query 5

#$&*

course Mth 271

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.

005. `query 5

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

Question: `qexplain why the slope of a depth vs. time trapezoid represents the average rate of change of the depth with respect to the time during the time interval represented

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

We use the change in depth as the rise and the change in time as the run. Rise over run for slope equals change in depth divided by change in time which find the rate of change of the depth.

confidence rating #$&*:

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

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

Given Solution:

`a the specific idea is that ave rate of depth change is [change in depth / change in time] ; rise represents change in depth and run represents change in time so slope = rise/run represents ave rate of depth change. **

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

Self-critique (if necessary):

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

Self-critique Rating:

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

Question: `qexplain why the area of a rate vs. time trapezoid for a given time interval represents the change in the quantity corresponding to that time interval

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The area of a trapezoid is found with the altitude and width. The altitude of a trapezoid represents the average velocity, while the width shows the change in time.

Average altitude * width = change in quanity

confidence rating #$&*:

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

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

Given Solution:

`a The average altitude represents the avg. velocity. The area of a trapezoid involves the altitude, which represents the avg. velocity, and the width, which represents the change in clock time.

When you multiply ave altitude by width you are representing ave vel * change in clock time, which gives change in position.

This reasoning isn't confined to velocities. For any rate vs. clock time graph, average altitude represents approximate average rate, which multiplied by the change in time (not by the time itself) gives you the change in quantity **

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

Self-critique (if necessary):

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

Self-critique Rating:

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

Question: `qtext problem 0.5 #10 add x/(2-x) + 2/(x-2)

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

x/2-x + 2/x-2

@&

x/2-x + 2/x-2

has a very different meaning than

x/(2-x) + 2/(x-2).

I know what you mean, but without appropriate signs of grouping your expressions are ambiguous.

We need to use strict order of operations throughout.

Your solution is in any case good.

*@

x/ -x+2 + 2/x-2

-1 [x/ (x-2)] + 2/ x-2

-x/ x-2 + 2/x-2

-x+2/x-2

-(x-2)/(x-2) = -1

confidence rating #$&*:

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

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

Given Solution:

`a common denominator could be [ (2-x)(x-2) ]. In this case we have

x / (2-x) + 2 / (x-2)

= [ (x-2) / (x-2) ] * [ x / (2-x) ] + [ (2-x) / (2-x) ] * [ 2 / (x-2) ]

= x(x-2) / [ (2-x)(x-2) ] + 2 (2-x) / [ (2-x)(x-2) ]

= [x(x-2) + 2(2-x) ] / [ (2-x)(x-2) ]

= [ x^2 - 2x + 4 - 2x ] / [ (2-x)(x-2) ]

= (x^2-4x+4) / [ -x^2+4x-4 ]

= (x-2)^2 / [-(x-2)^2]

= -1.

NOTE however that there is a SIMPLER SOLUTION:

We can note that x-2 = -(2-x) so that the original problem is -x/(x-2) + 2 /(x-2) = (-x + 2) / (x-2) = -(x-2)/(x-2) = -1. **

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

Self-critique (if necessary):

I got the same answer as you, but did not work out the problem the same. I really do not understand how you worked it out.

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

Self-critique Rating:

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

Question: `qtext problem 0.5 #50 cost = 6 x + 900,000 / x, write as single fraction and determine cost to store 240 units

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

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

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

Given Solution:

`a express with common denominator x:

[x / x] * 6x + 900,000 / x

= 6x^2 / x + 900,000 / x

= (6x^2 + 900,000) / x so

cost = (6x^2+900,000)/x

Evaluating at x = 240 we get cost = (6 * 240^2 + 900000) / 240 = 5190. **

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

Self-critique (if necessary):

I’m not really sure how you worked out this problem. I’m confused because on how to read it to understand it since It has to be typed.

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

Self-critique Rating:

@&

You haven't indicated how you solved the problem.

You did well on the typewriter notation exercise in Orientation. However you might want to review that.

You are advised to write out the expressions you see in the given solution on paper, in standard notation; don't try to read them in typewriter notation until you get used to it, and the way to get used to it is to write things out. It doesn't take long.

*@

"

Self-critique (if necessary):

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

Self-critique rating:

"

Self-critique (if necessary):

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

Self-critique rating:

#*&!

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