course Mth 158 I think I did okay on this, other than hitting enter when I doidn't mean to.
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20:00:31 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 --> The revenue function is R=demand multiplied by price = xp= (-20p+500)p= -20p^2+500p p= $25 The revenue if 20 units of product were sold would be -20*25^2+500(25)= $500
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20:01:00 ** 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 think I messed up somewhere, but I get the idea.
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20:06:11 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 --> Okay, P= some point (x,y) y=x^2-8 According to the Distance Formula: sqrt (0-x)^2+ (1-(x^2-8)= sqrt x^2+ (-7-x^2)^2= sqrt x^4-13x^2+49 d if x=0= 0^4-13(0)^2+49= sqrt 49= d= 7 units d if x=-1 (-1)^4-13(-1)^2+49 sqrt 64 d= 8 units
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20:06:35 ** 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 did more of this problem than what was asked for. Sorry!
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20:07:19 What are the values of d for x=0 and x = -1?
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RESPONSE --> I just answered that- oops. Okay, if x=0, d= sqrt 49= d= 7 units if x=-1, d= sqrt 64= 8 units
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20:07:28 ** 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
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20:09:35 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 --> The diameter of this circle is the same as the length of one side of the square. Diameter= 2r So the side of a square= 2r. Area of a circle= pi r^2. I have no idea where to go from here.
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20:11:26 ** 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 --> Okay, I see now- the area would be the side of the square (2r) just squared= 4r^2. The area of square not covered by circle is area of square minus area of circle= 4r^2-pi r^2= (4-pi)r^2. I don't know why I make things so complicated, this one was common sense.
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20:13:03 What is the expression for perimeter p as a function of the radius r of the circle?
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RESPONSE --> The perimeter of the square in terms of r would be P(r)= 8r because perimeter of the square is 4 times the length of one side, which is 2r in this case.
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20:13:08 ** The perimeter of the square is 4 times the length of a side which is 4 * 2r = 8r. **
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RESPONSE --> ok
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20:16:57 ** 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 --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. I hit enter before I meant to! Okay, if car #1 is 2 miles south of the intersection at 30mph, it would be 2+30t, and car #2 would be 3 miles east of the intersection at 40 mph, it would be = 3+40t. I see how to use the distance formula here- plug in the equations as coordinates= sqrt (2+30t)^2+(3+40t)^2= sqrt (2500t^2+360 t+13) This makes sense. Sorry about the keystrike error.
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