Assignment 13 15

#$&*

course Mth 158

7/9/13 8: 52 pm

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.

013. `* 13

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

Question: * 1.5.34 (was 1.5.24). How did you write the interval [0, 1) using an inequality with x? Describe your illustration using the number line.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

0

On the number line the closed point 0 (indicated by a bracket over that point on the line) connects with the soft point 1 (indicated by a parenthesis on that point)

confidence rating #$&*: 3

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

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

Given Solution:

* * My notes here show the half-closed interval [0, 1).

When sketching the graph you would use a filled dot at x = 0 and an unfilled dot at x = 1, and you would fill in the line from x = 0 to x = 1. **

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

Self-critique (if necessary): OK

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

Self-critique Rating: OK

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

Question: * 1.5.40 (was 1.5.30). How did you fill in the blank for 'if x < -4 then x + 4 ____ 0'?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

if x < -4 then x + 4 __<__ 0

confidence rating #$&*: 3

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

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

Given Solution:

* * if x<-4 then x cannot be -4 and x+4 < 0.

Algebraically, adding 4 to both sides of x < -4 gives us x + 4 < 0. **

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

Self-critique (if necessary): OK

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

Self-critique Rating: OK

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

Question: * 1.5.46 (was 1.5.36). How did you fill in the blank for 'if x > -2 then -4x ____ 8'?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

if x > -2 then -4x __<__ 8

Multiplied by a negative, so the sign reverses

confidence rating #$&*: 3

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

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

Given Solution:

* * if x> -2 then if we multiply both sides by -4 we get

-4x <8.

Recall that the inequality sign has to reverse if you multiply or divide by a negative quantity. **

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

Self-critique (if necessary): OK

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

Self-critique Rating: Ok

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

Question: * 1.5.58 (was 1.5.48). Explain how you solved the inquality 2x + 5 >= 1.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

2x + 5 >= 1

2x >= -4

x >=-2

confidence rating #$&*: 3

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

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

Given Solution:

* * Starting with

2x+5>= 1 we add -5 to both sides to get

2x>= -4, the divide both sides by 2 to get the solution

x >= -2. **

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

Self-critique (if necessary): OK

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

Self-critique Rating: OK

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

Question: * 1.5.64 (was 1.5.54). Explain how you solved the inquality 8 - 4(2-x) <= 2x.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

8 - 4(2-x) <= 2x

8 - 8 + 4x <= 2x

4x <= 2x

2x <= 0

x <=0

confidence rating #$&*: 3

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

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

Given Solution:

* * 8- 4(2-x)<= 2x. Using the distributive law:

8-8+4x<= 2x . Simplifying:

4x<=2x . Subtracting 2x from both sides:

2x<=0. Multiplying both sides by 1/2 we get

x<=-0 **

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

Self-critique (if necessary): OK

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

Self-critique Rating: OK

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

Question: * 1.5.76 (was 1.5.66). Explain how you solved the inquality 0 < 1 - 1/3 x < 1.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

0 < 1 - 1/3 x < 1

-1 < - 1/3 x < 0

3 > x > 0

confidence rating #$&*: 3

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

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

Given Solution:

* * Starting with

0<1- 1/3x<1 we can separate this into two inequalities, both of which must hold:

0< 1- 1/3x and 1- 1/3x < 1. Subtracting 1 from both sides we get

-1< -1/3x and -1/3x < 0. We solve these inequalitites separately:

-1 < -1/3 x can be multiplied by -3 to get 3 > x (multiplication by the negative reverses the direction of the inequality)

-1/3 x < 0 can be multiplied by -3 to get x > 0.

So our inequality can be written 3 > x > 0. This is not incorrect but we usually write such inequalities from left to right, as they would be seen on a number line. The same inequality is expressed as

0 < x < 3. **

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

Self-critique (if necessary): OK

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

Self-critique Rating: OK

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

Question: * 1.5.94 (was 1.5.84). Explain how you found a and b for the conditions 'if -3 < x < 3 then a < 1 - 2x < b.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

if -3 < x < 3 then a < 1 - 2x < b

-3 < x < 3 *-2

6 > -2x > -6 + 1

7 > 1 - 2x > -5

a = 7 b = -5

confidence rating #$&*: 3

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

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

Given Solution:

* * Adding 1 to each expression gives us

1 + 6 > 1 - 2x > 1 - 6, which we simplify to get

7 > 1 - 2x > -5. Writing in the more traditional 'left-toright' order:

-5 < 1 - 2x < 7. **

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

Self-critique (if necessary): Ok, though I could have rewritten it in the traditional order

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

Self-critique Rating: OK

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

Question: * 1.5.106 (was 1.5.96). Explain how you set up and solved an inequality for the problem. Include your inequality and the reasoning you used to develop the inequality. Problem (note that this statement is for instructor reference; the full statement was in your text) commision $25 + 40% of excess over owner cost; range is $70 to $300 over owner cost. What is range of commission on a sale?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The range is $25 + 40% (or .4) of the sale price minus the owner cost, which ranges between $200 and $3000 so I set up the formula like so--

25 + 200(.4) <= x <= 25 + 3000 (.4)

25 + 80 <= x <= 24 + 1200

105 <= x <= 1224

[$105, $1225]

confidence rating #$&*: 3

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

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

Given Solution:

* * If x = owner cost then

70 < x < 300.

.40 * owner cost is then in the range

.40 * 70 < .40 x < .40 * 300 and $25 + 40% of owner cost is in the range

25 + .40 * 70 < 25 + .40 x < 25 + .40 * 300 or

25 + 28 < 25 + .40 x < 25 + 120 or

53 < 25 + .40 x < 145. **

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

Self-critique (if necessary): ??? I do not understand this given solution. I thought we were looking for the range of commission paid in dollar amounts? I am not sure how the given solution formed its formula.

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

Self-critique Rating: 1

@&

The numbers in the problem might have changed without my noticing.

The range of owner costs in the given solution is $70 to $300.

For range $200 to $3000 your solution is good.

*@

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

Question: * 1.5.123 \ 112. Why does the inequality x^2 + 1 < -5 have no solution?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

x^2 + 1 < -5

x^2 < -4

x < 2i

There is no real solution because the sort of -4 yields an imaginary solution.

confidence rating #$&*: 3

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

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

Given Solution:

* * STUDENT SOLUTION: x^2 +1 < -5

x^2 < -4

x < sqrt -4

can't take the sqrt of a negative number

INSTRUCTOR COMMENT: Good.

Alternative: As soon as you got to the step x^2 < -4 you could have stated that there is no such x, since a square can't be negative. **

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

Self-critique (if necessary):

??? Is what I got still correct? I thought we were supposed to extract the imaginary roots, or are we only meant to do that in certain instances?

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

Self-critique Rating: 2"

@&

We know that the square root of a negative number isn't a real number, so it isn't necessary to actually find the imaginary roots, but that is certainly one way to get to the answer. So your solution is good.

*@

Self-critique (if necessary):

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

Self-critique rating:

Self-critique (if necessary):

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

Self-critique rating:

#*&!

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