#$&* course Mth 158 6:10PM 2/23/13 013. `* 13
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Given Solution: * * My notes here show the half-closed interval [0, 1). When sketching the graph you would use a filled dot at x = 0 and an unfilled dot at x = 1, and you would fill in the line from x = 0 to x = 1. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: * 1.5.40 (was 1.5.30). How did you fill in the blank for 'if x < -4 then x + 4 ____ 0'? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x = 4 < 0 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: -4x < 8 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * if x> -2 then if we multiply both sides by -4 we get -4x <8. Recall that the inequality sign has to reverse if you multiply or divide by a negative quantity. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: * 1.5.58 (was 1.5.48). Explain how you solved the inquality 2x + 5 >= 1. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2x + 5 >= 1 2x >= -4 divide both sides by 2: x >= -2 (-2, ∞) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 8- 4(2-x)<= 2x. Using the distributive law: 8-8+4x<= 2x . Simplifying: 4x<=2x . Subtracting 2x from both sides: 2x<=0. Multiplying both sides by 1/2 we get x<=-0 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I wasn't sure how to move on from 2x<=0. I see now. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * 1.5.76 (was 1.5.66). Explain how you solved the inquality 0 < 1 - 1/3 x < 1. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 0 < 1 - 1/3 x < 1 -1 < -1/3x < 0 3 > x > 0 0 < x < 3 (0 , 3) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: -3 < x < 3 Multiply by -2: 6 > -2x > -6 Add 1: 7 > 1 - 2x > -5 -5 < 1 - 2x < 7 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 ------------------------------------------------ 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 $300 over owner cost. What is range of commission on a sale? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I don't understand where the 70$ and 300$ come in. My textbook says the range of excess over owner cost in this question is 200$ to 3000$. x = salesperson's commission 25 + 0.4(200) <= x <= 25 + 0.4(3000) 105 <= x <= 1225 It can range from 105$ to 1225$ confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 $25 + 40% of owner cost is in the range 25 + .40 * 70 < 25 + .40 x < 25 + .40 * 300 or 25 + 28 < 25 + .40 x < 25 + 120 or 53 < 25 + .40 x < 145. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I don't understand where the 70$ and 300% come in. Could you explain???? And if it is simply a mistake of the wrong question or something of that sort, does my inequality work for the question I put up? ------------------------------------------------ Self-critique Rating: 2
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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): OK ------------------------------------------------ Self-critique Rating: OK" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!