course mth 158 I was just checking to see how long it usually takes to get grades back on our tests, and also where we can view those grades(blackboard?). Thank you.
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10:48:35 3.6.6. x = -20 p + 500, 0<=p<=25 What is the revenue function and what is the revenue if 20 units are sold?
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RESPONSE --> To find p... x = -20p + 500 20p = 500 - x p = (500 - x) / 20 R(x) = x ((500 - x) / 20) = x((500/20) - (x/20)) = -(1/20)x^2 + 25x R(20) = -(1/20)(20)^2 + 25(20) = -20 + 500 = $480 confidence assessment: 2
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10:50:34 ** revenue = demand * price = x * p = (-20 p + 500) * p = -20 p^2 + 500 p If price = 24 then we get R = -20 * 24^2 + 500 * 24 = 480. **
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RESPONSE --> The end results are the same, but I dont understand what else happened... my problem said to express the revenue if 20 units were sold... self critique assessment: 2
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11:07:13 3.6.10. P = (x, y) on y = x^2 - 8. Give your expression for the distance d from P to (0, -1)
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RESPONSE --> d = sqrt[(x-0)^2 + (y-(-1)^2] = sqrt[x^2 + (x^2 - 8 + 1)^2] = sqrt(x^2 + x^4 -7x^2 - 7x^2 + 49) sqrt(x^4 - 12x^2 + 49) d(x) = sqrt(x^4 - 12x^2 + 49) confidence assessment: 2
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11:08:05 ** P = (x, y) is of the form (x, x^2 - 8). So the distance from P to (0, -1) is sqrt( (0 - x)^2 + (-1 - (x^2-8))^2) = sqrt(x^2 + (-7-x^2)^2) = sqrt( x^2 + 49 - 14 x^2 + x^4) = sqrt( x^4 - 13 x^2 + 49). **
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RESPONSE --> I see I subtracted 2x^2 instead of x^2 from 14x^2. Oops. self critique assessment: 3
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11:10:42 What are the values of d for x=0 and x = -1?
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RESPONSE --> d(0) = sqrt(0^4 - 12(0)^2 + 49) = sqrt(49) = 7 d(-1) = sqrt((-1)^4 - 12(-1)^2 + 49) sqrt(38) = 5 sqrt(13) confidence assessment: 2
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11:12:32 If x = 0 we have sqrt( x^4 - 13 x^2 + 49) = sqrt(0^4 - 13 * 0 + 49) = sqrt(49) = 7. If x = -1 we have sqrt( x^4 - 13 x^2 + 49) = sqrt((-1)^4 - 13 * (-1) + 49) = sqrt( 64) = 8. Note that these results are the distances from the x = 0 and x = 1 points of the graph of y = x^2 - 8 to the point (0, -1). You should have a sketch of the function and you should vertify that these distances make sense. **
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RESPONSE --> Ah, again, I forgot to change the 12 to 13. I'm sorry self critique assessment: 2
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11:27:18 3.6. 18 (was and remains 3.6.18). Circle inscribed in square. What is the expression for area A as a function of the radius r of the circle?
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RESPONSE --> A(r) = (2r)(2r) = 4r confidence assessment: 2
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11:28:45 A circle inscribed in a square touches the square at the midpoint of each of the square's edges; the circle is inside the square and its center coincides with the center of the square. A diameter of the circle is equal in length to the side of the square. If the circle has radius r then the square has sides of length 2 r and its area is (2r)^2 = 4 r^2. The area of the circle is pi r^2. So the area of the square which is not covered by the circle is 4 r^2 - pi r^2 = (4 - pi) r^2. **
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RESPONSE --> Was I supposed to put the area of the square not covered by the circle? that wasn't one of the questions self critique assessment: 2
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11:29:19 What is the expression for perimeter p as a function of the radius r of the circle?
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RESPONSE --> P(r) = (2r)(4) = 8r confidence assessment: 2
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11:29:24 The perimeter of the square is 4 times the length of a side which is 4 * 2r = 8r. **
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RESPONSE --> ok self critique assessment: 3
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11:47:15 3.6.27 (was 3.6.30). one car 2 miles south of intersection at 30 mph, other 3 miles east at 40 mph Give your expression for the distance d between the cars as a function of time.
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RESPONSE --> The distance formula is h^2 = L^2 + L^2 for a triangle, and that is what this situation creates. d^2 = dc^2 + dc^2, d meaning the distance between the cars, and dc meaning the cars. d^2 = (2 - 30t)^2 + (3 - 40t)^2 d(t) = sqrt[(2 - 30t)^2 + (3 - 40t)^2] sqrt(4 - 120t + 900t^2 + 9 - 240t + 160t^2) sqrt(2,500t^2 - 360t + 5) confidence assessment: 2
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11:48:18 At time t the position of one car is 2 miles south, increasing at 30 mph, so its position function is 2 + 30 t. The position function of the other is 3 + 40 t. If these are the x and the y coordinates of the position then the distance between the cars is distance = sqrt(x^2 + y^2) = sqrt( (2 + 30 t)^2 + (3 + 40t)^2 ) = sqrt( 4 + 120 t + 900 t^2 + 9 + 240 t + 1600 t^2) = sqrt( 2500 t^2 + 360 t + 13). **
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RESPONSE --> I was thinking it was 4 -9 instead of -4 -9 for some reason. self critique assessment: 3
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