course Mth 163 I'm not sure if this is assignment 1 or not but I hope that is and I'm sending it right. end program½xÖ•wÝξ}Øz¸ÿïÍ‚ÓmÈU”
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17:05:19 `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 --> 1 *-2 3a+3b=9-------- -6a-6b=-18 2 6a+5b=16------ 6a+5b=16 -6b=-18 5b=16 subtract -b=-2 (divide by -1) b=2 substitute 2 for b in 1. 3a+3(2)=9 a=1 substitute both a and b in 2. 6(1)+5(2)=16 6+10=16 confidence assessment: 3
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17:06:19 `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 --> ok confidence assessment:
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17:06:34 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 --> self critique assessment:
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17:09:51 `q003. If y = 5x + 8, then for what value of x will we have y = 13?
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RESPONSE --> x will be 1. sub in the y value for y and solve for x. 13 = 5x + 8 -8 -8 5= 5x 1=x confidence assessment: 2
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17:10:11 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 --> ok self critique assessment: 3
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17:14:54 `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 --> (1,0) (3,5) (5,7) (7,8) confidence assessment: 1
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17:19:25 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 --> after drawing my graph I just don't understand how you got those points maybe I just didn't understand the question fully. self critique assessment:
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17:21:30 `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 --> (3,2) (1,1.5)(5,4)(7,5) confidence assessment:
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17:22:15 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 --> still I'm not seeing it. self critique assessment:
Deconstruct the given solution and explain in detail what you do and do not understand about every part.. &#
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17:25:36 `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 --> Because the measurements are not the same confidence assessment:
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17:27:54 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 --> I don't know how you got x=7.7 self critique assessment:
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17:29:48 `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 --> 4=m(-3)+b confidence assessment:
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17:30:11 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:
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17:31:00 `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 --> -2=5m+b confidence assessment:
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17:31:15 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:
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17:38:14 `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 --> *-1 --- -3 m + b = 4 --- 5 m + b =-2 3m- b=-4 5m+ b=-2 add. 8m=-6 divide by 8 m= -3/4 confidence assessment:
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17:38:22 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:
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17:40:15 `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 m=slope and b=y intercept confidence assessment:
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17:40:29 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:
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