course mth173 I finished up the test earlier today, Ms. Umbarger , my procter, should have faxed it in this evening, today was the end of our grading term and so was a little crazy. Rݛ١ͪʒ|assignment #010
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21:57:24 query problem 1.6.7 (was 1.10.16)cubic polynomial representing graph. What cubic polynomial did you use to represent the graph?
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RESPONSE --> cubic polynomial with zeros at (-3, 1, 4) y= k(x+3)(x-1)(x-4) there wasn't a y-intercept value so I couldn't find the value of K
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21:58:12 *&*& The given function has zeros at x = -2, 1, 5, so y = k (x+2)(x-1)(x-5). At x = 0 the function has value 2, so 2 = k (0+2)(0-1)(0-5), or 2 = 10 k. Thus k = .2 and the function is y = .2 ( x+2)(x-1)(x-5). **
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RESPONSE --> The answer had different 0's and a y-intercept but overall I think that the idea is the same.
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21:59:29 Query problem 1.6.15 (formerly 1.4.19) s = .01 w^.25 h^.75what is the surface area of a 65 kg person 160 cm tall?
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RESPONSE --> s = .01 w^.25 h^.75 w= 65 h= 160 s = .01 (65) ^.35 (160) ^.75 s= 1.28 m^2
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21:59:36 ** Substituting we get s = .01 *65^.25 *160^.75 = 1.277meters^2 **
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RESPONSE --> okay
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22:03:02 What is the weight of a person 180 cm tall whose surface area is 1.5 m^2?
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RESPONSE --> s = .01 w^.25 h^.75 1.5 = .01w^.25 (180)^.75 1.5 = .01 w^.25 * 49.14 .031 = .01 w^.25 3.05 = w^(1/4) 3.05^4 = w 86.54 = w
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22:03:14 ** Substituting the values we get 1.5 = .01 w^.25*180^.75 . Dividing both sides by 180: 1.5/180^.75 .01w^.25. Dividing both sides by .01: 3.05237 = w^.25 Taking the fourth power of both sides: w = 3.052^4 = 86.806 **
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RESPONSE --> okay
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22:05:33 For 70 kg persons what is h as a function of s?
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RESPONSE --> s = .01 w^.25 h^.75 s = .01 (70) ^.25 h^.75 s = .029 h^.75 (s/.029) = h^ (3/4) (s/.029)^(4/3) = h
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22:07:01 ** Substituting 70 for the weight we get s = .01 *70^.25 h^.75 s = .02893 h^.75 s/.02893 = h^.75. Taking the 1/.75 = 1.333... power of both sides: (s/.02893)^1.333 = h h = 110.7s^1.333... = 110.7 s^(4/3). **
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RESPONSE --> I didn't simplify the last step that you have there to ultimatly end up with 110.7 s^(4/3) = h
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22:08:24 query problem 1.6.20 flow at 5 cm/s thru rect cross section 3 cm by x cm. What is the expression for the volume emerging from the pipe in 1 second and how did you obtain it?
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RESPONSE --> every second 5cm worth of pipe volume exits (5(3x))/1 = cc / s flow volume
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22:09:32 ** The pipe has rectangular cross-section with dimensions 3 cm by x cm, so its cross-sectional area is 3 x. Every second the fluid filling a 5 cm section of the pipe exits the pipe. The volume of this section is the product of its length and its cross-section, or 5 cm * (3 x cm) = 15 x cm^2. If you express x in cm, then 15 x cm^2 will be in cm^3. **
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RESPONSE --> okay
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22:16:45 query problem 1.6.36 Box with square ends, length + girth < 108. Vol of max box with square of side s. What is your expression for the volume in terms of s?
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RESPONSE --> v = lwh but since the sides are square then ""box"" must be a cube so v= (l) (s^ 2) l + 4s <108 lmax= 108 - 4s v= (108 -4s) s^2 v= 108s^2 -45s^3
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22:19:43 ** If s is the side of the square base then the volume is length * s^2. Since length + girth < 108 and the girth is the perimeter of the square we have length + 4 s < 108 so that the max length is Lmax = 108 - 4s. Thus the max volume with side s is V = L * cross-sectional area = (108 - 4 s) * s^2 = 108 s^2 - 4 s^3. The graph of this V vs. s passes through (0, 0) and (27, 0), first increasing at a decreasing rate until it reaches a peak then decreasing at an increasing rate. Note for future reference that we can find the max possible volume: This max possible volume changes as s changes. To find the max possible volume note that the peak of the graph occurs where the graph levels off before beginning to decrease. The slope of the graph is given by the derivative of V with respect to s. find the derivative of V with respect to s, which we will denote V '. Since the derivative of x^3 is 3 s^2 and the derivative of s^2 is 2s we get V ' = 216 s - 12 s^2. This derivative also represents the rate at which volume changes with respect to s. The max of V will occur when V ' = 0 so we get 216 s - 12 s^2 = 0, which has solution s = 18. The peak of the graph occurs at s = 18, so the max volume occurs when s = 18. If s = 18 then L = 108 - 4 s = 108 - 4 * 18 = 108 - 72 = 36. Thus the volume is V = s^2 * L = 18^2 * 36 = 11664.**
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RESPONSE --> Th formula is the same, I didn't graph the equation but i understand that how you arrived at the volume by using the graph to evaluate the Smax variable.
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