course MTH 163 I believe I'm getting the hang of things. I'm still worried though that I'm not doing such a good job and that I have a bad grade in this class. Math is just NOT my friend :-(
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21:54:24 `q001. Note that this assignment has 10 questions Sketch the function y = x and describe your graph. Describe how the graphs of y = .5 x and y = 2 x compare with the graph of y = x.
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RESPONSE --> The graph of the function y = x is a diagonal line that increases from left to right and goes straight through the point (0, 0). The graph of y = .5x is similar to the graph of y = x, except the line has been moved up .5 but it still passes through (0, 0). The graph of the function y = 2x is similar to both y = x and y = .5x except that the line has been moved up 2 spaces; it still passes through (0, 0).
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21:58:07 The graph of y = x consists of a straight line through the origin, having slope 1. This line has basic points (0,0) and (1,1). The points of the graph of y = .5 x all lie twice as close to the x-axis as the points of the graph of y = x. The point (0,0) of the y = x graph is already on the x-axis, so the corresponding point on the graph of y = .5 x is also (0,0). The point (1,1) of the y = x graph lies 1 unit above the x-axis, so the corresponding point on the graph of y = .5 x will lie twice as close, or .5 units above the x-axis, so that the corresponding point is (1, .5). The graph of y = .5 x Thus passes through the points (0,0) and (1,.5). Of course this result could have been found by simply plugging 0 and 1 into the function y = .5 x, but the point here is to see that we can get the same result if we think of moving all points twice as close. This order thinking will be useful when dealing with more complex functions. Thinking along similar lines we expect the points of the graph of y = 2 x to all lie twice as far from the x-axis as the points of the function y = x. Thus the two basic points (0,0) and (1,1) of the y = x graph will correspond to the points (0,0) and (1,2) on the graph of y = 2 x.
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RESPONSE --> I forgot to include that the other point besides (0, 0) is (1, 1). But i understand why it would be in there. I also understand your answer; i was almost saying the same thing, just a little differently. My concept was the same, but my wording was different.
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21:59:46 `q002. If we were to sketch all the graphs of the form y = a x for which .5 < a < 2, what would our sketch look like?
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RESPONSE --> it would look the same as y = x, except each line would have different slopes. They would each still pass through (0, 0).
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22:01:34 If a =.5 then our function is y = .5 x and the basic points will be (0,0) and (1,.5), as seen in the preceding problem. Similarly if a = 2 then our function is y = 2 x, with basic points (0,0) and (1,2). For .5 < a < 2, our functions will lie between the graphs of y = .5 x and y = 2 x. Since these two functions have slopes .5 and 2, the slopes of all the graphs will lie between .5 and 2. We could represent these functions by sketching dotted-line graphs of y = .5 x and y = 2 x (the dotted lines indicating that these graphs are not included in the family, because the < sign does not include equality). We could then sketch a series of several solid lines through the origin and lying between the two dotted-line graphs.
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RESPONSE --> I understand what you're saying.
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22:07:20 `q003. Describe how the graphs of y = x - 2 and y = x + 3 compare with the graph of y = x. If we were to sketch all graphs of the form y = x + c for -2 < x < 3, what would our graph look like?
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RESPONSE -->
The graph of y = x - 2 passes through the points (0, -2) and (0, 2); the line is 2 units below the line of y = x. The graph of y = x + 3 passes through the points (0, -3) and (0, 3); this line is 3 units above the line of y = x. In the graphs for the form of y = x + c for -2 .................................................
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22:09:13
The graph of y = x - 2 lies at every point 2 units below the corresponding point on the graph of y = x, so this graph is parallel to the graph of y = x and 2 units lower. Similarly the graph of y = x + 3 lies parallel to the graph of y = x and 3 units higher.
To sketch the family y = x + c for -2 < x < 3, we first can draw dotted-line graphs of y = x - 2 and y = x + 3, then a series of several solid line graphs, all parallel to the graph of y = x, lying between the two dotted-line graphs.
STUDENT COMMENT: I got a little confused with y = x + c part, but I understand the first part completely.
** The instructions said to sketch all graphs of the form y = x + c for -2 < x < 3. So for example c could be -1, 0, 1 or 2, giving us the functions y = x - 1, y = x, y = x + 1 andy x+ 2. c could also be -1.9, or .432, or 2.9, giving us functions y = x - 1.9, y = x + .432, y = x + 2.9. c can be any number between -2 and 3.
These graphs are as described in the given solution. **
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RESPONSE -->
I believe I was correct. I understand.
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22:14:02
`q004. Describe how the graph of y = 2 x compares with the graph of y = x.
Describe how the graph of y = 2 x - 2 compares with the graph of y = 2 x.
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RESPONSE -->
The graph of y = 2x compares to the graph of y = x in that the graph of y = 2x goes through the origin, as that of the other one, but it has a slope of 2.
The graph of y = 2x - 2 compares to the graph of y = 2x in that the graph of y = 2 -2 is 2 units below the line of y = 2x. They are parallel to eachother.
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22:14:43
The graph of y = 2 x lies at every point twice as far the x-axis as that of y = x. This graph passes through the points (0,0) and (1, 2), i.e., passing through the origin with slope 2.
The graph of y = 2x - 2 will lie 2 units below the graph of y = 2 x. This graph will therefore have a slope of 2 and will pass-through the y axis at (0, -2).
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RESPONSE -->
I was correct I believe.
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22:16:57
`q005. Suppose we graph y = 2 x + c for all values of c for which -2 < c < 3. What with our graph look like?
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RESPONSE -->
If we were to graph all values of c for which -2 < c < 3, then the lines would be parallel to the line of y = 2x, except they would each have different slopes and not each would pass through the origin.
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22:18:46
Each graph will lie c units vertically from the graph of y = 2 x, therefore having slope 2 the passing through the y-axis at the point (0, c). The family of functions defined by y = 2 x + c will therefore consist of a series of straight lines each with slope 2, passing through the y-axis between (0, -2) and (0, 3).
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RESPONSE -->
I was wrong in that i said each line would have different slopes. I realize that they will each have the same slope of 2 since y = 2x.
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22:28:01
`q006. Sketch two points, not particularly close to one another, with one point in the second quadrant and the other in the first, with clearly different y values. Label the first point (x1, y1) and the second (x2, y2). Draw a straight line passing through both of these points and extending significantly beyond both. In terms of the symbols x1, x2, y1, and y2, what is slope of this straight line?
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RESPONSE -->
The line that i sketched passed through points (-2, 2) and (3, 1). The slope of this line would be -.2.
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22:28:41
The rise of a line is from y = y1 to y = y2, a rise of y2-y1. The run is similarly found to be x2-x1. The slope is therefore
slope = (y2-y1) / (x2-x1).
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RESPONSE -->
i was correct.
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22:32:23
`q007. On the sketch you made for the preceding problem, and add a point (x, y) on your straight line but not between the two points already labeled, and not too close to either. What is the slope from (x1, y1) to (x, y)?
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RESPONSE -->
The (x, y) point on my straight line that i chose was (-6, 3); the slope from (x1, y1) to (x, y) is .05
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22:35:45
The slope from (x1, y1) to (x, y) is
slope = rise/run = (y - y1) / (x - x1).
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RESPONSE -->
I was wrong. I understand that the function to find the slope of the two points is (y - y1) / (x - x1). In this case, the slope for my points should have been -.25.
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22:37:30
`q008. Should the slope from (x1, y1) to (x, y) be greater than, equal to or less than the slope from (x1, y1) to (x2, y2)?
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RESPONSE -->
The slope for my points of (x1, y1) to (x, y) is less than the slope of (x1, y1) to (x2, y2).
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22:38:48
The slope between any two points of a straight line must be the same. The two slopes must therefore be equal.
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RESPONSE -->
I understand what you're saying, I just don't understand how I got two different answers for slope. I got -.2 and -.25. maybe I miss calculated or something ???
If you got two different answers for the slope of the same straight line, there is an error somewhere--most likely just a little inaccuracy in estimating the coordinates of the points on your graph.
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22:39:49 `q009. The slope from (x1, y1) to (x, y) is equal to the slope from (x1, y1) to (x2, y2). If you set the expressions you obtained earlier for the slopes equal to one another, what equation do you get?
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RESPONSE --> i don't understand the question ???
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22:40:36 The slopes are (y2 - y1) / (x2 - x1) and (y - y1) / (x - x1). Setting the two slopes equal we obtain the equation (y - y1) / (x - x1) = (y2 - y1) / (x2 - x1).
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RESPONSE --> I understand the answer and what the question was saying now.
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22:42:45 `q010. Setting the two slopes equal you should have obtained (y - y1)/(x - x1) = (y2 - y1)/(x2 - x1). Solve this equation for y.
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RESPONSE --> I'm not exactly clear on how to do this. ??? I'm probably thinking to hard about it and know how to do it...
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22:45:11 Starting with (y - y1) / (x - x1) = (y2 - y1) / (x2 - x1), we wish to isolate y on the left-hand side. We therefore begin by multiplying both sides by (x - x1) to obtain (y - y1) = (y2 - y1) / (x2 - x1) * (x - x1). We could then add y1 to both sides to obtain y = (y2 - y1) / (x2 - x1) * (x - x1) + y1.
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RESPONSE --> yep, i was thinking to hard about it. I understand how to solve the problem for y.
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qв}{{}CށBǧ Student Name: assignment #011
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22:47:33 `q001. Note that this assignment has 11 questions How many squares one foot on a side would it take to construct a square two feet on a side?
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RESPONSE --> I don't really understand the problem, but my quess would be 2 feet, or twice as many squares as the one foot on a side. what is meant by ""foot on a side""??
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22:49:46 A common response is that it takes 2 one-ft. squares to make a 2-foot square. However, below thought shows that this isn't the case. If we put 2 one foot squares side by side we get a one-foot by two-foot rectangle, not a square. If we put a second such rectangle together with the first, so that we have 2 rows with 2 squares in a row, then we have a two-foot square. Thus we see that it takes 4 squares one foot on a side to make a square 2 ft. on a side.
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RESPONSE --> I understand the problem now. and how it would be 4 one-foot squares instead of two.
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22:51:09 `q002. How many cubes one foot on a side would it take to construct a cube two feet on a side?
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RESPONSE --> It would take 4 cubes one foot to construct a cube two feet on a side.
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22:51:50 We could begin by constructing two rows with two cubes in a row, which would sit on a square two feet by two feet. However this would not give is a cube two feet on a side, because at this point the figure we have constructed is only one foot high. So we have to add a second layer, consisting of two more rows with two cubes a row. Thus we have 2 layers, each containing 2 rows with 2 cubes in a row. Each layer has 4 cubes, so our two layers contain 8 cubes.
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RESPONSE --> I think I understand.
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22:53:11 `q003. How many squares one foot on a side would it take to construct a square three feet on a side?
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RESPONSE --> it would take 9 squares one foot on a side to construct a square three feet on a side.
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22:53:30 We would require three rows, each with 3 squares, for a total of 9 squares.
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RESPONSE --> i was correct
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22:56:17 `q004. How many cubes one foot on a side would take to construct a cube three feet on a side?
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RESPONSE --> it would take 54 cubes one foot on a side to construct a cube three feet on a side. ???
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22:59:19 This would require three layers to make a cube three feet high. Each layer would have to contain 3 rows each with three cubes. Each layer would contain 9 cubes, so the three-layer construction would contain 27 cubes.
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RESPONSE --> I can see why it would be 27 cubes since 9 * 3 is 27, but I don't understand why it would only be 27 cubes, when to make a cube, it would have to be 6-sided and each side have 9 cubes. ???
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23:00:41 `q005. Suppose one of the Egyptian pyramids had been constructed of cubical stones. Suppose also that this pyramid had a weight of 100 million tons. If a larger pyramid was built as an exact replica, using cubical stones made of the same material but having twice the dimensions of those used in the original pyramid, then what would be the weight of the larger pyramid?
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RESPONSE --> The weight would be twice as much as the first pyramid, therefore it would weigh 200 million tons. ???
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23:02:47 Each stone of the larger pyramid has double the dimensions of each stone of the smaller pyramid. Since it takes 8 smaller cubes to construct a cube with twice the dimensions, each stone of the larger pyramid is equivalent to eight stones of the smaller. Thus the larger pyramid has 8 times the weight of the smaller. Its weight is therefore 8 * 100 million tons = 800 million tons.
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RESPONSE --> I don't understand the concept of each stone of the larger pyramid is equivalent to eight stones of the smaller one. where did the 8 come from??
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23:05:16 `q006. Suppose that we wished to paint the outsides of the two pyramids described in the preceding problem. How many times as much paint would it take to paint the larger pyramid?
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RESPONSE --> It would take 8 times as much paint to paint the larger pyramid. ???
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23:06:53 The outside of each pyramid consists of square faces of uniform cubes. Since the cubes of the second pyramid have twice the dimension of the first, their square faces have 4 times the area of the cubes that make up the first. There is therefore 4 times the area to paint, and the second cube would require 4 times the paint
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RESPONSE --> I was wrong, but I think I get it now...
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23:08:32 `q007. Suppose that we know that y = k x^2 and that y = 12 when x = 2. What is the value of k?
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RESPONSE --> the value of k is 3
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23:09:11 To find the value of k we substitute y = 12 and x = 2 into the form y = k x^2. We obtain 12 = k * 2^2, which we simplify to give us 12 = 4 * k. The dividing both sides by 410 reversing the sides we easily obtain k = 3.
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RESPONSE --> i was correct
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23:10:06 `q008. Substitute the value of k you obtained in the last problem into the form y = k x^2. What equation do you get relating x and y?
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RESPONSE --> since k = 3 then the equation would be y = 3x^2.
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23:10:25 We obtained k = 3. Substituting this into the form y = k x^2 we have the equation y = 3 x^2.
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RESPONSE --> i was correct.
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23:11:48 `q009. Using the equation y = 3 x^2, determine the value of y if it is known that x = 5.
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RESPONSE --> The value of y = 75 since x = 5.
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23:12:05 If x = 5, then the equation y = 3 x^2 give us y = 3 (5)^2 = 3 * 25 = 75.
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RESPONSE --> i was correct.
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23:17:03 `q010. If it is known that y = k x^3 and that when x = 4, y = 256, then what value of y will correspond to x = 9? To determine your answer, first determine the value of k and substitute this value into y = k x^3 to obtain an equation for y in terms of x. Then substitute the new value of x.
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RESPONSE --> If x = 4 when y = 256, then the value of k would be 4 because 4^3 is 64 and 64 * 4 is 256. So is x = 9 and k = 4, then the value of y will be 2916 because 9^3 is 729 and 4 * 729 is 2916.
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23:17:55 To we first substitute x = 4, y = 256 into the form y = k x^3. We obtain the equation 256 = k * 4^3, or 256 = 64 k. Dividing both sides by 64 we obtain k = 256 / 64 = 4. Substituting k = 4 into the form y = k x^3, we obtain the equation y = 4 x^3. We wish to find the value of y when x = 9. We easily do so by substituting x equal space 9 into our new equation. Our result is y = 4 * 9^3 = 4 * 729 = 2916.
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RESPONSE --> i was correct.
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23:22:12 `q011. If it is known that y = k x^-2 and that when x = 5, y = 250, then what value of y will correspond to x = 12?
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RESPONSE --> If x = 5 when y = 250 in the equation y = k x^2, then k = 10. and if k = 10 and x = 12, then y will equal 1440. because 12^2 = 144 and 144 * 10 is 1440.
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23:24:08 Substituting x = 5 and y = 250 into the form y = k x^-2 we obtain 250 = k * 5^-2. Since 5^-2 = 1 / 5^2 = 1/25, this becomes 250 = 1/25 * k, so that k = 250 * 25 = 6250. Thus our form y = k x^-2 becomes y = 6250 x^-2. When x = 12, we therefore have y = 6250 * 12^-2 = 6250 / 12^2 = 6250 / 144 = 42.6, approximately.
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RESPONSE --> I don't understand where i went wrong and why k = 6250 instead of 10.
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