course Mth 151 ???????v??}???assignment #001001. Only assignment: prelim asst qa prelim
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00:27:26 `questionNumber 10001 `q001. Part 1 includes six activities. If you have completed an activity, just enter the answer 'completed'. This question is appearing in the Question box. The box to the right is the Answer box, where you will type in your answers to the questions posed here. To use this program you read a question, then enter your answer in the Answer box and click on Enter Answer. In your answers give what is requested, but don't go into excruciating detail. Try to give just enough that the instructor can tell that you understand an item. After entering an answer click on Next Question/Answer above the Question box. Do you understand these instructions?
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RESPONSE --> Yes
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00:29:05 `questionNumber 10001 This program has created the folder c:\vhmthphy on your hard drive. Browse to that folder and locate the file whose name begins with SEND. The name of this file will also include your name, as you gave it to the program, and the file will show as a Text file. Never tamper with a SEND file in any way. It contains internal codes as if these codes are tampered with you won't get credit for the assignment. However you are welcome to copy this file to another location and view it, make changes, etc. Just be sure that when requested to do so you send the instructor the original, tamper-free file. State in the Answer box whether or not you have been able to locate the SEND file. Don't send the SEND file yet. Note that more questions/instructions remain in the q_a_prelim.
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RESPONSE --> Yes
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00:31:30 `questionNumber 10002 `q002. Note that every time you click on Enter Answer the program writes your response to your SEND file. Even if the program disappears all the information you have entered with the Enter Answer button will remain in that file. This program never 'unwrites' anything. Even if this program crashes your information will still be there in the SEND file. Explain this in your own words.
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RESPONSE --> The program auto-saves whenever I click the ""Enter Response"" button. The information is never unwritten by the program even if it crashes.
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00:32:09 `questionNumber 10002 Any time the instructor does not post a response to your access site by the end of the following day, you should resubmit your work using the Submit Work form, and be sure at the beginning to indicate that you are resubmitting, and also indicate the date on which you originally submitted your work. If you don't know where your access site is or how to access it, go to http://www.vhcc.edu/dsmith/_vti_bin/shtml.dll/request_access_code.htm and request one now. You can submit the q_a_prelim without your access code, but other assignments should contain your code.
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RESPONSE --> I already have my access code.
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00:34:01 `questionNumber 10003 `q003. If you are working on a VHCC computer, it is probably set up in such a way as to return to its original configuration when it is rebooted. To avoid losing information it is suggested that you back up your work frequently, either by emailing yourself a copy or by using a key drive or other device. This is a good idea on any computer. Please indicate your understanding of this suggestion.
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RESPONSE --> If I work from a VHCC computer I need to backup my work before I leave the computer.
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??????????assignment #001 001. typewriter notation qa initial problems 01-17-2009 ?l??Z^?`??~????assignment #001 001. typewriter notation qa initial problems 01-17-2009
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12:51:03 `questionNumber 10000 `q001. Explain the difference between x - 2 / x + 4 and (x - 2) / (x + 4). The evaluate each expression for x = 2.
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RESPONSE --> The difference in the two expressions is that : (x-2)/(x+4) has parentheses which means whatever is in parentheses needs to be worked on first. evaluation for x - 2 / x + 4 when x=2 gives me 2-2/2+4 which = 0/6 . (x - 2) / (x + 4) with x=2 gives me (2-2)/(2+4) which = (0)/(6).The answer for both expression is 0 confidence assessment: 1
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12:57:26 `questionNumber 10000 2 ^ x + 4 indicates that you are to raise 2 to the x power before adding the 4. 2 ^ (x + 4) indicates that you are to first evaluate x + 4, then raise 2 to this power. If x = 2, then 2 ^ x + 4 = 2 ^ 2 + 4 = 2 * 2 + 4 = 4 + 4 = 8. and 2 ^ (x + 4) = 2 ^ (2 + 4) = 2 ^ 6 = 2*2*2*2*2*2 = 64.
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RESPONSE --> ok self critique assessment: 0
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13:15:31 `questionNumber 10000 `q003. What is the numerator of the fraction in the expression x - 3 / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x? What is the denominator? What do you get when you evaluate the expression for x = 2?
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RESPONSE --> The numerator is x-3 the denominator is (2x-5)^2 * 3x + 1 ] - 2 + 7x. When i evaluate the expression for x=2. 2 - 3/[ (2(2)-5)^2 * 3(2) + 1 ] - 2 + 7(2) the 2 - 3/{(4*4-25) *7} -2+7*2 then 2 - 3/ {64}=-2+14 then 2 - 3/-66+14 equals -1/52 confidence assessment: 1
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13:27:53 `questionNumber 10000 `q004. Explain, step by step, how you evaluate the expression (x - 5) ^ 2x-1 + 3 / x-2 for x = 4.
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RESPONSE --> ok confidence assessment: 0
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???{?z???|????? assignment #001 001. typewriter notation qa initial problems 01-18-2009
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13:57:36 `questionNumber 10000 `q001. Explain the difference between x - 2 / x + 4 and (x - 2) / (x + 4). The evaluate each expression for x = 2.
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RESPONSE --> The difference between the two expressions is the parentheses.With the order of operations the exression in parenthesis say whatever is in the parentheses is to be worked first.On the expression without parenthesis the division would be worked first.So: x - 2 / x + 4 x=2 2-2/2+4 2/2=1 and -2/4=-.5 1+-.5= 1.5 with (x - 2) / (x + 4) (2-2)/(2+4) (2-2)=0 (2+4)=6 so 0/6=0 confidence assessment: 1
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????y??T?????y assignment #001 001. typewriter notation qa initial problems 01-18-2009
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18:17:33 `questionNumber 10000 `q001. Explain the difference between x - 2 / x + 4 and (x - 2) / (x + 4). The evaluate each expression for x = 2.
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RESPONSE --> The order of operations are that the parenthensis are worked first.where as the expression without parenthesis the division would be worked first. x - 2 / x + 4 2-2/2+4 2-1+4 5 and (x - 2) / (x + 4) (2-2)/(2+4) 0/6 0 confidence assessment: 2
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???????????assignment #001 001. typewriter notation qa initial problems 01-18-2009
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18:24:05 `questionNumber 10000 `q001. Explain the difference between x - 2 / x + 4 and (x - 2) / (x + 4). The evaluate each expression for x = 2.
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RESPONSE --> The order of operations calls for whatever is in parenthesis to be worked first.In the expression without parenthesis the division is worked first. x - 2 / x + 4 2-2/2+4 2-1+4 5 and (x - 2) / (x + 4) (2-2)/(2+4) 0/6 confidence assessment: 2
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18:25:33 `questionNumber 10000 `q002. Explain the difference between 2 ^ x + 4 and 2 ^ (x + 4). Then evaluate each expression for x = 2. Note that a ^ b means to raise a to the b power. This process is called exponentiation, and the ^ symbol is used on most calculators, and in most computer algebra systems, to represent exponentiation.
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RESPONSE --> ok confidence assessment: 3
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18:26:20 `questionNumber 10000 2 ^ x + 4 indicates that you are to raise 2 to the x power before adding the 4. 2 ^ (x + 4) indicates that you are to first evaluate x + 4, then raise 2 to this power. If x = 2, then 2 ^ x + 4 = 2 ^ 2 + 4 = 2 * 2 + 4 = 4 + 4 = 8. and 2 ^ (x + 4) = 2 ^ (2 + 4) = 2 ^ 6 = 2*2*2*2*2*2 = 64.
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RESPONSE --> there appears to be an error im getting the answer to the question before the question itself self critique assessment: 3
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18:35:18 `questionNumber 10000 `q003. What is the numerator of the fraction in the expression x - 3 / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x? What is the denominator? What do you get when you evaluate the expression for x = 2?
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RESPONSE --> numerator: x - 3 denominator: [ (2x-5)^2 * 3x + 1 ] - 2 + 7x [(2*2-5)^2*3(2)+1}-2+7(2) {(-1)^2*6+1} 1*6+1 6+1 7-2+14 5+14 19 -1/19 confidence assessment: 2
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18:41:29 `questionNumber 10000 The numerator is 3. x isn't part of the fraction. / indicates division, which must always precede subtraction. Only the 3 is divided by [ (2x-5)^2 * 3x + 1 ] and only [ (2x-5)^2 * 3x + 1 ] divides 3. If we mean (x - 3) / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x we have to write it that way. The preceding comments show that the denominator is [ (2x-5)^2 * 3x + 1 ] Evaluating the expression for x = 2: - 3 / [ (2 * 2 - 5)^2 * 3(2) + 1 ] - 2 + 7*2 = 2 - 3 / [ (4 - 5)^2 * 6 + 1 ] - 2 + 14 = evaluate in parenthese; do multiplications outside parentheses 2 - 3 / [ (-1)^2 * 6 + 1 ] -2 + 14 = add inside parentheses 2 - 3 / [ 1 * 6 + 1 ] - 2 + 14 = exponentiate in bracketed term; 2 - 3 / 7 - 2 + 14 = evaluate in brackets 13 4/7 or 95/7 or about 13.57 add and subtract in order. The details of the calculation 2 - 3 / 7 - 2 + 14: Since multiplication precedes addition or subtraction the 3/7 must be done first, making 3/7 a fraction. Changing the order of the terms we have 2 - 2 + 14 - 3 / 7 = 14 - 3/7 = 98/7 - 3/7 = 95/7. COMMON STUDENT QUESTION: ok, I dont understand why x isnt part of the fraction? And I dont understand why only the brackets are divided by 3..why not the rest of the equation? INSTRUCTOR RESPONSE: Different situations give us different algebraic expressions; the situation dictates the form of the expression. If the above expression was was written otherwise it would be a completely different expression and most likely give you a different result when you substitute. If we intended the numerator to be x - 3 then the expression would be written (x - 3) / [(2x-5)^2 * 3x + 1 ] - 2 + 7x, with the x - 3 grouped. If we intended the numerator to be the entire expression after the / the expression would be written x - 3 / [(2x-5)^2 * 3x + 1 - 2 + 7x ].
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RESPONSE --> self critique assessment:
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18:47:44 `questionNumber 10000 `q004. Explain, step by step, how you evaluate the expression (x - 5) ^ 2x-1 + 3 / x-2 for x = 4.
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RESPONSE --> (x - 5) ^ 2x-1 + 3 / x-2 (4-5)^2(4)-1+3/4-2 (-1)^2(4)-1+3/4-2 parethsis first 1(4)-1+3/4-2then exponents 4-1+3/4-2 then multiply 4-1+.75-2 then division 3+.75-2then work left to right add /sub 3.75-2 1.75 confidence assessment: 1
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18:48:51 `questionNumber 10000 We get (4-5)^2 * 4 - 1 + 3 / 1 - 4 = (-1)^2 * 4 - 1 + 3 / 4 - 2 evaluating the term in parentheses = 1 * 4 - 1 + 3 / 4 - 2 exponentiating (2 is the exponent, which is applied to -1 rather than multiplying the 2 by 4 = 4 - 1 + 3/4 - 2 noting that 3/4 is a fraction and adding and subtracting in order we get = 1 3/4 = 7 /4 (Note that we could group the expression as 4 - 1 - 2 + 3/4 = 1 + 3/4 = 1 3/4 = 7/4). COMMON ERROR: (4 - 5) ^ 2*4 - 1 + 3 / 4 - 2 = -1 ^ 2*4 - 1 + 3 / 4-2 = -1 ^ 8 -1 + 3 / 4 - 2. INSTRUCTOR COMMENTS: There are two errors here. In the second step you can't multiply 2 * 4 because you have (-1)^2, which must be done first.?Exponentiation precedes multiplication. ? Also it isn't quite correct to write -1^2*4 at the beginning of the second step. If you were supposed to multiply 2 * 4 the expression would be (-1)^(2 * 4).? Note also that the -1 needs to be grouped because the entire expression (-1) is taken to the power.?-1^8 would be -1 because you would raise 1 to the power 8 before applying the - sign, which is effectively a multiplication by -1.?......!!!!!!!!................................... RESPONSE --> ok self critique assessment: 2
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18:50:06 `questionNumber 10000 *&*& Standard mathematics notation is easier to see. On the other hand it's very important to understand order of operations, and students do get used to this way of doing it. You should of course write everything out in standard notation when you work it on paper. It is likely that you will at some point use a computer algebra system, and when you do you will have to enter expressions through a typewriter, so it is well worth the trouble to get used to this notation. Indicate your understanding of the necessity to understand this notation.
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RESPONSE --> I understand i need to get used to it because i may use it in the future. self critique assessment: 2
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18:52:04 `questionNumber 10000 `q005. At the link http://www.vhcc.edu/dsmith/genInfo/introductory problems/typewriter_notation_examples_with_links.htm (copy this path into the Address box of your Internet browser; alternatively use the path http://vhmthphy.vhcc.edu/ > General Information > Startup and Orientation (either scroll to bottom of page or click on Links to Supplemental Sites) > typewriter notation examples and you will find a page containing a number of additional exercises and/or examples of typewriter notation.Locate this site, click on a few of the links, and describe what you see there.
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RESPONSE --> I see ""line format"" representing the traditional format I'm used to. confidence assessment: 2
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18:53:01 `questionNumber 10000 You should see a brief set of instructions and over 30 numbered examples. If you click on the word Picture you will see the standard-notation format of the expression. The link entitled Examples and Pictures, located in the initial instructions, shows all the examples and pictures without requiring you to click on the links. There is also a file which includes explanations. The instructions include a note indicating that Liberal Arts Mathematics students don't need a deep understanding of the notation, Mth 173-4 and University Physics students need a very good understanding,
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RESPONSE --> ok self critique assessment: 2
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18:53:28 `questionNumber 10000 while students in other courses should understand the notation and should understand the more basic simplifications. There is also a link to a page with pictures only, to provide the opportunity to translated standard notation into typewriter notation.
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RESPONSE --> ok self critique assessment: 2
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??????Yy????y???assignment #001 001. Areas qa areas volumes misc 01-18-2009
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19:13:36 `questionNumber 10000 `q001. There are 11 questions and 7 summary questions in this assignment. What is the area of a rectangle whose dimensions are 4 m by 3 meters.
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RESPONSE --> a=4*3 a=12 confidence assessment: 1
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19:16:56 `questionNumber 10000 A 4 m by 3 m rectangle can be divided into 3 rows of 4 squares, each 1 meter on a side. This makes 3 * 4 = 12 such squares. Each 1 meter square has an area of 1 square meter, or 1 m^2. The total area of the rectangle is therefore 12 square meters, or 12 m^2. The formula for the area of a rectangle is A = L * W, where L is the length and W the width of the rectangle. Applying this formula to the present problem we obtain area A = L * W = 4 m * 3 m = (4 * 3) ( m * m ) = 12 m^2. Note the use of the unit m, standing for meters, in the entire calculation. Note that m * m = m^2.
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RESPONSE --> ok self critique assessment: 2
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19:20:25 `questionNumber 10000 `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?
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RESPONSE --> a=.5bxh a=.5(4)*3 a=2*3 a=6 confidence assessment: 2
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19:22:30 `questionNumber 10000 A right triangle can be joined along its hypotenuse with another identical right triangle to form a rectangle. In this case the rectangle would have dimensions 4.0 meters by 3.0 meters, and would be divided by any diagonal into two identical right triangles with legs of 4.0 meters and 3.0 meters. The rectangle will have area A = L * W = 4.0 m * 3.0 m = 12 m^2, as explained in the preceding problem. Each of the two right triangles, since they are identical, will therefore have half this area, or 1/2 * 12 m^2 = 6.0 m^2. The formula for the area of a right triangle with base b and altitude h is A = 1/2 * b * h.
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RESPONSE --> ok self critique assessment: 2
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19:23:43 `questionNumber 10000 `q003. What is the area of a parallelogram whose base is 5.0 meters and whose altitude is 2.0 meters?
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RESPONSE --> a=h*h a=5*2 a+10 confidence assessment: 2
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19:24:52 `questionNumber 10000 A parallelogram is easily rearranged into a rectangle by 'cutting off' the protruding end, turning that portion upside down and joining it to the other end. Hopefully you are familiar with this construction. In any case the resulting rectangle has sides equal to the base and the altitude so its area is A = b * h. The present rectangle has area A = 5.0 m * 2.0 m = 10 m^2.
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RESPONSE --> ok self critique assessment: 2
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19:29:05 `questionNumber 10000 `q004. What is the area of a triangle whose base is 5.0 cm and whose altitude is 2.0 cm?
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RESPONSE --> a=1/5b*h a=1/25*2 a=10*2 a=20 m^2 confidence assessment: 2
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19:29:42 `questionNumber 10000 It is possible to join any triangle with an identical copy of itself to construct a parallelogram whose base and altitude are equal to the base and altitude of the triangle. The area of the parallelogram is A = b * h, so the area of each of the two identical triangles formed by 'cutting' the parallelogram about the approriate diagonal is A = 1/2 * b * h. The area of the present triangle is therefore A = 1/2 * 5.0 cm * 2.0 cm = 1/2 * 10 cm^2 = 5.0 cm^2.
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RESPONSE --> ok self critique assessment: 2
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19:36:05 `questionNumber 10000 `q005. What is the area of a trapezoid with a width of 4.0 km and average altitude of 5.0 km?
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RESPONSE --> a=1/2(a+b)h a=1/2(4)5 a=2*5 a=10km^2 confidence assessment: 1
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19:37:06 `questionNumber 10000 Any trapezoid can be reconstructed to form a rectangle whose width is equal to that of the trapezoid and whose altitude is equal to the average of the two altitudes of the trapezoid. The area of the rectangle, and therefore the trapezoid, is therefore A = base * average altitude. In the present case this area is A = 4.0 km * 5.0 km = 20 km^2.
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RESPONSE --> ok i understand now. self critique assessment: 2
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19:43:48 `questionNumber 10000 `q006. What is the area of a trapezoid whose width is 4 cm in whose altitudes are 3.0 cm and 8.0 cm?
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RESPONSE --> a=1/2(3+8)4 1/2(11)*4 5.5*4 a=22cm^2 confidence assessment: 2
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19:44:02 `questionNumber 10000 The area is equal to the product of the width and the average altitude. Average altitude is (3 cm + 8 cm) / 2 = 5.5 cm so the area of the trapezoid is A = 4 cm * 5.5 cm = 22 cm^2.
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RESPONSE --> ok self critique assessment: 3
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19:46:58 `questionNumber 10000 `q007. What is the area of a circle whose radius is 3.00 cm?
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RESPONSE --> a=3.14*3^2 a=3.14*9 a=28.26 cm^2 confidence assessment: 2
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19:47:42 `questionNumber 10000 The area of a circle is A = pi * r^2, where r is the radius. Thus A = pi * (3 cm)^2 = 9 pi cm^2. Note that the units are cm^2, since the cm unit is part r, which is squared. The expression 9 pi cm^2 is exact. Any decimal equivalent is an approximation. Using the 3-significant-figure approximation pi = 3.14 we find that the approximate area is A = 9 pi cm^2 = 9 * 3.14 cm^2 = 28.26 cm^2, which we round to 28.3 cm^2 to match the number of significant figures in the given radius. Be careful not to confuse the formula A = pi r^2, which gives area in square units, with the formula C = 2 pi r for the circumference. The latter gives a result which is in units of radius, rather than square units. Area is measured in square units; if you get an answer which is not in square units this tips you off to the fact that you've made an error somewhere.
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RESPONSE --> ok self critique assessment: 2
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19:50:04 `questionNumber 10000 `q008. What is the circumference of a circle whose radius is exactly 3 cm?
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RESPONSE --> c=2 pi r c=2(3.14)3 c=6.28(3) c=18.84 cm confidence assessment: 2
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19:50:34 `questionNumber 10000 The circumference of this circle is C = 2 pi r = 2 pi * 3 cm = 6 pi cm. This is the exact area. An approximation to 3 significant figures is 6 * 3.14 cm = 18.84 cm. Note that circumference is measured in the same units as radius, in this case cm, and not in cm^2. If your calculation gives you cm^2 then you know you've done something wrong.
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RESPONSE --> ok self critique assessment: 2
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19:53:15 `questionNumber 10000 `q009. What is the area of a circle whose diameter is exactly 12 meters?
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RESPONSE --> a=pi r^2 a=pi 6(half diamtr=radius)^2 a=pi 36 a=113.04 m^2 confidence assessment: 2
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19:53:29 `questionNumber 10000 The area of a circle is A = pi r^2, where r is the radius. The radius of this circle is half the 12 m diameter, or 6 m. So the area is A = pi ( 6 m )^2 = 36 pi m^2. This result can be approximated to any desired accuracy by using a sufficient number of significant figures in our approximation of pi. For example using the 5-significant-figure approximation pi = 3.1416 we obtain A = 36 m^2 * 3.1416 = 113.09 m^2.
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RESPONSE --> ok self critique assessment: 2
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19:56:35 `questionNumber 10000 `q010. What is the area of a circle whose circumference is 14 `pi meters?
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RESPONSE --> What is a pi meter?I dont know the answer or even how to begin to answer this question. confidence assessment: 0
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19:58:09 `questionNumber 10000 We know that A = pi r^2. We can find the area if we know the radius r. We therefore attempt to use the given information to find r. We know that circumference and radius are related by C = 2 pi r. Solving for r we obtain r = C / (2 pi). In this case we find that r = 14 pi m / (2 pi) = (14/2) * (pi/pi) m = 7 * 1 m = 7 m. We use this to find the area A = pi * (7 m)^2 = 49 pi m^2.
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RESPONSE --> ok.I will probably need more work on this concept. self critique assessment: 0
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20:13:19 `questionNumber 10000 `q011. What is the radius of circle whose area is 78 square meters?
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RESPONSE --> a=pi*r^2 78=pi*r^2 ^2*24.84=^2r 4.98=r confidence assessment: 1
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20:13:51 `questionNumber 10000 Knowing that A = pi r^2 we solve for r. We first divide both sides by pi to obtain A / pi = r^2. We then reverse the sides and take the square root of both sides, obtaining r = sqrt( A / pi ). Note that strictly speaking the solution to r^2 = A / pi is r = +-sqrt( A / pi ), meaning + sqrt( A / pi) or - sqrt(A / pi). However knowing that r and A are both positive quantities, we can reject the negative solution. Now we substitute A = 78 m^2 to obtain r = sqrt( 78 m^2 / pi) = sqrt(78 / pi) m.{} Approximating this quantity to 2 significant figures we obtain r = 5.0 m.
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RESPONSE --> ok self critique assessment: 2
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20:18:18 `questionNumber 10000 `q012. Summary Question 1: How do we visualize the area of a rectangle?
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RESPONSE --> by multiplying the the top and bottom? confidence assessment: 0
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20:19:36 `questionNumber 10000 We visualize the rectangle being covered by rows of 1-unit squares. We multiply the number of squares in a row by the number of rows. So the area is A = L * W.
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RESPONSE --> ok self critique assessment: 0
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20:23:36 `questionNumber 10000 `q013. Summary Question 2: How do we visualize the area of a right triangle?
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RESPONSE --> the area would be one unit squares from base multplied by half and then multiplied by one unit squares of height confidence assessment: 0
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20:24:01 `questionNumber 10000 We visualize two identical right triangles being joined along their common hypotenuse to form a rectangle whose length is equal to the base of the triangle and whose width is equal to the altitude of the triangle. The area of the rectangle is b * h, so the area of each triangle is 1/2 * b * h.
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RESPONSE --> ok self critique assessment: 0
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20:24:20 `questionNumber 10000 `q014. Summary Question 3: How do we calculate the area of a parallelogram?
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RESPONSE --> Area = b? confidence assessment: 3
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20:24:34 `questionNumber 10000 The area of a parallelogram is equal to the product of its base and its altitude. The altitude is measured perpendicular to the base.
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RESPONSE --> ok self critique assessment: 3
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20:24:48 `questionNumber 10000 `q015. Summary Question 4: How do we calculate the area of a trapezoid?
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RESPONSE --> Area = ?a+b)h confidence assessment: 3
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20:25:04 `questionNumber 10000 We think of the trapezoid being oriented so that its two parallel sides are vertical, and we multiply the average altitude by the width.
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RESPONSE --> ok self critique assessment: 2
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20:25:41 `questionNumber 10000 `q016. Summary Question 5: How do we calculate the area of a circle?
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RESPONSE --> Area = pi r2 confidence assessment: 3
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20:25:52 `questionNumber 10000 We use the formula A = pi r^2, where r is the radius of the circle.
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RESPONSE --> ok self critique assessment: 3
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20:28:34 `questionNumber 10000 `q017. Summary Question 6: How do we calculate the circumference of a circle? How can we easily avoid confusing this formula with that for the area of the circle?
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RESPONSE --> Circumference=2 pi r Area = pi r2 if the circle is a cherry pie then the circumference is the rim of the pie plate and the area is the amount of filling in the pie confidence assessment: 3
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20:28:56 `questionNumber 10000 We use the formula C = 2 pi r. The formula for the area involves r^2, which will give us squared units of the radius. Circumference is not measured in squared units.
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RESPONSE --> ok self critique assessment: 2
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20:29:40 `questionNumber 10000 `q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> I have notated areas i did not understand for future reference. confidence assessment: 1
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?????????? assignment #004 004. Liberal Arts Mathematics qa initial problems 01-18-2009
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21:53:03 `questionNumber 40000 `q001. Consider the statement 'If that group of six-year-olds doesn't have adult supervision, they won't act in an orderly manner.' Under which of the following circumstances would everyone have to agree that the statement is false? The group does have supervision and they do act in an orderly manner. The group doesn't have supervision and they don't act in an orderly manner. The group doesn't have supervision and they do act in an orderly manner. The group does have supervision and they don't act in an orderly manner.
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RESPONSE --> The group does have supervision and they do act in an orderly manner. This statement refers to when children DO have supervision, and could be true or false. The group doesn't have supervision and they don't act in an orderly manner. This statement is true. The group doesn't have supervision and they do act in an orderly manner. This statement is false because it contradicts the original statement and must be false. The group does have supervision and they don't act in an orderly manner. This statement refers to when children DO have supervision, and could be true or false. confidence assessment: 3
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21:53:58 `questionNumber 40000 The statement says that if the group doesn't have supervision, they will not act in an orderly manner. So if they don't have supervision and yet do act in an orderly manner the statement is contradicted. If the group does have supervision, the statement cannot be contradicted because condition of the statement, that the group doesn't have supervision, does not hold. The statement has nothing to say about what happens if the group does have supervision. Of course if the group doesn't have supervision and doesn't act in orderly manner this is completely consistent with the statement. Therefore the only way to statement can be considered false is the group doesn't have supervision and does act in an overly manner. Note that what we know, or think we know, about childrens' behavior has nothing at all to do with the logic of the situation. We could analyze the logic of a statement like 'If the Moon is made of green cheese then most six-year-olds prefer collard greens to chocolate ice cream'. Anything we know about the composition of the Moon or the tastes of children has nothing to do with the fact that the only way this statement could be shown false would be for the Moon to be made of green cheese and most six-year-olds to prefer the ice cream.
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RESPONSE --> OK self critique assessment: 3
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21:59:12 `questionNumber 40000 `q002. List the different orders in which the letters a, b and c could be arranged (examples are 'acb' and 'cba'). Explain how you know that your list contains every possible order.
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RESPONSE --> the list has 6 possible orders where orders=A and x= the number of variables. A=x^2-x abc acb bca bac cba cab confidence assessment: 3
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22:01:49 `questionNumber 40000 The only reliable way to get all possible orders is to have a system which we are sure the list every order without missing any. Perhaps the simplest way to construct all possible orders is to list then alphabetically. We start with abc. There is only one other order that starts with a, and it is obtained by switching the last two letters to get acb. The next alphabetical order must start with b. The first possible listing starting with b must follow b with a, leaving c for last. The orders therefore bac. The only other order starting with b is bca. The next order must start with c, which will be followed by a to give us cab. The next order is obtained by switching the last two letters to get cba. This exhausts all possibilities for combinations of the three letters a, b and c. Our combinations are, in alphabetical order, abc, acb, bac, bca, cab, cba.
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RESPONSE --> ok self critique assessment: 3
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22:04:49 `questionNumber 40000 `q003. One collection consists of the letters a, c, d and f. Another collection consists of the letters a, b, d and g. List the letters common to both collections. List the letters which appear in at least one of the collections. List the letters in the first half of the alphabet which do not appear in either of the collections.
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RESPONSE --> List the letters common to both collections. a, d List the letters which appear in at least one of the collections. a,b,c,d,g,f List the letters in the first half of the alphabet which do not appear in either of the collections. e,h,i,j,k,l,m confidence assessment: 3
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22:06:28 `questionNumber 40000 To letters a and d each appear in both collections. No other letter does. The letters a, c, d, and f appear in the first collection, so they all in at least one of the collections. In addition to letters b and g appear in the second collection. Therefore letters a, b, c, d, f and g all appear in at least one of the collections. We consider the letters in the first half of the alphabet, in alphabetical order. a, b, c and d all appear in at least one of the collections, but the letter e does not. The letters f and g also appear in at least one of the collections, but none of the other letters of the alphabet do. The first half of the alphabet ends at m, so the list of letters in the first half of the alphabet which do not occur in at least one of the collections is e, h, i, j, k, l, m.
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RESPONSE --> self critique assessment: 3
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22:09:03 `questionNumber 40000 `q004. Give the next element in each of the following patterns and explain how you obtained each: 2, 3, 5, 8, 12, ... 3, 6, 12, 24, ... 1, 3, 4, 7, 11, 18, ...
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RESPONSE --> 2, 3, 5, 8, 12, ... 20, 2+3=5+3=8 and so on 3, 6, 12, 24, ...48, answer*2 1, 3, 4, 7, 11, 18, ... 29, adding answer with last answer same formula as first pattern confidence assessment: 3
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22:11:04 `questionNumber 40000 The pattern of the sequence 2, 3, 5, 8, 12, ... can be seen by subtracting each number from its successor. 3-2 = 1, 5-3 = 2, 8-5 = 3, 12-8 = 4. The sequence of differences is therefore 1, 2, 3, 4, ... . The next difference will be 5, indicating that the next number must be 12 + 5 = 17. The pattern of the sequence 3, 6, 12, 24, ... can be discovered by dividing each number into its successor. We obtain 6/3 = 2, 12/6 = 2, 24/12 = 2. This shows us that we are doubling each number to get the next. It follows that the next number in the sequence will be the double of 24, or 48. The pattern of the sequence 1, 3, 4, 7, 11, 18, ... is a little obvious. Starting with the third number in the sequence, each number is the sum of the two numbers proceeding. That is, 1 + 3 = 4, 3 + 4 = 7, 4 + 7 = 11, and 7 + 11 = 18. It follows that the next member should be 11 + 18 = 29.
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RESPONSE --> ok self critique assessment: 3
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22:13:35 `questionNumber 40000 `q005. The number 18 can be 'broken down' into the product 9 * 2, which can then be broken down into the product 3 * 3 * 2, which cannot be broken down any further . Alternatively 18 could be broken down into 6 * 3, which can then be broken down into 2 * 3 * 3. Show how the numbers 28 and 34 can be broken down until they can't be broken down any further. Show that there at least two different ways to break down 28, but that when the breakdown is complete both ways end up giving you the same numbers.
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RESPONSE --> 28 2*14 2*2*7 or 4*7 2*2*7 34 2*17 confidence assessment: 3
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22:14:21 `questionNumber 40000 A good system is to begin by attempting to divide the smallest possible number into the given number. In the case of 34 we see that the number can be divided by 2 give 34 = 2 * 17. It is clear that the factor 2 cannot be further broken down, and is easy to see that 17 cannot be further broken down. So the complete breakdown of 34 is 2 * 17. To breakdown 28 we can again divide by 2 to get 28 = 2 * 14. The number 2 cannot be further broken down, but 14 can be divided by 2 to give 14 = 2 * 7, which cannot be further broken down. Thus we have 28 = 2 * 2 * 7. The number 28 could also the broken down initially into 4 * 7. The 4 can be further broken down into 2 * 2, so again we get 28 = 2 * 2 * 7. It turns out that the breakdown of a given number always ends up with exactly same numbers, no matter what the initial breakdown.
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RESPONSE --> self critique assessment: 3
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22:18:48 `questionNumber 40000 `q006. Give the average of the numbers in the following list: 3, 4, 6, 6, 7, 7, 9. By how much does each number differ from the average?
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RESPONSE --> average = 6 differance= 3=3 4=2 6=0 6=0 7=1 7=1 9=3 confidence assessment: 3
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22:21:02 `questionNumber 40000 To average least 7 numbers we add them in divide by 7. We get a total of 3 + 4 + 6 + 6 + 7 + 7 + 9 = 42, which we then divide by 7 to get the average 42 / 7 = 6. We see that 3 differs from the average of 6 by 3, 4 differs from the average of 6 by 2, 6 differs from the average of 6 by 0, 7 differs from the average of 6 by 1, and 9 differs from the average of 6 by 3. A common error is to write the entire sequence of calculations on one line, as 3 + 4 + 6 + 6 + 7 + 7 + 9 = 42 / 7 = 6. This is a really terrible habit. The = sign indicates equality, and if one thing is equal to another, and this other today third thing, then the first thing must be equal to the third thing. This would mean that 3 + 4 + 6 + 6 + 7 + 7 + 9 would have to be equal to 6. This is clearly not the case. It is a serious error to use the = sign for anything but equality, and it should certainly not be used to indicate a sequence of calculations.
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RESPONSE --> self critique assessment: 3
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22:22:21 `questionNumber 40000 `q007. Which of the following list of numbers is more spread out, 7, 8, 10, 10, 11, 13 or 894, 897, 902, 908, 910, 912? On what basis did you justify your answer?
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RESPONSE --> 894, 897, 902, 908, 910, 912 is more spread out 912-894=18 13-7=6 confidence assessment: 3
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22:22:39 `questionNumber 40000 The first set of numbers ranges from 7 to 13, a difference of only 6. The second set ranges from 894 to 912, a difference of 18. So it appears pretty clear that the second set has more variation the first. We might also look at the spacing between numbers, which in the first set is 1, 2, 0, 1, 2 and in the second set is 3, 5, 6, 2, 2. The spacing in the second set is clearly greater than the spacing in the first. There are other more sophisticated measures of the spread of a distribution of numbers, which you may encounter in your course.
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RESPONSE --> self critique assessment: 3
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22:27:06 `questionNumber 40000 `q008. 12 is 9 more than 3 and also 4 times 3. We therefore say that 12 differs from 3 by 9, and that the ratio of 12 to 3 is 4. What is the ratio of 36 to 4 and by how much does 36 differ from 4? If 288 is in the same ratio to a certain number as 36 is to 4, what is that number?
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RESPONSE --> 36 to 4 differs by 32, 36-4=32 ratio is 9, 36=9*4 288 288/9=32 or 288/36*4=32 confidence assessment: 3
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22:27:26 `questionNumber 40000 Just as the ratio of 12 to 3 is 12 / 3 = 4, the ratio of 36 to 4 is 36 / 4 = 9. 36 differs from 4 by 36 - 4 = 32. Since the ratio of 36 to 4 is 9, the number 288 will be in the same ratio to a number which is 1/9 as great, or 288 / 9 = 32. Putting this another way, the question asks for a 'certain number', and 288 is in the same ratio to that number as 36 to 4. 36 is 9 times as great as 4, so 288 is 9 times as great as the desired number. The desired number is therefore 288/9 = 32.
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RESPONSE --> self critique assessment: 3
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22:29:34 `questionNumber 40000 `q009. A triangle has sides 3, 4 and 5. Another triangle has the identical shape of the first but is larger. Its shorter sides are 12 and 16. What is the length of its longest side?
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RESPONSE --> 20, the triangles have a ratio of 4 to 1 confidence assessment: 3
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22:29:43 `questionNumber 40000 ** You need to first see that that each side of the larger triangle is 4 times the length of the corresponding side of the smaller. This can be seen in many ways, one of the most reliable is to check out the short-side ratios, which are 12/3 = 4 and 16/4 = 4. Since we have a 4-to-1 ratio for each set of corresponding sides, the side of the larger triangle that corresponds to the side of length 5 is 4 * 5 = 20. **
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RESPONSE --> self critique assessment: 3
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???f?m?y???? assignment #001 001. Sets Liberal Arts Mathematics I 01-24-2009
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12:26:42 `questionNumber 10000 `q001. Note that there are 4 questions in this assignment. `q001. Let A stand for the collection of all whole numbers which have at least one even digit (e.g., 237, 864, 6, 3972 are in the collection, while 397, 135, 1, 9937 are not). Let A ' stand for the collection of all whole numbers which are not in the collection A. Let B stand for the collection { 3, 8, 35, 89, 104, 357, 4321 }. What numbers do B and A have in common? What numbers do B and A' have in common?
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RESPONSE --> B and A have in common the numbers 8,89,104, and 4321 because they are all in B and have a even digit. B and A' have in common the numbers 3,35, and 357 because they are in B and not in A confidence assessment: 1
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12:27:16 `questionNumber 10000 Of the numbers in B, 8, 89, 104, 4321 each have at least one even digit and so are common to both sets. 3 is odd, both of the digits in the number 35 are odd, as are all three digits in the number 357. Both of these numbers are therefore in A ' .
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RESPONSE --> good :) self critique assessment: 3
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12:30:09 `questionNumber 10000 `q002. I have in a room 8 people with dark hair brown, 2 people with bright red hair, and 9 people with light brown or blonde hair. Nobody has more than one hair color. Is it possible that there are exactly 17 people in the room?
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RESPONSE --> No its not possible because 8+2+9=19 so it's impossible for only 17 people to be in the room. confidence assessment: 1
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12:30:33 `questionNumber 10000 If we assume that dark brown, light brown or blonde, and bright red hair are mutually exclusive (i.e., someone can't be both one category and another, much less all three), then we have at least 8 + 2 + 9 = 19 people in the room, and it is not possible that we have exactly 17.
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RESPONSE --> ok self critique assessment: 3
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12:32:18 `questionNumber 10000 `q003. I have in a room 6 people with dark hair and 10 people with blue eyes. There are only 14 people in the room. But 10 + 6 = 16, which is more than 14. How can this be?
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RESPONSE --> Hair color and eye color are not mutually exclusive so two of people with dark hair must also have blue eyes. confidence assessment: 2
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12:32:36 `questionNumber 10000 The key here is that there is nothing mutully exclusive about these categories-a person can have blue eyes as well as dark hair. So if there are 2 people in the room who have dark hair and blue eyes, which is certainly possible, then when we add 10 + 6 = 16 those two people would be counted twice, once among the 6 blue-eyed people and once among the 10 dark-haired people. So the 16 we get would be 2 too high. To get the correct number we would have to subtract the 2 people who were counted twice to get 16 - 2 = 14 people.
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RESPONSE --> ok self critique assessment: 3
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12:34:10 `questionNumber 10000 `q004. In a set of 100 child's blocks 60 blocks are cubical and 40 blocks are cylindrical. 30 of the blocks are red and 20 of the red blocks are cubical. How many of the cylindrical blocks are red?
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RESPONSE --> If 30 of the bloxks are red and twenty of them are cubicle then that leaves ten red cylindrical blocks. confidence assessment: 2
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12:34:19 `questionNumber 10000 Of the 30 red blocks 20 are cubical, so the rest must be cylindrical. This leaves 10 red cylindrical blocks.
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RESPONSE --> ok self critique assessment: 3
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??hztx?????????? assignment #002 002. Representing Sets Liberal Arts Mathematics I 01-24-2009
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12:43:37 `questionNumber 20000 `q001Note that there are 2 questions in this assignment. `q001. We can represent the collection consisting of the letters a, b, c, d, e, f by a circle in which we write these letters. If we have another collection consisting of the letters a, c, f, g, k, we could represent it also by a circle containing these letters. If both collections are represented in the same diagram, then since the two collections have certain elements in common the two circles should overlap. Sketch a diagram with two overlapping circles. The two circles will create four regions (click below on 'Next Picture'). The first region is the region where the circles overlap. The second region is the one outside of both circles. The third region is the part of the first circle that doesn't include the overlap. The fourth region is the part of the second circle that doesn't include the overlap. Number these regions with the Roman numerals I (the overlap), II (first circle outside overlap), III (second circle outside overlap) and IV (outside both circles). Let the first circle contain the letters in the first collection and let the second circle contain the letters in the second collection, with the letters common to both circles represented in the overlapping region. Which letters, if any, go in region I, which in region II, which in region III and which in region IV?
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RESPONSE --> I=ACF II= ABCDEF III= ACFGK IV= ABCDEFGK confidence assessment: 2
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12:47:08 `questionNumber 20000 The letters a, c and f go in the overlapping region, which we called Region I. The remaining letters in the first collection are b, d, and e, and they go in the part of the first circle that does not include the overlapping region, which we called Region II. The letters g and k go in the part of the second circle that does not include the overlapping region (Region III). There are no letters in Region IV. Click below on 'Next Picture' for a picture.
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RESPONSE --> Okay I see the letters can only be on the diagram once and I was repeating them in each category. self critique assessment: 0
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12:55:06 `questionNumber 20000 `q002. Suppose that we have a total of 35 people in a room. Of these, 20 have dark hair and 15 have bright eyes. There are 8 people with dark hair and bright eyes. Draw two circles, one representing the dark-haired people and the other representing the bright-eyed people. Represent the dark-haired people without bright eyes by writing this number in the part of the first circle that doesn't include the overlap (region II). Represent the number of bright-eyed people without dark hair by writing this number in the part of the second circle that doesn't include the overlap (region III). Write the appropriate number in the overlap (region I). How many people are included in the first circle, and how many in the second? How many people are included in both circles? How many of the 35 people are not included in either circle?
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RESPONSE --> The 1st circle is dark hair and has 20 total people 12 exclusivly The 2nd circle is bright eyes and has 15 total people 7 exclusivly The amount of people in both circles(overlapping) is 8 total but the amount of people in I,II, & III is 27 total. 8 people out of the 35 (IV) are not included in I,II, or III confidence assessment: 2
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12:56:03 `questionNumber 20000 Of the 20 dark-haired people in the preceding example, 8 also have bright eyes. This leaves 12 dark-haired people for that part of the circle that doesn't include the overlap (region I). The 8 having both dark hair and bright eyes will occupy the overlap (region I). Of the 15 people with bright eyes, 8 also have dark hair so the other 7 do not have dark hair, and this number will be represented by the part of the second circle that doesn't include the overlap (region III). We have accounted for 12 + 8 + 7 = 27 people. This leaves 35-27 = 8 people who are not included in either of the circles. The number 8 can be written outside the two circles (region IV) to indicate the 8 people who have neither dark hair nor bright eyes (click below on 'Next Picture').
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RESPONSE --> ok self critique assessment: 3
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?????????x??? assignment #004 004. Subsets; One-to-One Correspondences. Liberal Arts Mathematics I 02-03-2009 ??????????????assignment #005 005. Infinite Sets Liberal Arts Mathematics I 02-03-2009
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00:37:19 `questionNumber 50000 `q001. Note that there are 8 questions in this assignment. The set { 1, 2, 3, ... } consists of the numbers 1, 2, 3, etc.. The etc. has no end. This set consists of the familiar counting numbers, which most of us have long known to be unending. This is one example of an infinite set. Another is the set of even positive numbers { 2, 4, 6, ... }. This set is also infinite. There is an obvious one-to-one correspondence between these sets. This correspondence could be written as [ 1 <--> 2, 2 <--> 4, 3 <--> 6, ... ], where the ... indicates as before that the pattern should be clear and that it continues forever. Give a one-to-one correspondence between the sets { 1, 2, 3, ... } and the set { 1, 3, 5, ... } of odd numbers.
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RESPONSE --> 1 <--> 1, 2 <--> 3, 3 <--> 5, ... confidence assessment: 3
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00:40:30 `questionNumber 50000 `q002. Writing [ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... ] for the correspondence between { 1, 2, 3, ... } and { 1, 3, 5, ... } isn't bad, but the pattern here might be a bit less clear to the reader than the correspondence [ 1 <--> 2, 2 <--> 4, 3 <--> 6, ... ] given for { 1, 2, 3, ... } and { 2, 4, 6, ... }. That is because in the latter case it is clear that we are simply doubling the numbers in the first set to get the numbers in the second. It might not be quite as clear exactly what the rule is in the correspondence [ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... ], except that we know we are pairing the numbers in the two sets in order. Without explicitly stating the rule in a form as clear as the doubling rule, we can't be quite as sure that our rule really works. How might we state the rule for the correspondence [ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... ] as clearly as the 'double-the-first-number' rule for [ 1 <--> 2, 2 <--> 4, 3 <--> 6, ... ]?
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RESPONSE --> where the first set is x and the second set y, then 2x-1=y confidence assessment: 3
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00:40:48 `questionNumber 50000 We might say something like 'choose the next odd number'. That wouldn't be too bad. Even clearer would be to note that the numbers 1, 3, 5, ... are each 1 less than the 'double-the-counting-number' numbers 2, 4, 6. So our rule could be the 'double-the-first-number-and-subtract-1' rule. If we double each of the numbers 1, 2, 3, ... and subtract 1, we get 1, 3, 5, ... .
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RESPONSE --> ok self critique assessment: 2
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00:45:20 `questionNumber 50000 `q003. The 'double-the-number' rule for the correspondence [ 1 <--> 2, 2 <--> 4, 3 <--> 6, ... ] could be made even clearer. First we we let n stand for the nth number of the set {1, 2, 3, ... }, like 10 stands for the 10th number, 187 stands for the 187th number, so whatever it is and long as n is a counting number, n stands for the nth counting number. Then we note that the correspondence always associates n with 2n, so the correspondence could be written0 [ 1 <--> 2, 2 <--> 4, 3 <--> 6, ... , n <--> 2n, ... ]. This tells us that whatever counting number n we choose, we will associate it with its double 2n. Since we know that any even number is a double of the counting number, of the form 2n, this rule also tells us what each even number is associated with. So we can argue very specifically that this rule is indeed a 1-to-1 correspondence. In terms of n, how would we write the rule for the correspondence [ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... ]?
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RESPONSE --> n<->2n-1 confidence assessment: 3
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00:45:34 `questionNumber 50000 The rule for this correspondence is 'double and subtract 1', so n would be associated with 2n - 1. The correspondence would thus be [ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... , n <--> 2n-1, ... ]. Note how this gives a definite formula for the rule, removing all ambiguity. No doubt is left as to how to figure which number goes with which.
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RESPONSE --> self critique assessment: 3
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00:54:06 `questionNumber 50000 `q004. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 5, 10, 15, ... }.
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RESPONSE --> n<->5n confidence assessment: 3
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00:54:13 `questionNumber 50000 It should be clear that each element of the second set is 5 times as great as the corresponding element the first set. The rule would therefore be n <--> 5n.
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RESPONSE --> self critique assessment: 3
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00:58:26 `questionNumber 50000 `q005. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 7, 12, 17, ... }.
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RESPONSE --> n<->5n+2 confidence assessment: 3
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00:58:39 `questionNumber 50000 First we note that the numbers in the second set are going up by 5 each time. This indicates that we will probably somehow have to use 5n in our formula. Just plain 5n gives us 5, 10, 15, ... . It's easy to see that these numbers are each 2 less than the numbers 7, 12, 17, ... . So if we add 2 to 5n we get the numbers we want. Thus the rule is n <--> 5n+2, or in a bit more detail [ 1 <--> 7, 2 <--> 12, 3 <--> 17, ..., n <--> 5n+2, ... ].
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RESPONSE --> self critique assessment: 3
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01:01:47 `questionNumber 50000 `q006. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 3, 10, 17, ... }.
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RESPONSE --> n<->7n-4 confidence assessment: 3
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01:01:54 `questionNumber 50000 The numbers in the second set are going up by 7 each time, so we will probably 7n in our formula. Just plain 7n gives us 7, 14, 21, ... . These numbers are each 4 greater than the numbers 3, 10, 17, ... . So if we subtract 4 from 7n we get the numbers we want. Thus the rule is n <--> 7n-4, or [ 1 <--> 3, 2 <--> 10, 3 <--> 17, ..., n <--> 7n-4, ... ].
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RESPONSE --> self critique assessment: 3
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01:06:16 `questionNumber 50000 `q007. Surprisingly, the numbers { 1, 2, 3, ... } can be put into correspondence with the set of all integer-divided-by-integer fractions (the set of all such fractions is called the set of Rational Numbers). This set would include the fractions 1/2, 1/3, 1/4, ..., as well as fractions like 38237 / 819872 and 232987 / 3. It is a bit surprising that this set could be in 1-1 correspondence with the counting numbers, because just the fractions 1/2, 1/3, 1/4, ... can be put into one-to-one correspondence with the set {1, 2, 3, ... }, and these fractions are less than a drop in the bucket among all possible fractions. These fractions all have numerator 1. The set will also contain the fractions of numerator 2: 2/1, 2/2, 2/3, 2/4, ... . And the fractions with numerator 3: 3/1, 3/2, 3/3, 3/4, ... . We could go on, but the idea should be clear. It certainly seems like there should be more fractions than counting numbers. But it isn't so, as you will see in the lectures and the text. Give a one-to-one correspondence between just the fractions 1/2, 1/3, 1/4, ... and the counting numbers {1, 2, 3, ... }.
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RESPONSE --> 1/2<->1,1/3<->2,1/4<->3,... confidence assessment: 3
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01:06:40 `questionNumber 50000 The correspondence would be [ 1 <--> 1/2, 2 <--> 1/3, 3 <--> 1/4, ... ]. The denominator of the fraction is always 1 greater than the counting number. So if the counting number is n, the denominator the corresponding fraction is n + 1. We would therefore write the correspondence as n <--> 1 / (n+1), or in a bit more detail [ 1 <--> 1/2, 2 <--> 1/3, 3 <--> 1/4, ... , n <--> 1/(n+1), ... ].
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RESPONSE --> self critique assessment: 3
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01:15:16 `questionNumber 50000 `q008. How might we write a one-to-one correspondence between the set {1, 2, 3, ... } of counting numbers and the set union { 1/2, 1/3, 1/4, ... } U { 2/2, 2/3, 2/4, ... } of fractions with numerator 1 or 2?
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RESPONSE --> 1<->1/2,2/2, 2<->1/3,2/3, 3<->1/4,2/4,..., n<->1/(n+1),2/(n+1),... confidence assessment: 1
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01:15:48 `questionNumber 50000 We could alternate between the two sets, using odd numbers with fractions whose numerator is 1 and even numbers with fractions whose numerator is 2. The correspondence would be [ 1 <--> 1/2, 2 <--> 2/2, 3 <--> 1/3, 4 <--> 2/3, 5 <--> 1/4, 6 <--> 2/4, ... ]. It would be a little bit tricky, but not all that difficult, to write this rule in terms of n. However, we won't go into that here.
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RESPONSE --> oh ok self critique assessment: 3
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????????z???????assignment #004 004. Subsets; One-to-One Correspondences. Liberal Arts Mathematics I 02-08-2009
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13:40:11 `questionNumber 40000 `qNote that there are 4 questions in this assignment. `q001. From the collection of letters a, b, c, d, e, how many smaller collections having at least one element may be formed?
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RESPONSE --> (a),(b),(c),(d),(e) (ab), (ac), (ad), (ae) (bc), (bd), (be) (cd), (ce) Fiveteen smaller collections can be formed confidence assessment: 1
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13:40:56 `questionNumber 40000 We will list the original collection by placing its elements between braces: { a, b, c, d, e }. The collection {a, b, c, d} is a smaller collection obtained by eliminating e from the original collection. Similarly we can eliminate d or c or b or a to get the 4-element collections {a, b, c, e}, {a, b, d, e}, { a, c, d, e} and {b, c, d, e}. Alternatively we could simply include either a or b or c or d or e in a 1-element collection, obtaining {a}, {b}, {c}, {d} and {e}. It should be clear that these are the only ways to form collections of 1 or 4 elements. To form a collection of 2 elements we could include a and one other element, obtaining { a, b}, { a, c }, { a, d } and { a, e }. Or we could include b and one other element (excluding a, since we already have the collection { a, b } which is identical to the collection { b, a } since it has exactly the same elements). We obtain { b, c }, { b, d } and { b, e }. {}Or we could include c and one other element (other than a or b, since these have already been listed) to obtain { c, d } and { c, e }. Finally we could include d and the only other element left, e, to get { d, e}. This gives us a complete listing of the 10 sets we can form with 2 of the original elements. This leaves us the 3-element sets, which can be formed by excluding the 2-element sets. Working in reverse order, we can exclude { d, e } to get { a, b, c }, or { c, e } to get { a, b, d }, etc.. The remaining sets we get in this fashion are { a, b, e}, { a, c, d }, { a, c, e}, { a, d, e}, { b, c, d}, {b, c, e}, {b, d, e}, {c, d, e}. We thus have 10 three-element sets. The total number of smaller sets containing at least one element is therefore 5 + 5 + 10 + 10 = 30.
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RESPONSE --> I misread the question and thought they were just 2 element sets.. okay it makes sense now self critique assessment: 2
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13:42:52 `questionNumber 40000 `q002. A one-to-one correspondence between two sets is a rule that associates each element of the each with exactly one element of the other. A natural one-to-one correspondence between the sets { a, b, c } and { 1, 2, 3 } would be to associate a with 1, b with 2, c with 3. This correspondence might be represented as [ a <--> 1, b <--> 2, c <--> 3 ]. This isn't the only possible one-to-one correspondence between these sets. Another might be [ a <--> 2, b <--> 1, c <--> 3 ]. In each case, every element of each set is associated with exactly one element of the other. Another correspondence between the sets might be [ a <--> 3, b<-->2, c<-->3 ]. This correspondence is not one-to-one. In what way does it fail to be a one-to-one correspondence (remember that a one-to-one correspondence is one in which every element of each set is associated with exactly one element of the other).
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RESPONSE --> Because none of the letters correspond with the number 1 confidence assessment: 2
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13:43:05 `questionNumber 40000 [ a <--> 3, b<-->2, c<-->3 ] fails to be a one-to-one correspondence for two reasons. In the first place, 3 is associated with a and with c, and every element of each set is to be associated with exactly one element of the other. 3 is associated with two elements of the other set. It also fails because the element 1 of the second set is not associated with anything in the first set.
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RESPONSE --> ok self critique assessment: 2
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13:46:12 `questionNumber 40000 `q003. There are four possible one-to-one correspondences between the sets {a, b, c} and {1, 2, 3} which were not described in the preceding exercise. One of them would be [ a <--> 3, b <--> 2, c <--> 1 ]. What are the other three possible one-to-one correspondences?
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RESPONSE --> {a <-->1, b<--> 2, c<--> 3} {a<--> 2, b<--> 3, c <-->1} {a<--> 3, b<--> 1, c<--> 2} confidence assessment: 2
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13:46:32 `questionNumber 40000 If we designate the correspondence [ a <--> 1, b <--> 2, c <--> 3 ] as the '123' correspondence, [a <--> 2, b <--> 1, c <--> 3 ] as the '213' correspondence and [a <--> 3, b <--> 2, c <--> 1 ] as the '321' correspondence, in each case listing the numbers associated with a, b, c in that order, we see that the remaining three correspondences could be designated 132, 231 and 312. These correspondences could of course be written out as [ a <--> 1, b <--> 3, c <--> 2 ], [ a <--> 2, b <--> 3, c <--> 1 ] and [ a <--> 3, b <--> 1, c <--> 2 ]. Note that 123, 132, 213, 231, 312, 321 represent the six ways of rearranging the digits 1, 2, 3 into a 3-digit number, listed in increasing order.
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RESPONSE --> ok self critique assessment: 2
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13:47:15 `questionNumber 40000 `q004. Explain why it is not possible to put the sets { a, b, c} and {1, 2, 3, 4} into a one-to-one correspondence.
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RESPONSE --> Because there are more numbers than letters and you would not be able to properly correspond the sets one-on-one confidence assessment: 2
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13:48:43 `questionNumber 40000 One set has 3 elements and the other has 4 elements. A 1-to-1 correspondence has to match each element of each set with exactly one element of the other. It would not be possible to find four different elements of the first set to match with the four elements of the second.
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RESPONSE --> ok self critique assessment: 2
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