course Mth 158 assignment #027027. `query 27 College Algebra 08-06-2008
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10:48:47 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 --> R(x)=-20p+500 If 20 units were sold r(20)=-20(20)+500 r(20)=100 confidence assessment: 2
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10:49:10 ** 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 --> I understand self critique assessment: 3
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11:00:25 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 --> Distance from (x, x^2-8) to (0, -1) Using distance formula =sqrt (0-x)^2 - (1-(x^2+8))^2 =sqrt -x^2- (1-(x^2+8))^2 confidence assessment: 1
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11:01:27 ** 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 --> Looks like I should have went a little further to get a better answer. self critique assessment: 3
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11:04:52 What are the values of d for x=0 and x = -1?
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RESPONSE --> If x=0 sqrt0^4-13(0)^2+49=sqrt 49=7 If x=-1 sqrt -1^4-13(-1)^2=49=sqrt64=8 confidence assessment: 2
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11:05:04 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 --> ok self critique assessment: 3
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11:09:04 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 --> f(a)=(2r)^2=4r^2 confidence assessment: 1
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11:09:24 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 --> ok self critique assessment: 3
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11:09:56 What is the expression for perimeter p as a function of the radius r of the circle?
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RESPONSE --> f(p)=4r^2-pir^2 confidence assessment: 1
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11:10:07 08-06-2008 11:10:07 The perimeter of the square is 4 times the length of a side which is 4 * 2r = 8r. **
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NOTES -------> ok
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11:10:08 The perimeter of the square is 4 times the length of a side which is 4 * 2r = 8r. **
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RESPONSE --> self critique assessment:
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11:10:11 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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11:19:15 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 --> x would be at 3+40t y would be at 2+30t =sqrt(3+40t)^2 + (2+30t)^2 =sqrt (9+240t+1600t^2+4+120t+900t^2) =sqrt 2500t^2+360t+13 confidence assessment: 2
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11:19:22 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 --> ok self critique assessment: 3
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