#$&* course Mth 158 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: * * My notes here show the half-closed interval [0, 1). When sketching the graph you would use a filled dot at x = 0 and an unfilled dot at x = 1, and you would fill in the line from x = 0 to x = 1. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):0 ------------------------------------------------ Self-critique rating #$&*0 ********************************************* 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: Since x must be less than -4, x has a value that when 4 is added to it, no matter what it will be less than 0. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * if x<-4 then x cannot be -4 and x+4 < 0. Algebraically, adding 4 to both sides of x < -4 gives us x + 4 < 0. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):0 ------------------------------------------------ Self-critique rating #$&*0 ********************************************* 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: Since x has a value greater than -2, any value assigned to x must be greater than -2, which when multiplied by -4 will always be less than 8. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * if x> -2 then if we multiply both sides by -4 we get -4x <8. Recall that the inequality sign has to reverse if you multiply or divide by a negative quantity. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):0 ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: * 1.5.58 (was 1.5.48). Explain how you solved the inquality 2x + 5 >= 1. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First, I would subtract 5 from the left side of the equation and perform the same operation to the second side of the equation, giving me, 2x>=-4 Then, I would divide 2 from both sides of the equation, giving me x>=-2 confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Starting with 2x+5>= 1 we add -5 to both sides to get 2x>= -4, the divide both sides by 2 to get the solution x >= -2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):0 ------------------------------------------------ Self-critique rating #$&*0 ********************************************* Question: * 1.5.64 (was 1.5.54). Explain how you solved the inquality 8 - 4(2-x) <= 2x. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First distribute -4 to 2-x to get 8-8-4x<2x Subtract 8 from 8 to get 4x<=2x Then subtract 2x to get 2x<=0. Then divide by 2 to get X<=0 confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 8- 4(2-x)<= 2x. Using the distributive law: 8-8+4x<= 2x . Simplifying: 4x<=2x . Subtracting 2x from both sides: 2x<=0. Multiplying both sides by 1/2 we get x<=-0 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):0 ------------------------------------------------ Self-critique rating #$&*0 ********************************************* Question: * 1.5.76 (was 1.5.66). Explain how you solved the inquality 0 < 1 - 1/3 x < 1. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First, separate these inequalitites into two separate equations. 0< 1-1/3x and 1- 1/3x< 1 Similarly in equations, subtract 1 to get -1<-1/3x and -1/3x<0 Separate these equations and solve separately. First, -1<-1/3x Multiply -3 from the right side and also on the left to get 3 > x In the second equation, perform as follows -1/3x<0 Mulitply by -3 to get x > 0 So 0
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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):0 ------------------------------------------------ Self-critique rating #$&*0 ********************************************* Question: * 1.5.94 (was 1.5.84). Explain how you found a and b for the conditions 'if -3 < x < 3 then a < 1 - 2x < b. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First, add 1 to both equations then, 1+ 6>1 - 2x>1-6 7>1-2x>-5 -5<1-2x< 7 This cannot be simplified any further. confidence rating #$&* 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):0 ------------------------------------------------ Self-critique rating #$&*0 ********************************************* 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: First, designate x as being the owners cost. Since x greater than 70 and less than 300, the equation will be set up as follows. 70 < x < 300 The percent .40 * used as the owners cost fits this criteria, therefore the equation .40*70<.40 x< .40*300 and $25 + 40% 25+.40*70< 25+.40x< 25+.40 25+28<25+.40x< 25+120 53<25+.40 x<145 confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 $25 + 40% of owner cost is in the range 25 + .40 * 70 < 25 + .40 x < 25 + .40 * 300 or 25 + 28 < 25 + .40 x < 25 + 120 or 53 < 25 + .40 x < 145. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):0 ------------------------------------------------ Self-critique rating #$&*0 ********************************************* Question: * 1.5.123 \ 112. Why does the inequality x^2 + 1 < -5 have no solution? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First, subtract 1 from the equation to get X^2<-4 Take the square root of both sides of the equation to get you x
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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):0 ------------------------------------------------ Self-critique rating #$&*0 "