course Mth 163 I just wanted to make sure that I completed everything required for orientation.Thanks
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11:02:32 `questionNumber 10000 `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 --> 3a + 3b = 9 6a + 5b = 16 1. First I found a multiple to knock out one of the letters and to make them into a + and - equation (-2). -2 (3a+3b)=-2(9) 6a + 5b = 16 Work the equation- -6a-6b=-18 6a + 5b = 16 Combine equation 0a-b=-2 b=2 Then you plug b=2 into the equation 3a+3(2)=9 3a=9-6 3a=3 3a/3=3/3 a=1 So a=1 and b=2 confidence assessment: 2
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11:03:16 `questionNumber 10000 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 --> I got it! self critique assessment: 3
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11:08:23 `questionNumber 10000 `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 --> 4a + 5b = 18 6a + 9b = 30 Since 5 and 9 are prime I multipy them by themselves and a negative to solve for one equation. 9(4a+5b)=9(18) -5(6a=9b)=-5(30) 36a+45b=162 -30a-45b=-150 b cancels itself so we solve for a. 6a=12 6a/6=12/6 a=2 Then apply a=2 into equation. 4(2)+5b=18 5b=18-8 5b=10 b=2 You can plug into other equation to check but a & b =2 confidence assessment: 2
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11:08:43 `questionNumber 10000 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 --> Awesome self critique assessment: 3
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11:09:47 `questionNumber 10000 `q003. If y = 5x + 8, then for what value of x will we have y = 13?
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RESPONSE --> y = 5x + 8 13=5x+8 13-8=5x 5=5x 1=x confidence assessment: 3
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11:09:54 `questionNumber 10000 We first substitute y = 13 into the equation y = 5 x + 8 to obtain 13 = 5 x + 8. Subtracting 8 from both equations and reversing the equality we obtain 5 x = 5, which we easily solve to obtain x = 1.
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RESPONSE --> self critique assessment: 3
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11:11:48 `questionNumber 10000 `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 --> The approximate corresponding coordinates to my graph are x=1 y=-2, x=3 y=5, x=5 y=7, x=7 y=8. The answers to x= 1,3, and 7 were given in the graphing section. confidence assessment: 3
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11:12:38 `questionNumber 10000 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 --> My response was approximately right but it was closer to seven since it was just an estimated drawing. self critique assessment: 3
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11:14:06 `questionNumber 10000 `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 --> The x coordinates would be 1,1.5,3,5,and 1. This is with my freehand graph. confidence assessment: 2
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11:14:39 `questionNumber 10000 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 --> I was really close. self critique assessment: 3
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11:17:11 `questionNumber 10000 `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 --> Selling price=x profit=y If the item sold for $4 the profit would be approx. .06 cents. If the profit is 7 cents the selling price would be approx. $5 This graph is not a realistic model b/c it is an estimate and not actual profits vs. selling price. confidence assessment: 2
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11:17:18 `questionNumber 10000 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 --> self critique assessment: 3
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11:18:26 `questionNumber 10000 `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 --> y = m x + b 4=m(-3) +b confidence assessment: 3
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11:18:34 `questionNumber 10000 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 --> self critique assessment: 3
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11:19:05 `questionNumber 10000 `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 --> y = m x + b -2=m(5)+b confidence assessment: 3
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11:19:11 `questionNumber 10000 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 --> self critique assessment: 3
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11:22:43 `questionNumber 10000 `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 --> 4=-3m+b (-) -2=5m+b 6=-8m Divide all by -8 -3/4=m or -.75 Put m=-3/4 into equation 4=(-3/4)(-3) +b 4=2.25 +b 1.75=b or 7/4=b m=-3/4 and b=7/4 Then you can test them on the other equation. confidence assessment: 2
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11:22:48 `questionNumber 10000 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 --> self critique assessment: 3
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11:23:53 `questionNumber 10000 `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 = m x + b y= -3/4x+7/4 I think the significance is it represents the line we graphed earlier. confidence assessment: 2
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11:24:08 `questionNumber 10000 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 self critique assessment: 2
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