Homework 101

#$&*

course Mth 271

10/7/2010Sorry for the late work. I have internet now at my house but for some reason blackboard would not load. Since I do most of my work at school, I decided to email myself all the hyperlinks and therefore I could not access my code or links.

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

.

Your solution, attempt at solution.

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

010. `query 10

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Zeros can be found at x = -2, 1 , 5 so y = k(x+2)(x-1)(x-5). So at x = 0 the function has a value of 2 so that would equal to .2. Thus the new equation for this question is y = .2(x+2)(x-1)(x-5)

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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):

------------------------------------------------

self-critique rating #$&*:

*********************************************

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:

s = .01 * 65^.25 * 160^.75 = 1.28meters^2

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** Substituting we get

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

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

self-critique rating #$&*:

*********************************************

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 = .01 w^.25* 180^.75 divide both sides by 180^.75

0.0305 = 0.01w^.25, divide both sides by 0.01

3.05237 = w^.25, when I got to this point I had to look at the given solution I got stumped on the final part of this solution.

So the final answer is w = 86.806

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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):

For the final part of this problem I looked at the given solution because everything I did wasn’t working out.

------------------------------------------------

self-critique rating #$&*:

*********************************************

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

If I substitute 70 for the weight then the equation would look like s = .01 *70^.25 h^.75, I then solved 0.01 * 70^.25 which gave me 0.02893. So the new equation is s = .02893 h^.75, I then divided both sides by 0.02893. I then div

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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 necessar

------------------------------------------------

self-critique rating #$&*:

"

&#This looks good. Let me know if you have any questions. &#

#$&*