course MTH 158 After the last question I got a message reading: run time error subscript out of range.
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20:12:01 query 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 --> If x = -20p + 500 then solve for p 20p + x = 500 p = -x/20 + 25 R(x) = x*p x((-x/20) + 25) = (-x^2/20) + 25x if 20 units are sold you are finding f(20) ((-20)^2/20) + 25 * 20 = $480
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20:13:16 ** 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 --> yes, that is much simpler
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20:18:57 query 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 --> working this out from the beginning: d=sqrt[(x-0)^2 + (y + 1)^2] substituiting x^2 - 8 for y d = sqrt [x^2 + (x^2 -8 + 1)^2] (x^2 - 8 + 1)^2 = x^4 - 8x^2 + x^2 - 8x^2 + 64 - 8 + x^2 - 8 + 1 next simplifying: d = sqrt (x^4 - 13x^2 + 57) I am unsure if I did this correctly. I originally worked it out substituting x^2 - 8 under sqrt as follows: (x^2 - 8)^2 +(1)^2 We will see if I did it correctly.
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20:21:17 ** 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 used (0,-1) for points x1, y1 so we have worked it out differently. Did I do it correctly?
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20:22:41 What are the values of d for x=0 and x = -1?
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RESPONSE --> d(0) = sqrt(0^4 - 13*0^2 + 57) = sqrt (57) d(1) = sqrt (-1^4 - 13 * -1^2 + 57) = sqrt (69)
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20:23:17 ** 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 --> again, I had a different answer.
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20:26:39 query 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 --> Area of square = l * w let x = l and y = w r is a line on the graph y=r for function A(r) we woud get: xy A(r) = x * r
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20:33:40 ** 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 --> When I looked at the pic I split it into a graph with all 4 quadrants and x,y coordinates. My answer does not make sense after reading the correct answer but that is where I was coming from. The key for me in this explanation is that the diameter of the circle is equal in length to the side of the square.
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20:35:57 What is the expression for perimeter p as a function of the radius r of the circle?
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RESPONSE --> If the length of the side is 2r: P=2l + 2w 2(2r) + 2(2r) = 4r + 4r = 8r
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20:36:26 ** The perimeter of the square is 4 times the length of a side which is 4 * 2r = 8r. **
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RESPONSE --> right
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20:38:02 query 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 --> d(t) = sqrt {2500t^2 + 360t + 13)
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20:39:16 ** 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
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