#$&* course Mth 173 Question: `q query problem 1.6.7 5th ed; 1.6.12 4th; 1.6.9 (was 1.10.16)cubic polynomial representing graph. What cubic polynomial did you use to represent the graph?
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Given Solution: *&*& The given function has zeros at x = -2, 1, 5, so y = k (x+2)(x-1)(x-5). At x = 0 the function has value 2, so 2 = k (0+2)(0-1)(0-5), or 2 = 10 k. Thus k = .2 and the function is y = .2 ( x+2)(x-1)(x-5). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q Query problem 1.6.14 5th ed; 1.6.15 4th ed; 1.6.15 (formerly 1.4.19) s = .01 w^.25 h^.75what is the surface area of a 65 kg person 160 cm tall? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: .01*65^.25*160^.75=1.277meters^2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Substituting we get s = .01 *65^.25 *160^.75 = 1.277meters^2 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q What is the weight of a person 180 cm tall whose surface area is 1.5 m^2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1.5=.01w^.25*180^.75 1.5/180^.75=.01w^.25 .0305=.01w^.25 .0305/.01=w^.25 3.05237=w^.25 w=3.052^4 w=86.806 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Substituting the values we get 1.5 = .01 w^.25*180^.75 . Dividing both sides by 180: 1.5/180^.75 .01w^.25. Dividing both sides by .01: 3.05237 = w^.25 Taking the fourth power of both sides: w = 3.052^4 = 86.806 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q For 70 kg persons what is h as a function of s? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: h=111s^(4/3) confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Substituting 70 for the weight we get s = .01 *70^.25 h^.75 s = .02893 h^.75 s/.02893 = h^.75 34.60 s = h^.75 Taking the 1/.75 = 4/3 power of both sides: (34.60 s)^(4/3) = h h = 111 s^(4/3), approximately ** STUDENT QUESTION Ok don’t understand where the 4/3 comes from but do understand that you need this to establish the resultant. INSTRUCTOR RESPONSE To solve •34.60 s = h^.75 for h you need to take the 1/.75 power of both sides, which gives you (34.60 s)^(1/.75) = (h^.75)^(1/.75). The right-hand side becomes h^(.75 * (1/.75) ) = h^1 = h, so we have h = (34.60 s)^(1 / .75). Since 1 / .75 reduces to 4/3, we have h = (34.70 s)^(4/3). 1 / .75 = 4/3 for the same reason $1.00 / $.75 = 4/3 (ratio of four quarters to three quarters, where by 'quarter' I mean the coin we most commonly put into vending machines). More formally 1 / .75 means 1.00 / .75 = 100 / 75. Dividing numerator and denominator by 25 reduces this to 4/3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!