course MTH 158 x?i???????[?|?????assignment #012
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12:26:52 **** query 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.
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RESPONSE --> [0,1) 0 0 is less than or equal to 0 and less than 1 confidence assessment: 3
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12:27:07 10-07-2007 12:27:07 ** 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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12:28:18 1.5.40 (was 1.5.30). How did you fill in the blank for 'if x < -4 then x + 4 ____ 0'?
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RESPONSE --> by working out the problem x+4___0, I found that x is less than -4 so x+4<0 confidence assessment: 3
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12:28:25 ** 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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RESPONSE --> ok self critique assessment: 3
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12:29:17 1.5.46 (was 1.5.36). How did you fill in the blank for 'if x > -2 then -4x ____ 8'?
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RESPONSE --> by working out the problem, -4x___8, I found that x> -2, confidence assessment: 3
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12:29:31 10-07-2007 12:29:31 **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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12:31:05 1.5.58 (was 1.5.48). Explain how you solved the inquality 2x + 5 >= 1.
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RESPONSE --> by working out the problem: 2x+5>=1 2x+5-5>=1-5 2x>=-4 2x/2>=-4/2 x>=-2 confidence assessment: 3
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12:31:12 ** 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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RESPONSE --> ok self critique assessment: 3
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12:33:15 1.5.64 (was 1.5.54). Explain how you solved the inquality 8 - 4(2-x) <= 2x.
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RESPONSE --> first, distribute: 8-8+4x<=2x 4x<=2x 4x-2x<=2x-2x 2x<=0 x<=0 confidence assessment: 3
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12:33:21 ** 8- 4(2-x)<= 2x. Using the distributive law: 8-8+4x<= 2x. Simplifying: 4x<=2x. Subtracting 2x from both sides: 2x<= 0 x<=0 **
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RESPONSE --> ok self critique assessment: 3
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12:45:26 1.5.76 (was 1.5.66). Explain how you solved the inquality 0 < 1 - 1/3 x < 1.
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RESPONSE --> i found that 0<=1-1/3x and 1-1/3x<1 -1<=-1/3x -1/3x<0 3>=x x>0 3>=x>0 confidence assessment: 2
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12:45:36 10-07-2007 12:45:36 ** 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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12:51:10 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.
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RESPONSE --> solving each inequality a<1-2x and 1-2x a-1<-2x -2x
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12:51:14 10-07-2007 12:51:14 ** 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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12:52:15 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?
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RESPONSE --> i am not sure confidence assessment: 0
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12:52:20 10-07-2007 12:52:20 ** 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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12:53:37 1.5.122. Why does the inequality x^2 + 1 < -5 have no solution?
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RESPONSE --> because inequalities are real numbers, and exponents and squares are not. confidence assessment: 1
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12:53:46 10-07-2007 12:53:46 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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