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.
Your solution:
Confidence Assessment:
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. **
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Question: * 1.5.40 (was 1.5.30). How did you fill in the blank for 'if x < -4 then x + 4 ____ 0'?
Your solution:
Confidence Assessment:
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. **
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Question: * 1.5.46 (was 1.5.36). How did you fill in the blank for 'if x > -2 then -4x ____ 8'?
Your solution:
Confidence Assessment:
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. **
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Question: * 1.5.58 (was 1.5.48). Explain how you solved the inquality 2x + 5 >= 1.
Your solution:
Confidence Assessment:
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. **
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Question: * 1.5.64 (was 1.5.54). Explain how you solved the inquality 8 - 4(2-x) <= 2x.
Your solution:
Confidence Assessment:
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 **
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Question: * 1.5.76 (was 1.5.66). Explain how you solved the inquality 0 < 1 - 1/3 x < 1.
Your solution:
Confidence Assessment:
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. **
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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.
Your solution:
Confidence Assessment:
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. **
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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?
Your solution:
Confidence Assessment:
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. **
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Question: * 1.5.123 \ 112. Why does the inequality x^2 + 1 < -5 have no solution?
Your solution:
Confidence Assessment:
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. **
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