 `q006. Suppose that every week you go to the jewler and obtain a certain number of grams of pure gold, which you then place in an old sock and bury in your backyard. Assume that buried gold lasts a long, long time, that the the gold remains undisturbed, and that no other source adds gold to your backyard. 1. If you graph the rate at which gold is accumulating from week to week vs. tne number of weeks since Jan 1, 2000, will the points on your graph lie on a level straight line, a rising straight line, a falling straight line, a line which rises faster and faster, a line which rises but more and more slowly, a line which falls faster and faster, or a line which falls but more and more slowly? 2. Answer the same question assuming that every week you bury 1 more gram than you did the previous week. 3. Answer the same question assuming that every week you bury half the amount you did the previous week.
......!!!!!!!!...................................
RESPONSE --> 1. The points lie on a rising straight line. 2. The points will lie on a rising line, because you are adding more gold from week to week. 3. The points will lie on a rising line, but it will change to a decreasing rate because you are adding less to the pile. confidence assessment: 3
.................................................
......!!!!!!!!...................................
13:39:56 This set of questions is different from the preceding set. This question now asks about a graph of rate vs. time, whereas the last was about the graph of quantity vs. time. Question 1: This question concerns the graph of the rate at which gold accumulates, which in this case, since you buy the same amount eact week, is constant. The graph would be a horizontal straight line. Question 2: Each week you buy one more gram than the week before, so the rate goes up each week by 1 gram per week. You thus get a risingstraight line because the increase in the rate is the same from one week to the next. Question 3. Since half the previous amount will be half of a declining amount, the rate will decrease while remaining positive, so the graph remains positive as it decreases more and more slowly. The rate approaches but never reaches zero.
......!!!!!!!!...................................
RESPONSE --> 2. I see now that adding a constant 1 gram a week, the line would still be straight and increase at the same rate each time. self critique assessment: 2
.................................................
......!!!!!!!!...................................
13:46:31 ``q007. If the depth of water in a container is given, in centimeters, by 100 - 2 t + .01 t^2, where t is clock time in seconds, then what are the depths at clock times t = 30, t = 40 and t = 60? On the average is depth changing more rapidly during the first time interval or the second?
......!!!!!!!!...................................
RESPONSE --> Depth at t=30 is 49, t=40 is 36 and t=60 is 16 t=30 100-2*(30)+.01*(60)^2 100-60+.01*(900) 100-60+9 49 t=40 100-2*(40)+.01(40)^2 100-80+.01(1600) 20+16 36 t=60 100-2*(60)+.01*(60)^2 100-120+.01*(3600) -20+36 16 The depth is changing more rapidly between the second interval because it drops more drastically than in the first one. self critique assessment: 2
.................................................
......!!!!!!!!...................................
13:47:01 At t = 30 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 30 + .01 * 30^2 = 49. At t = 40 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 40 + .01 * 40^2 = 36. At t = 60 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 60 + .01 * 60^2 = 16. 49 cm - 36 cm = 13 cm change in 10 sec or 1.3 cm/s on the average. 36 cm - 16 cm = 20 cm change in 20 sec or 1.0 cm/s on the average.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
13:55:31 `q008. If the rate at which water descends in a container is given, in cm/s, by 10 - .1 t, where t is clock time in seconds, then at what rate is water descending when t = 10, and at what rate is it descending when t = 20? How much would you therefore expect the water level to change during this 10-second interval?
......!!!!!!!!...................................
RESPONSE --> t=10 10-.1*(10 10-1 9 t=20 10-.1*(20) 10-2 8 I would expect the water level to change by 1 cm in this interval (9-8=1) confidence assessment: 2
.................................................
......!!!!!!!!...................................
13:57:03 At t = 10 sec the rate function gives us 10 - .1 * 10 = 10 - 1 = 9, meaning a rate of 9 cm / sec. At t = 20 sec the rate function gives us 10 - .1 * 20 = 10 - 2 = 8, meaning a rate of 8 cm / sec. The rate never goes below 8 cm/s, so in 10 sec the change wouldn't be less than 80 cm. The rate never goes above 9 cm/s, so in 10 sec the change wouldn't be greater than 90 cm. Any answer that isn't between 80 cm and 90 cm doesn't fit the given conditions.. The rate change is a linear function of t. Therefore the average rate is the average of the two rates, or 9.5 cm/s. The average of the rates is 8.5 cm/sec. In 10 sec that would imply a change of 85 cm.
......!!!!!!!!...................................
RESPONSE --> I see now that the rate doesn't fall from 8 cm/s and exceed 9cm/s. So the average rate would be 9+8/2=8.5 self critique assessment: 2
.................................................
?????z?l?Z?? assignment #001 001. Rates qa rates 01-15-2009
......!!!!!!!!...................................
14:04:52 `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 --> When using this program, first click on next question/answer. After reviewing the question, come to this box (the right hand box) and State the answer here in detail. Then click ""Enter response"" Now click Next question again to reveal the answer. If you got anything wrong, or learned anything, this is the time to state in detail what you learned or why you missed an answer. If you want to save any part as your notes, click ""Save as notes"" and it will create a folder in the vhmthphy folder on your computer with the name of this assignment with it. confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:05:03 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
.................................................
......!!!!!!!!...................................
14:06:11 `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 $50 in 5 hr, then at what rate are you earning money?
......!!!!!!!!...................................
RESPONSE --> 50/5=10 You are making 10$/ hr because if you make 50$ in 5 hours, you divide 50 by 5 to get the dollar amount for one hour. confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:06:37 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 --> confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:08:11 `q003.If you make $60,000 per year then how much do you make per month?
......!!!!!!!!...................................
RESPONSE --> 60000/12=5000 There are 12 months in each year. For these 12 months you made $60,000. To arrive at the rate of pay for 1 month, you have to divide the total amount of 1 year 60,000 by the number of months in a year 12 which gives you $5,000 a month. confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:08:39 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 --> ok confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:09:51 `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 --> It would be more appropriate to state that on average, the business makes $5000 a month because we are only given one month, and it would be impossible to gauge all other months by this amount. confidence assessment: 2
.................................................
......!!!!!!!!...................................
14:10:45 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 --> I didn't write the other possibilities that could occur in all months a business in service, but I do see that outside elements contribute to a business's income. confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:12:14 `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 --> 300/6= 50 mph We are covering approximately 50 miles per hour. This is arrived at an average rate because it is assumed that the car would sometimes go above 50 and others below 50. No car can travel the exact amount of 50 mph for the full 6 hours. confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:13:22 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 --> I also see that the outside elements can cause a car to slow or speed up at any given time. I also see that we divide the total distance covered by a time amount of an hour to arrive at our average rate. confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:15:53 `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 --> 60/1200=.05 We take the total amount of gasoline used (60 gallons) and divide that by total miles traveled (1200) to arrive at a average usage rate of .05 gallons per mile. confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:17:11 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 see and understand that this problem didn't ask for any time measurement, which almost got me confused, until I realized they wanted to gallons as opposed to miles, minutes etc... confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:17:59 `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 --> The addition has already been done for us. We are given the total amounts, and all there is left to do is divide to arrive at the average. confidence assessment: 2
.................................................
......!!!!!!!!...................................
14:18:52 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 --> I understand that the reason we are using average amounts is because we are using examples where the idea that there is a constant is highly unlikely. confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:24:41 `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 --> Because the first group increased lifting strength ave. 147 lb. We take 147 and divide by the total days in a year (365) to arrive at an ave rate of .40 lb per day. In the second group, we take the ave 162 lb and divide also by the days in a year (365) to arrive at an ave of .44 lb per day. confidence assessment: 2
.................................................
......!!!!!!!!...................................
14:27:09 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 --> I didn't understand how to calculate the rate at first solving the problem, but now I see the solution. Because the second group lifted 15 more pounds that group one, and did 40 more pushups per day; we take the amount of weight difference and divide that by the difference in pushups to arrive at the pounds per pushup that was gained in the process. confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:33:33 `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 second group lifted an extra 20 lb of weight each day compared to the first group, and as a result lifeted a total of 17 more lb total. By dividing the total weight more (17lb) by the daily weight (20 lb) we arrive at an extra .85 lb each day. confidence assessment: 1
.................................................
......!!!!!!!!...................................
14:34:21 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 --> The difference was actually .85 lifting per pound of extra weight, which I didn't see it my working of the problem but I see now. confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:39:45 `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 runner passed the 100 meter mark in 12 seconds The runner passed the 200 meter mark 22 seconds after the start. We can deduce that the runner ran an additional 100 meters in 10 seconds because it took the runner 12 to reach the first 100 meters. With this information, we can divide the total meters ran between these positions by the total time taken between the two positions to arrive at the answer. 100 m/ 10 sec= 10 meters per second between the position. confidence assessment: 1
.................................................
......!!!!!!!!...................................
14:40:23 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 --> I can see that this is an average rate because the runner will not carry the constant speed throughout the race, it will most likely flucuate. confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:42:06 `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 --> From the information provided, we can average the two rates the runner was passing in 100 meter incriments to arrive at an average distance of 9 + 8 / 2= 8.5 seconds per 100 meters. confidence assessment: 1
.................................................
......!!!!!!!!...................................
14:44:06 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 --> I can see now that by taking the ave meter per second ( 10 = 9 / 2= 9.5 m/s) and divide that by 100 meters to arrive at the speed of 10.5 seconds ( 100/9.5) confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:44:54 `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 --> Before we had the total number needed to divide to get the average. This time, we were given both times that we were required to add in order to divide. confidence assessment: 2
.................................................
......!!!!!!!!...................................
14:45:47 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 --> I understand now that the previous examples were not that of rates. This example we were given 2 rates in need of averaging instead of the amount of change already. confidence assessment: 2
.................................................
??????x?I??? assignment #001 001. Areas qa areas volumes misc 01-15-2009
......!!!!!!!!...................................
14:51:34 `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 --> The area of a rectangle is length * width. Therefore area = 4 * 3 = 12 confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:53:31 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 understand now that instead of just 12, we have to say 12m^2 because we can divide a rectangle into 12 individual squares, and together would be 12 square meters self critique assessment: 2
.................................................
......!!!!!!!!...................................
14:57:06 `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?
......!!!!!!!!...................................
RESPONSE --> The area of a right triangle is 1/2 * base * height. Therefore the area of this triangle = 1/2 * 4 * 3 = 6m^2. confidence assessment: 2
.................................................
......!!!!!!!!...................................
14:58:58 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 --> I can see now that we can use the area formula for a rectangle because we can join two identical right triangles to form a rectangle with the area of 12 m^2, and can take half of that because each triangle is half of the rectangle to arrive at the answer. self critique assessment: 2
.................................................
......!!!!!!!!...................................
15:00:18 `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 also length * width. This would translate into A= 5 *2=10m^2. The altitude is the height, while the base is the length of the parallelogram. confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:00:59 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 --> I also understand that by cutting of the end of a parallelogram, we have a rectangle remaining, which is also the same formula. self critique assessment: 2
.................................................
......!!!!!!!!...................................
15:02:06 `q004. What is the area of a triangle whose base is 5.0 cm and whose altitude is 2.0 cm?
......!!!!!!!!...................................
RESPONSE --> The formula for a triangle is 1/2 * base * height. A= 1/2 * 5 *2=2.5*2=5m^2 confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:03:23 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 --> I can understand that joining any triangle with its copy will form a parallelogram, which formula is b*h. To solve for 1 triangle we simply take 1/2 base and multiply by the height. self critique assessment: 2
.................................................
......!!!!!!!!...................................
15:05:03 `q005. What is the area of a trapezoid with a width of 4.0 km and average altitude of 5.0 km?
......!!!!!!!!...................................
RESPONSE --> We can find the area of this trapezoid by multiplying the width of 4.0 km by the average altitude of 5.0 km to arrive at an area of 20.0 km^2 confidence assessment: 1
.................................................
......!!!!!!!!...................................
15:06:05 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 --> I see that by reconstructing the trapezoid by the width and the average of 2 altitudes will lend to a formula that can be solved for the trapezoid. self critique assessment: 2
.................................................
......!!!!!!!!...................................
15:07:41 `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 --> By averaging the 2 altitudes of this trapezoid (3 + 8 / 2=5.5 cm) We then can multiply the width 4 cm by the ave. altitude of 5.5 cm to arrive at an area of 22 cm^2. We can do this by reconstructing the trapezoid into a rectangle and plugging in the formula. confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:07:56 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 --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:11:18 `q007. What is the area of a circle whose radius is 3.00 cm?
......!!!!!!!!...................................
RESPONSE --> The area of a circle is pi*r2. Because we have the radius in this equation we can solve as follows. A= pi * r^2 A= pi* 3.00cm^2 A= pi * 9 A= 28.27 if pi is used on a calculator, but the answer comes to 28.26 if 3.14 is used. confidence assessment: 2
.................................................
......!!!!!!!!...................................
15:12:12 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 --> I see now that the answer 9 pi cm^2 is more exact that 28.26 because this is an approximation. 9 pi cm^2 is the most exact answer we can get. self critique assessment: 2
.................................................
......!!!!!!!!...................................
15:13:24 `q008. What is the circumference of a circle whose radius is exactly 3 cm?
......!!!!!!!!...................................
RESPONSE --> The circumference of a circle in 2* pi * r. Thus C= 2 * pi * 3 cm C= 6 pi cm We use pi because again it is the most exact form we can use. confidence assessment: 2
.................................................
......!!!!!!!!...................................
15:14:05 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 --> I understand that circumference does not include cm^2. If this arises, I see something would be wrong. self critique assessment: 2
.................................................
......!!!!!!!!...................................
15:15:51 `q009. What is the area of a circle whose diameter is exactly 12 meters?
......!!!!!!!!...................................
RESPONSE --> Because diameter is all that is given in this problem. We first have to find the radius. The diameter of a circle is twice the radius. Using this info. We can see that R= 12 m / 2= 6 m. Now having the radius, we can solve for area. A= pi * r^2 A= pi * 6 ^2 A = 12 pi m^2 We use pi to get the most exact answer possible. confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:16:39 The area of a circle is A = pi r^2, where r is the radius. The radius of this circle is half the 12 m diameter, or 6 m. So the area is A = pi ( 6 m )^2 = 36 pi m^2. This result can be approximated to any desired accuracy by using a sufficient number of significant figures in our approximation of pi. For example using the 5-significant-figure approximation pi = 3.1416 we obtain A = 36 m^2 * 3.1416 = 113.09 m^2.
......!!!!!!!!...................................
RESPONSE --> I see what I did wrong, 6 ^2 is 36, not 12. I also see that we can use any decimal point in the pi sequence to get a more accurate account. self critique assessment: 2
.................................................
......!!!!!!!!...................................
15:19:44 `q010. What is the area of a circle whose circumference is 14 `pi meters?
......!!!!!!!!...................................
RESPONSE --> Because the equation for circumference is pi * diameter, we can deduce that the diameter of this circle is 14 meters. The radius of a circle is half of the diameter. So we take half of 14 m to get a radius of 7 m. Now we can find the area of a circle. A= pi * r^2 A= pi * 7^2 A= 49 pi cm^2 confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:20:38 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 also can see we can find the radius of the circle by dividing C by 2 pi. self critique assessment: 2
.................................................
......!!!!!!!!...................................
15:24:04 `q011. What is the radius of circle whose area is 78 square meters?
......!!!!!!!!...................................
RESPONSE --> The area of a circle is A=pi * r^2 78 = pi * r^2 78/pi = r^2 24.83= r^2 square root of 24.83 = r 4.98= r confidence assessment: 1
.................................................
......!!!!!!!!...................................
15:26:49 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 --> I solved up to r = sqrt(A/pi) but instead of leaving letters, I plugged the numbers, I see now how this works. I did come up with 4.98, but I can see we can round up to 5 to get the right answer. self critique assessment: 2
.................................................
......!!!!!!!!...................................
15:27:29 `q012. Summary Question 1: How do we visualize the area of a rectangle?
......!!!!!!!!...................................
RESPONSE --> We can visualize the area of a rectangle by taking the base (how long) and multiplying by the height ( how tall/altitude) confidence assessment: 2
.................................................
......!!!!!!!!...................................
15:27:48 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 --> Also we visualize as 1-unit squares. self critique assessment: 2
.................................................
......!!!!!!!!...................................
15:28:33 `q013. Summary Question 2: How do we visualize the area of a right triangle?
......!!!!!!!!...................................
RESPONSE --> The area of a right triangle can be double with another right triangle at the hypotenuse to form a rectangle that we can solve by l * w, and taking half of the answer. confidence assessment: 2
.................................................
......!!!!!!!!...................................
15:29:10 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 --> The length of this rectangle is also equal to the base and height of the triangle. self critique assessment: 2
.................................................
......!!!!!!!!...................................
15:29:54 `q014. Summary Question 3: How do we calculate the area of a parallelogram?
......!!!!!!!!...................................
RESPONSE --> The area of a parallelogram can be calculated by cutting off the protrduing end to equal a rectangle thats height and base are equal to the parallelogram and multiplying l * w. confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:30:00 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 --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:30:36 `q015. Summary Question 4: How do we calculate the area of a trapezoid?
......!!!!!!!!...................................
RESPONSE --> The area of a trapezoid can be calculated by averaging 2 of its base and multiplying by its height. confidence assessment: 2
.................................................
......!!!!!!!!...................................
15:30:55 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 --> Average altitude instead of base. self critique assessment: 2
.................................................
......!!!!!!!!...................................
15:31:20 `q016. Summary Question 5: How do we calculate the area of a circle?
......!!!!!!!!...................................
RESPONSE --> The area of a circle is pi multiplied by the radius squared. confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:31:27 We use the formula A = pi r^2, where r is the radius of the circle.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:32:33 `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 --> The circumference of a circle is pi * diameter, or 2 * pi * r. We can easily avoid confusion by noticing that the circumference is in units, where the area is square units. confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:32:43 We use the formula C = 2 pi r. The formula for the area involves r^2, which will give us squared units of the radius. Circumference is not measured in squared units.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:33:30 `q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
......!!!!!!!!...................................
RESPONSE --> I have taken notes at each question to ensure I have refreshed my memories of these formulas, also I have studied my responses as opposed to those given to me to reconcile any differences I had. confidence assessment: 3
.................................................
??????j????????assignment #002 002. Volumes qa areas volumes misc 01-15-2009
......!!!!!!!!...................................
20:42:57 `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 determine the volume of a rectangular solid, we simply multiply length * width * height V= l * w * h V= 3cm * 5cm * 7cm V=105 cm confidence assessment: 2
.................................................
......!!!!!!!!...................................
20:44:36 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 --> I can see now that we can further break the formula into A * h, because the l * w of the rectangular solid is the area of a solid rectangle. self critique assessment: 2
.................................................
......!!!!!!!!...................................
20:46:11 `q002. What is the volume of a rectangular solid whose base area is 48 square meters and whose altitude is 2 meters?
......!!!!!!!!...................................
RESPONSE --> Because the area of a rectangular solid is equal to the l * w of its sides, we can deduce that the volume of a rectangular solid is area * h V= A * h V= 48 m^2 * 2 m V= 96 m^3 confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:46:25 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 --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:49:04 `q003. What is the volume of a uniform cylinder whose base area is 20 square meters and whose altitude is 40 meters?
......!!!!!!!!...................................
RESPONSE --> The formula for the volume is A * h, or pi *r^2. Since we have the area in this problem, the volume of the cylinder would be as follows: V= A * h V= 20 m^2 * 40m V= 800 * m* m* m V= 800 m^3 confidence assessment: 2
.................................................
......!!!!!!!!...................................
20:49:51 V = A * h applies to uniform cylinders as well as to rectangular solids. We are given the altitude h and the base area A so we conclude that V = A * h = 20 m^2 * 40 m = 800 m^3. The relationship V = A * h applies to any solid object whose cross-sectional area A is constant. This is the case for uniform cylinders and uniform prisms.
......!!!!!!!!...................................
RESPONSE --> I understand now that the relationship of Volume and area applies to any object that the area is constant. self critique assessment: 2
.................................................
......!!!!!!!!...................................
20:53:18 `q004. What is the volume of a uniform cylinder whose base has radius 5 cm and whose altitude is 30 cm?
......!!!!!!!!...................................
RESPONSE --> This equation calls for using volume = pi*r^2 * h because area is not given. This can be solved 2 ways. First we will find the circles area. A= pi*r^2 A= pi* 5cm^2 A= 25 * pi We now can determine volume by the equation: V= A * h V= 25cm* pi * 30 cm V=750* pi cm^3 confidence assessment: 2
.................................................
......!!!!!!!!...................................
20:53:52 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 --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:58:22 `q005. Estimate the dimensions of a metal can containing food. What is its volume, as indicated by your estimates?
......!!!!!!!!...................................
RESPONSE --> The estimate I say is 12cm around by 6 cm height by 3 cm wide This can be answered by using the formula, V= A * h We now need to discover the area of the cylinder. We use 6 as the radius, because the diameter is 12 and radius is half of the dieameter which is 6 cm. A= pi * r^2 A= pi * 6^2 A= 36 * pi cm^2 Now we can use Area to determine volume. V= A * h V= 36 * pi cm^2 * 6 cm V= 216 cm^3 confidence assessment: 2
.................................................
......!!!!!!!!...................................
20:59:04 People will commonly estimate the dimensions of a can of food in centimeters or in inches, though other units of measure are possible (e.g., millimeters, feet, meters, miles, km). Different cans have different dimensions, and your estimate will depend a lot on what can you are using. A typical can might have a circular cross-section with diameter 3 inches and altitude 5 inches. This can would have volume V = A * h, where A is the area of the cross-section. The diameter of the cross-section is 3 inches so its radius will be 3/2 in.. The cross-sectional area is therefore A = pi r^2 = pi * (3/2 in)^2 = 9 pi / 4 in^2 and its volume is V = A * h = (9 pi / 4) in^2 * 5 in = 45 pi / 4 in^3. Approximating, this comes out to around 35 in^3. Another can around the same size might have diameter 8 cm and height 14 cm, giving it cross-sectional area A = pi ( 4 cm)^2 = 16 pi cm^2 and volume V = A * h = 16 pi cm^2 * 14 cm = 224 pi cm^2.
......!!!!!!!!...................................
RESPONSE --> My estimates weren't the same as these, but the steps I performed were the same as these. self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:02:06 `q006. What is the volume of a pyramid whose base area is 50 square cm and whose altitude is 60 cm?
......!!!!!!!!...................................
RESPONSE --> The formula for a pyramid is 1/3 * b* h. B represents the area of the pyramid, while h represents the altitude. From this, we can determine Volume V= 1/3 * b * h V= 1/3 * 50 cm^2 * 60 cm V= 1/3 * 3000 cm^3 V= 1000 cm^3 confidence assessment: 2
.................................................
......!!!!!!!!...................................
21:03:03 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 --> I do understand that we can't use V= A ^ h because it doens't have a uniform side we can use. self critique assessment: 2
.................................................
......!!!!!!!!...................................
21:06:18 `q007. What is the volume of a cone whose base area is 20 square meters and whose altitude is 9 meters?
......!!!!!!!!...................................
RESPONSE --> The volume of a cone is 1/3 * pi * r^2 * h. We can also use V= 1/3* A * h because we have a uniform side to used for the area, which happens to be the bottome of the cone. We can solve as follows. V= 1/3 * A * h V= 1/3 * 20 m^2 * 9 m V= 1/3 * 180 m^3 V=60 m^3 confidence assessment: 2
.................................................
......!!!!!!!!...................................
21:06:41 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: 3
.................................................
......!!!!!!!!...................................
21:08:49 `q008. What is a volume of a sphere whose radius is 4 meters?
......!!!!!!!!...................................
RESPONSE --> The formula for volume of a sphere is 4/3 * pi * r^3 From this information, we can determine the volume as follows: V= 4/3 * pi * 4m^3 V= 4/3 * pi * 64m^3 V= 85 1/3 pi m^3 confidence assessment: 2
.................................................
......!!!!!!!!...................................
21:09:30 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 --> I got the correct answer, but I had 256/3 as a mixed number instead of a fraction. self critique assessment: 2
.................................................
......!!!!!!!!...................................
21:13:48 `q009. What is the volume of a planet whose diameter is 14,000 km?
......!!!!!!!!...................................
RESPONSE --> A planet would be classified as a sphere, so the formula would be: First, since we have diameter, we must divide in half to get the radius. 14,000/2=7000 km V= 4/3 * pi * r^3 V= 4/3 * pi * 7000km^3 V= 4/3 * pi * 343,000,000,000km^3 V=4,573,333,333,000 pi km^3 confidence assessment: 1
.................................................
......!!!!!!!!...................................
21:15:34 The planet is presumably a sphere, so to the extent that this is so the volume of this planet is V = 4/3 pi r^3, where r is the radius of the planet. The diameter of the planet is 14,000 km so the radius is half this, or 7,000 km. It follows that the volume of the planet is V = 4/3 pi r^3 = 4/3 pi * (7,000 km)^3 = 4/3 pi * 343,000,000,000 km^3 = 1,372,000,000,000 / 3 * pi km^3. This result can be approximated to an appropriate number of significant figures.
......!!!!!!!!...................................
RESPONSE --> I also see the answer can be 1,372,000,000,000/3 but I answered by going ahead and dividing by 3. self critique assessment: 2
.................................................
......!!!!!!!!...................................
21:16:23 `q010. Summary Question 1: What basic principle do we apply to find the volume of a uniform cylinder of known dimensions?
......!!!!!!!!...................................
RESPONSE --> If we have all dimensions, we can find the area because we have one uniformed side to find the area, and then multiply the height to arrive at volume. V=A * h confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:16:29 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 --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:17:46 `q011. Summary Question 2: What basic principle do we apply to find the volume of a pyramid or a cone?
......!!!!!!!!...................................
RESPONSE --> We can also use the V= A * h , but we have to also multiply by 1/3 V= 1/3 * A * h confidence assessment: 2
.................................................
......!!!!!!!!...................................
21:17:55 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 --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:18:26 `q012. Summary Question 3: What is the formula for the volume of a sphere?
......!!!!!!!!...................................
RESPONSE --> The formula for a sphere is as follows V= 4/3 * pi * r^2 confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:18:47 The volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere.
......!!!!!!!!...................................
RESPONSE --> I recognize that r is the radius of the sphere. self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:19:28 `q013. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
......!!!!!!!!...................................
RESPONSE --> I have also wrote down all formulas and the notes that go along with them so I can study and be sure my mind is regreshed around the idea of them for future use. confidence assessment: 3
.................................................
????????T???v?? assignment #003 003. Misc: Surface Area, Pythagorean Theorem, Density qa areas volumes misc 01-15-2009
......!!!!!!!!...................................
21:21:56 `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?
......!!!!!!!!...................................
RESPONSE --> The area of a rectangular solid is length * width. A= l * w A= 3 m * 4m A 12m^2 confidence assessment: 2
.................................................
......!!!!!!!!...................................
21:24:40 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.
......!!!!!!!!...................................
RESPONSE --> I can now visualize the surface area of the six sided rectangular solid and see that we must add all volumes together to come up with the entire surgace area. self critique assessment: 2
.................................................
......!!!!!!!!...................................
21:29:11 `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?
......!!!!!!!!...................................
RESPONSE --> The surface area of a cylinder is equal to 2 * pi * r * h * 2 * pi * r^2 SA= 2 * pi * 5m * 12m * 2 * pi * 5^2 SA= 4* pi^2 * 5m * 25 SA= 500 pi^2 m^3 confidence assessment: 1
.................................................
......!!!!!!!!...................................
21:34:02 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.
......!!!!!!!!...................................
RESPONSE --> I can see now that if you cut the side of a cylinder it would be equal to a rectangle, so the area becomes Circumference * altitude. And the total area would be the area of all sides + 2 (area of the base) confidence assessment: 2
.................................................
......!!!!!!!!...................................
21:35:23 `q003. What is surface area of a sphere of diameter three cm?
......!!!!!!!!...................................
RESPONSE --> The surface area of a sphere is 4 * pi * r^2 SA= 4 * pi * 3cm^2 SA= 4 * pi * 9 cm^2 SA= 36 pi cm ^2 confidence assessment: 2
.................................................
......!!!!!!!!...................................
21:36:54 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.
......!!!!!!!!...................................
RESPONSE --> I still don't quite understand why the radius is 3 cm/2 Otherwise, I understand the formula. self critique assessment: 1
.................................................
......!!!!!!!!...................................
21:37:17 `q004. What is hypotenuse of a right triangle whose legs are 5 meters and 9 meters?
......!!!!!!!!...................................
RESPONSE --> The hypotneuse is 9 meter side. confidence assessment: 1
.................................................
......!!!!!!!!...................................
21:39:27 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.
......!!!!!!!!...................................
RESPONSE --> I understnad th Pythagorean Theorem states that c^2= a^2 + b^2. I also understand that to solve, we wait until the operations are done inside of the parenthesis before taking the square root. self critique assessment: 2
.................................................
......!!!!!!!!...................................
21:42:40 `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?
......!!!!!!!!...................................
RESPONSE --> The Pythagorean theory states that C^2 = A^2 + B^2 C^2= 6m^2 + 4m^2 C= sqrt(36m^2+ 16m^2) C= sqrt ( 52m^2) C= 7.21m confidence assessment: 2
.................................................
......!!!!!!!!...................................
21:43:37 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.
......!!!!!!!!...................................
RESPONSE --> I understand where I messed up, I assumed that the number we needed finding was c, instead of a or b. Other than that mistake, the steps I understand. self critique assessment: 2
.................................................
??????????? assignment #003 003. Misc: Surface Area, Pythagorean Theorem, Density qa areas volumes misc 01-15-2009
......!!!!!!!!...................................
21:48:44 `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?
......!!!!!!!!...................................
RESPONSE --> The area of a rectangular solid is length * width. A= l * w A= 3 m * 4m A 12m^2 I can now visualize the surface area of the six sided rectangular solid and see that we must add all volumes together to come up with the entire surgace area. confidence assessment: 2
.................................................
......!!!!!!!!...................................
21:48:56 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.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:53:47 `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?
......!!!!!!!!...................................
RESPONSE --> The equation for Density is mass divided by volume. We must first find the Volume of the rectangular solid which is l * w * h V= l * w * h V= 4cm * 7cm * 12 cm V=336cm^3 Now we can determine Density D= m/v D= 700 grams/ 336cm^3 confidence assessment: 1
.................................................
......!!!!!!!!...................................
21:54:25 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).
......!!!!!!!!...................................
RESPONSE --> I see now that we just divide the two and see that there are 2.06 grams for every cm^3 self critique assessment: 2
.................................................
......!!!!!!!!...................................
21:57:28 `q007. What is the mass of a sphere of radius 4 meters if its average density is 3,000 kg/cubic meter?
......!!!!!!!!...................................
RESPONSE --> Because we have density, we can also use the density formula to determine the mass of the sphere. D= M/V M= D* V We must now find the volume of the sphere. V=4/3 * pi * r^3 V =4/3 * 4^3 V= 4/3 * 64 * pi V= 85 1/3 pi kg^3 M= 3,000kg/m^3 * 85 1/3 pi kg^3 confidence assessment: 1
.................................................
......!!!!!!!!...................................
21:58:04 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.
......!!!!!!!!...................................
RESPONSE --> I didn't know the last step until now. I should have just multiplied my answer to get the correct answer. self critique assessment: 2
.................................................
......!!!!!!!!...................................
21:58:14 `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?
......!!!!!!!!...................................
RESPONSE --> confidence assessment: 0
.................................................
......!!!!!!!!...................................
22:00:19 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.
......!!!!!!!!...................................
RESPONSE --> I didn't know how to approach this problem but now I see how I should have. We first calculate the mass for each piece. Then calculate the total mass for the 2 objects. We then take total mass and divide that by total volume to arrive at the average density. self critique assessment: 2
.................................................
......!!!!!!!!...................................
22:02:39 `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?
......!!!!!!!!...................................
RESPONSE --> We would take the density of both the cannon balls and the sand and average them. Total Density=2100 kg/m^3 + 8000 kg/m^3 Total Density= 10100 kg/ m^3 confidence assessment: 0
.................................................
......!!!!!!!!...................................
22:05:01 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..
......!!!!!!!!...................................
RESPONSE --> I can see now that I should have calculated the mass by multiplying the density by the volume for both the sand and the cannonballs. Then I should have take the total mass by adding the two answers together, and divided that by the total volume by adding the two together, and the answer would have been right. self critique assessment: 2
.................................................
......!!!!!!!!...................................
22:08:12 `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?
......!!!!!!!!...................................
RESPONSE --> We can find the volume of the oil slick by multiplying the coverage of the oil slick by the depth of the oil slick. V= 1,700,000 m^3 * 0.015 m V= 25500 m^4 Now, the mass can be calculated. M= D * V M= 860 kg/m^3 * 25,000 m^3 M= 21,500,000 kg/m^3 confidence assessment: 1
.................................................
......!!!!!!!!...................................
22:09:28 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.
......!!!!!!!!...................................
RESPONSE --> I don't understand how the volume went from 25,500 to 24,400. But other than that I understood the steps and got them all right. self critique assessment: 1
.................................................
......!!!!!!!!...................................
22:10:36 `q011. Summary Question 1: How do we find the surface area of a cylinder?
......!!!!!!!!...................................
RESPONSE --> We find the surface area of a cylinder by multiply 2*l*w * 2*l*h * 2*w*h confidence assessment: 2
.................................................
......!!!!!!!!...................................
22:11:56 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.
......!!!!!!!!...................................
RESPONSE --> I thought the question was for a rectangular Solid. I understand that the surface area of a cylinder is unrolled to form a rectangle that equals the C * A of a cylinder. Therefore the surface area = C * A self critique assessment: 2
.................................................
......!!!!!!!!...................................
22:12:17 `q012. Summary Question 2: What is the formula for the surface area of a sphere?
......!!!!!!!!...................................
RESPONSE --> The surface area of a sphere is 4 * pi * r^2 confidence assessment: 3
.................................................
......!!!!!!!!...................................
22:12:23 The surface area of a sphere is A = 4 pi r^2, where r is the radius of the sphere.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:13:18 `q013. Summary Question 3: What is the meaning of the term 'density'.
......!!!!!!!!...................................
RESPONSE --> Density is also the mass per unit you are measuring confidence assessment: 2
.................................................
......!!!!!!!!...................................
22:13:54 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'
......!!!!!!!!...................................
RESPONSE --> I understand it is calculated by dividing total mass by total volume. self critique assessment: 2
.................................................
......!!!!!!!!...................................
22:14:56 `q014. Summary Question 4: If we know average density and mass, how can we find volume?
......!!!!!!!!...................................
RESPONSE --> We can manipulate the Density formula to become: V= density/mass confidence assessment: 1
.................................................
......!!!!!!!!...................................
22:15:40 Since mass = ave density * volume, it follows by simple algebra that volume = mass / ave density.
......!!!!!!!!...................................
RESPONSE --> I mixed up ave density and mass I now understand that volume= mass/ ave. density. self critique assessment: 2
.................................................
......!!!!!!!!...................................
22:16:09 `q015. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
......!!!!!!!!...................................
RESPONSE --> I have detailed notes of each question on paper so that I can go back over them and work on a complete understanding of the concepts covered here. confidence assessment: 3
.................................................