010 query 10

course MTH 173

27June2010 at 2109

010. `query 10

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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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Your solution:

f(x)=kx^p

x=0 at points: -2, 1, 5 on the graph so

k(x+2)(x-1)(x-5)=

y intercept=2

2=k(2*-1*-5)

=k=2/10=1/5

=1/5 (x^2+x-2)(x-5)= (x^3+ x^2-2x-5x^2-5x+10)1/5

=1/5(x^3-4x^2-7x+10)

confidence rating #$&* 3

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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). **

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Self-critique (if necessary):ok

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Self-critique rating #$&*3

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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?

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Your solution:

s = .01 w^.25 h^.75

=.01 (65^.25)(160^.75)=

1.27737cm^2 is surface area

confidence rating #$&* 3

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Given Solution:

** Substituting we get

s = .01 *65^.25 *160^.75 = 1.277meters^2 **

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Self-critique (if necessary): I gave answer in cm should have been in m

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Self-critique rating #$&*3

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Question: `q What is the weight of a person 180 cm tall whose surface area is 1.5 m^2?

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Your solution:

s = .01 w^.25 h^.75

s/(.01h^.75)= w^.25

w^.25=1.5/(.01*180^.75)—substute in variables

w^.25=3.05237—solve for w

multiply both sides by ^4 to get w by itself

w=approx. 86.8056kg

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 **

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Self-critique (if necessary):olk

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Self-critique rating #$&*3

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Question: `q For 70 kg persons what is h as a function of s?

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Your solution:

s = .01 w^.25 h^.75

s/(.01w^.25) =h^.75

s/(.01(70^.25)=h^.75

s/.028925)=h^.75

multiply both sides by^ 4/3 to get rid of ^.75

h=(s/.028925)^(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.

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Self-critique (if necessary):???????????I don’t under stand how you got 34.60 s = h^.75 from s/.02893 = h^.75…. I understand the 4/3 part????????????????

1 / .02893 = 34.60

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Self-critique rating #$&*3

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&#Good work. See my notes and let me know if you have questions. &#

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