course 䠝|ތqXassignment #001
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00:36:15 `q001. Note that this assignment has 10 questions Solve the following system of simultaneous linear equations: 3a + 3b = 9 6a + 5b = 16.
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RESPONSE --> 6a+5b=16 -6a-6b=18 b=2 6a=6 a=1 confidence assessment: 2
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00:36:34 The system 3a + 3b = 9 6a + 5b = 16 can be solved by adding an appropriate multiple of one equation in order to eliminate one of the variables. Since the coefficient of a in the second equation (the coefficient of a in the second equation is 6)) is double that in the first (the coefficient of a in the first equation is 3), we can multiply the first equation by -2 in order to make the coefficients of a equal and opposite: -2 * [ 3a + 3b ] = -2 [ 9 ] 6a + 5b = 16 gives us -6a - 6 b = -18 6a + 5b = 16 . Adding the two equations together we obtain -b = -2, or just b = 2. Substituting b = 2 into the first equation we obtain 3 a + 3(2) = 9, or 3 a + 6 = 9 so that 3 a = 3 and a = 1. Our solution is therefore a = 1, b = 2. This solution is verified by substituting these values into the second equation, where we get 6 * 1 + 5 * 2 = 6 + 10 = 16.
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RESPONSE --> ok self critique assessment: 3
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00:41:06 `q002. Solve the following system of simultaneous linear equations using the method of elimination: 4a + 5b = 18 6a + 9b = 30.
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RESPONSE --> 3(4a+5b)=54 -2(6a+9b=-60 -3b=-6 b=2 6a+18=30 6a=12 a=2 confidence assessment: 2
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00:41:18 In the system 4a + 5b = 18 6a + 9b = 30 we see that the coefficients of b are relatively prime and so have a least common multiple equal to 5 * 9, whereas the coefficients 4 and 6 of a have a least common multiple of 12. We could therefore 'match' the coefficients of a and b by multiplying the first equation by 9 in the second by -5 in order to eliminate b, or by multiplying the first equation by 3 and the second by -2 in order to eliminate a. Choosing the latter in order to keep the number smaller, we obtain 3 * [4a + 5b ] = 3 * 18 -2 * [ 6a + 9b ] = -2 * 30, or 12 a + 15 b = 54 -12 a - 18 b = -60. Adding the two we get -3 b = -6, so b = 2. Substituting this value into the first equation we obtain 4 a + 5 * 2 = 18, or 4 a + 10 = 18, which we easily solve to obtain a = 2. Substituting this value of a into the second equation we obtain 6 * 2 + 9 * 2 = 30, which verifies our solution.
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RESPONSE --> ok self critique assessment: 3
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00:42:23 `q003. If y = 5x + 8, then for what value of x will we have y = 13?
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RESPONSE --> 5x=5 x=1 confidence assessment: 3
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00:47:06 `q004. Sketch a set of coordinate axes representing y vs. x, with y on the vertical axis and x on the horizontal axis, and plot the points (1, -2), (3, 5) and (7, 8). Sketch a smooth curve passing through these three points. On your curve, what are the y coordinates corresponding to x coordinates 1, 3, 5 and 7? Estimate these coordinates as accurately as you can from your graph. Retain your sketch for use in future assignments.
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RESPONSE --> x=1 y=-2 x=3 y=3 x=5 y=8.5 x=7 y=8 confidence assessment: 2
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00:49:09 The x coordinates 1, 3 and 7 match the x coordinates of the three given points, the y coordinates will be the y coordinates -2, 3 and 8, respectively, of those points. At x = 5 the precise value of x, for a perfect parabola, would be 8 1/3, or about 8.333. Drawn with complete accuracy a parabola through these points will peak between x = 3 in and x = 7, though unless you have a very fine sense of the shape of a parabola your sketch might well peak somewhere to the right of x = 7. The peak of the actual parabola will occur close to x = 6, and the value at x = 7 will be just a bit greater than 8, perhaps 8.5 or so. If your peak was to the right of x = 7, your x = 5 value will be lass than 7.
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RESPONSE --> i drew my graph carefully but it was still of by a hair at the peak point self critique assessment: 2
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00:52:57 `q005. Using your sketch from the preceding exercise, estimate the x coordinates corresponding to y coordinates 1, 3, 5 and 7. Also estimate the x values at which y is 0.
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RESPONSE --> x=1.6666 x=2.3333 x=5 x=7.6666 confidence assessment: 2
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00:53:06 The easiest way to estimate your points would be to make horizontal lines on your graph at y = 1, 3, 5 and 7. You would easily locate the points were these lines intersect your graph, then estimate the x coordinates of these points. For the actual parabola passing through the given points, y will be 1 when x = 1.7 (and also, if your graph extended that far, near x = 10). y = 3 near x = 2.3 (and near x = 9.3). y = 5 at the given point (3, 5), where x = 3. y = 7 near x = 4 (and also near x = 7.7).
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RESPONSE --> ok self critique assessment: 3
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01:00:44 `q006. Suppose the graph you used in the preceding two exercises represents the profit y on an item, with profit given in cents, when the selling price is x, with selling price in dollars. According to your graph what would be the profit if the item is sold for 4 dollars? What selling price would result in a profit of 7 cents? Why is this graph not a realistic model of profit vs. selling price?
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RESPONSE --> I think that a 7 cent profit would be made from the selling price of 4 dollars. A real graph of profit vs. selling would not look like this, it would appear much striaghter. confidence assessment: 1
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01:01:39 To find the profit for a selling price of x = 4 dollars, we would look at the x = 4 point on the graph. This point is easily located by sketching a vertical line through x = 4. Projecting over to the y-axis from this point, you should have obtained an x value somewhere around 7. The profit is the y value, so to obtain the selling price x corresponding to a profit of y = 7 we sketch the horizontal line at y = 7, which as in a preceding exercise will give us x values of about 4 (or x = 7.7, approx.).
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RESPONSE --> This is a little confusing to me I'm still not sure if i did it correctly. self critique assessment: 1
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01:03:01 `q007. On another set of coordinate axes, plot the points (-3, 4) and (5, -2). Sketch a straight line through these points. We will obtain an approximate equation for this line: First substitute the x and y coordinates of the first point into the form y = m x + b. What equation do you obtain?
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RESPONSE --> -3m+b=-2 confidence assessment: 3
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01:03:19 Substituting x = -3 and y = 4 into the form y = m x + b, we obtain the equation 4 = -3 m + b. Reversing the sides we have -3 m + b = 4.
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RESPONSE --> ok self critique assessment: 3
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01:04:07 `q008. Substitute the coordinates of the point (5, -2) into the form y = m x + b. What equation do you get?
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RESPONSE --> 5m+b=-2 confidence assessment: 3
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01:04:15 Substituting x = 5 and y = -2 into the form y = m x + b, we obtain the equation -2 = 5 m + b. Reversing the sides we have 5 m + b = -2
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RESPONSE --> ok self critique assessment: 3
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01:09:08 `q009. You have obtained the equations -3 m + b = 4 and 5 m + b = -2. Use the method of elimination to solve these simultaneous equations for m and b.
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RESPONSE --> 6m=-8 m=-.75 2.25+b=4 b=1.75 5*-.75+1.75=-2 =-2 confidence assessment: 2
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01:09:19 Starting with the system -3 m + b = 4 5 m + b = -2 we can easily eliminate b by subtracting the equations. If we subtract the first equation from the second we obtain -8 m = 6, with solution m = -3/4. Substituting this value into the first equation we obtain (-3/4) * -3 + b = 4, which we easily solve to obtain b = 7/4. To check our solution we substitute m = -3/4 and b = 7/4 into the second equation, obtaining 5 ( -3/4) + 7/4 = -2, which gives us -8/4 = -2 or -2 = -2.
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RESPONSE --> ok self critique assessment: 2
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01:12:20 `q010. Substitute your solutions b = 7/4 and m = -3/4 into the original form y = m x + b. What equation do you obtain? What is the significance of this equation?
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RESPONSE --> y=-.75x+1.75 This appears to be the equation of the line from the sketched line on problem number seven. confidence assessment: 2
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01:12:34 Substituting b = 7/4 and m = -3/4 into the form y = m x + b, we obtain the equation y = -3/4 x + 7/4. This is the equation of the straight line through the given points (-3, 4) and (5, -2).
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RESPONSE --> ok good self critique assessment: 3
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{Zφ̱ assignment #001 001. `query1 Precalculus I 06-15-2007
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01:15:40 Query Introduction to General Themes; Examples (no summary needed) What were some of the things in this introduction that you found interesting or surprising?
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RESPONSE --> I like the way the course is set up and i really liked the visual teaching like the water experiment. confidence assessment: 3
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01:17:46 Query Introductory Flow Experiment (no summary needed) Is the depth changing at a regular rate, at a faster and faster rate, or at a slower and slower rate? Support your conclusion.
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RESPONSE --> They change at slower and slower rates. The depth change decreases between each interval of time. confidence assessment: 2
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01:18:18 ** If you time the water at equal time intervals you should find that the depth changes by less and less with each new interval. If you timed the depth at equal intervals of depth you should find that each interval takes longer than the one before it. } Either way you would conclude that water flows from the hole at a decreasing rate. The reason is that as the water depth decreases the pressure forcing the water out of the hole decreases. **
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RESPONSE --> I found this to be very obvious. self critique assessment: 2
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01:23:22 What does the graph of depth vs. clock time look like? Is it increasing or decreasing? Does the rate of increase or decrease speed up or slow down? Does your graph intercept the y axis? Does it intercept the x axis? How would you describe its overall shape?
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RESPONSE --> It looks like the left half of a parabola. the vertex should either be at zero or ten centimeters, depending on where one was trying to measure from. The actual hole was ten centimeters from the base of the tube. confidence assessment: 2
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