#$&* course MTH 152 Please note each comment/question in bold print. ------------------------------------------------
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Given Solution:: ** We can list the possibilities, or we can analyze the numbers. First we list: Using letters for the names, there are 12 possibilities (note that the secretary must be either k or e, the others chosen from a, b, d. The secretary, president and treasurer are listed in said order): eab, ead, eba, ebd, eda, edb, kab, kad, kba, kbd, kda, kdb. Next we analyze the numbers: There are two women, so two possibilities for the first person selected. The other two will be selected from among the three men, so there are 3 possibilities for the second person chosen, leaving 2 possibilities for the third. The number of possibilities is therefore 2 * 3 * 2 = 12. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: question 10.1.12 / 11.1.12 In how many ways can the total of two dice equal 5? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: - The basic die comprises of 6 sides (one considers a 3D diagram of a cube) numbered 1 - 6. The possibilities could be: - 1 and 4 - 4 and 1 - 2 and 3 - 3 and 2 - So four combinations. One assumes that because two die are used, the reverse order also counts here (in other words, (1,4) and (4, 1) are technically the same, but relate to each die being rolled?) confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:: ** Listing possibilities on first then second die you can get 1,4, or 2,3 or 3,2 or 4,1. There are four ways. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* question: In how many ways can the total of two dice equal 11? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: - Since each die comprises of a number 1-6, the number of ways the sum can equal 11 are fewer. - (6, 5) - (5,6) - So two ways to roll an 11. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:: ** STUDENT SOLUTION AND INSTRUCTOR RESPONSE: There is only 1 way the two dice can equal 11 and that is if one lands on 5 and the other on 6 INSTRUCTOR RESPONSE: There's a first die and a second. You could imagine that they are painted different colors to distinguish them. You can get 5 on the first and 6 on the second, or vice versa. So there are two ways. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: question 10.1.32 (previously 11.1.36) 5-pointed star, number of complete triangles How many complete triangles are there in the star and how did you arrive at this number? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: - Note that in absence of a text book, the star drawn by me was formed using five line segments. Just to communicate how I drew this, consider the second quadrant on a Cartesian plane in which the y-scale extends 8 positive units, and the x-scale extends 11 positive units. The points for my triangle would be formed by: - Point one: (6, 7) - Point two: (10, 5) - Point three: (9, 1) - Point four: (3, 1) - Point five: (2,5) Where line segments are: Line Segment 1 (point 1 - point 3) Line Segment 2 (point 3 - point 5) Line Segment 3 (point 5 - point 2) Line Segment 4 (point 2 - point 4) Line Segment 5 (point 4 - point 1) - A 5 point star is sketched on graphing paper, then the triangles will be drawn separate for easier visualization, as well, each ‘section’ is labeled by number to determine which triangles go where. (For this reason, this will explain why subsequent numbers cannot form a triangle. Imagine the top point being point one, and continuing around clockwise in increasing number. Point 1 and point 2 cannot make a triangle, but point 1 and point 3 can. - Keeping the above point in mind, one can make a total of 10 triangles. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:: ** If you look at the figure you see that it forms a pentagon in the middle (if you are standing at the very center you would be within this pentagon). Each side of the pentagon is the side of a unique triangle; the five triangles formed in this way are the 'spikes' of the star. Each side of the pentagon is also part of a longer segment running from one point of the start to another. This longer segment is part of a larger triangle whose vertices are the two points of the star and the vertex of the pentagon which lies opposite this side of the pentagon. There are no other triangles, so we have 5 + 5 = 10 triangles. STUDENT COMMENT I am sorry but I cannot see but 8 triangles the 5 spikes , but only 3 larger triangles that incorporate two spikes and the vertex just as you discussed. What am I missing????? INSTRUCTOR RESPONSE: I know your work well, and you are seldom wrong, so I went back to take another look, just to be sure. The figure fools the eye, and it fooled mine enough that I was just about convinced. I clearly saw just three larger triangles and couldn't make myself see five. So I started out my response believing that I had a long-standing error in the solution to this problem. And it took me awhile to convince myself that I was really right in the first place. Here's the reasoning that led me back to my original solution. My eye still doesn't really want to believe it, but I can draw the picture, and if I look at one vertex of the pentagon at a time, I can see it. I'm more convinced than ever that we can't believe our eyes. Here goes. You might want to draw the picture in order to follow the labels: If the points of the 'star' are A, B, C, D and E, in order as we go around the 'star', then each of these points is connected to exactly two of the others, and no point is connected to either of its 'nearest neighbors'. So A is connected to C and D B is connected to D and E C is connected to A and E D is connected to A and B E is connected to B and C. This gives us 10 line segments, namely AC, AD, BD, BE, CE, CA, DA, DB, EB and ED, each potentially the side of a triangle. However these 10 segments are redundant, as follows: AC AD BD BE CA (same segment as AC) CE DA (same segment as AD) DB (same segment as BD) EB (same segment as BE) EC (same segment as CE). Thus we have only five segments connecting the points of the star. Each of these segments forms the longest side of an iscosceles triangle, two of whose vertices are the two points of the 'star', and the third of which is a vertex of the pentagon. The pentagon has five vertices. If you look at each vertex in turn, you will see that it is the vertex of a triangle. So in addition to the five 'points' on the star, there are five triangles, one for each vertex. This raises the total number of triangles to 10. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Good to go! ------------------------------------------------ Self-critique Rating: Great, finally glad to have some sort of geometry related question. question 10.1.36 (formerly 11.1.40 ) 4 x 4 grid of squares, how many squares in the figure? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: - Throughout this problem, one will assume ‘square’ relates to the geometric definition beyond a simple quadrilateral, but whose sides are all an equal number of units. (Exempli gratia, each side has to be a matching number of 1, 2, 3, 4, and so on.) - 4 * 4 - 16 units. But, one must also count the various degrees of the size of the squares too. - The first squares counted are the smallest. A total of 16 squares are counted by simple multiplication. - The next squares increase by a factor of 1. Consider a square whose numbers list as follows: - - - 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 - - - Where each number symbolizes one square; consider the square 1, 2, 5, 6 as the 17th square aside from each other 16 individual squares. - Next one goes down one unit to consider 5, 6, 9, and 10 as square number 18. - Square number 19 would be 9, 10, 13, and 14. - Square number 20 would be 2, 3, 6, and 7. - Square number 21 would be 6, 7, 10, and 11. - Square number 22 would be 10, 11, 14, and 15. - Square number 23 would be 3, 4, 7, and 8. - Square number 24 would be 7, 8, 11, and 12. - Square number 25 would be 11, 12, 15, and 16. - The next set of squares would increase in number, following the same pattern of numerical movement. - - - 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 - - - Square number 26 would be 1, 2, 3, 5, 6, 7, 9, 10, and 11. - Square number 27 would be 5, 6, 7, 9, 10, 11, 13, 14, and 15. - Square number 28 would be 2, 3, 4, 6, 7, 8, 10, 11, and 12. - Square number 29 would be 6, 7, 8, 10, 11, 12, 14, 15 and 16. - Square number 30 would be 1, 2, 3, … ,and 16. - Therefore, the cubic interval 4 x 4 where one considers four columns and four rows, contains 30 squares. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:: ** I count 16 small 1 x 1 squares, then 9 larger 2 x 2 squares (each would be made up of four of the small squares), 4 even larger 3 x 3 squares (each made up of nin small squares) and one 4 x 4 square (comprising the whole grid), for a total of 30 squares. Do you agree? ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok. See my notes regarding cubic intervals, perhaps other students would find this summarization clarifying! If you see this as a good example for students and wished to use that as a student comment in the future, I would certainly agree.
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Given Solution:: **STUDENT SOLTION AND INSTRUCTOR COMMENT: There are 4 ways 30 can be written as the sum of two prime numbers: • 30 = 29 + 1 (instructor note: this is not a sum of two primes) • 30 = 19 + 11 • 30 = 23 + 7 • 30 = 17 + 13 INSTRUCTOR COMMENT: Good, but 1 isn't a prime number. It only has one divisor. The rest of your answers are correct. All sums give you 30, and 7, 11, 13, 17, 19 and 23 are all prime numbers.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): See my methods above and critique as necessary. ------------------------------------------------ Self-critique Rating: ok question 10.1.58 (previously 11.1.60) four adjacent switches; how many settings if no two adj can be off and no two adj can be on YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: - confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:: ** There are a total of 16 settings but only two have the given property of alternating off and on. If the first switch is off then the second is on so the third is off so the fourth is on. If the first is off then the second is on and the third is off so the fourth is on. So the two possibilities are off-on-off-on and on-off-on-off. If we use 0 to represent ‘off’ and 1 to represent ‘on’ these possibilities they are written 0101 and 1010. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): See my response to the final question regarding comments/additional questions concerning this assignment. ------------------------------------------------ Self-critique Rating: question Add comments on any surprises or insights you experienced as a result of this assignment. ** STUDENT COMMENT: No surprises and it's early so i'm reaching for insight as a child reaches for a warm bottle of milk Your comments or questions: - I feel as if I can’t accurately surmise the last question (10.1.58 to be specific) in which each adjacent switch can be on or off. Does one assume there are four switches in a row where one switch turns on the light in a room, and the other turns it off? Some previous student comments and questions: I would like the answers to all the problems I worked in Assignment 11.1. I was surprised that you only ask for a few. I could not answer 11.1. 63 - What is a Cartesain plane? I could not find it in the text. INSTRUCTOR RESPONSE: The question you ask about the Cartesian Plane is a good one and I’ll be glad to answer below, but first let me address your request for answers to all questions. I ask for selected answers so you can submit work quickly and efficiently. I don't provide answers to all questions, since the text provides answers to most of the odd-numbered questions. Between those answers and the comments provided here, most students get enough feedback to be confident in the rest of their work. Another reason I don’t provide answers to all questions is that want students to learn to work ‘forward’ through the problems, which won’t be the case if all answers are provided. If they were most students would fall into the habit of 'working backward' from the answer to the solution. If you have a question on a specific text problem or on anything else, you should submit it using the Submit Question form from the General Information page http://vhcc2.vhcc.edu/dsmith/geninfo/. A direct link to the question form is http://vhcc2.vhcc.edu/dsmith/forms/question_form.htm . To answer your question about the Cartesian Plane: The Cartesian Plane can be thought of as a plane defined by an x axis and a y axis, on which you can specify points by their coordinates. For example the point (5, 9) can be found by starting at the origin (0, 0) and moving 5 units along the positive x axis, then moving 9 units in the direction parallel to the positive y axis. The above is probably sufficient for the work you are doing at this point in your course. A more specific definition: The x axis and the y axis are mutually perpendicular (i.e., at right angles with one another). The x axis is traditionally oriented in the horizontal direction, the y axis in the vertical direction, and is right-handed. The idea of right-handedness is defined in 3 dimensions, but if the x axis is directed ‘toward the right’ and the y axis is ‘up’, then the system will be right-handed. If the x axis was directed ‘toward the right’ and the y axis was ‘down’ the system would be left’-handed’. In a right-handed system, if the y axis is rotated 90 degrees in what we perceive as the ‘right-hand’ direction for rotation (the direction in which you turn a standard right-handed screw or bolt to tighten it), it will then coincide with the x-axis. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&! " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&!