#$&* course Mth 158 5/11 2 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: * * My notes here show the half-closed interval [0, 1). When sketching the graph you would use a filled dot at x = 0 and an unfilled dot at x = 1, and you would fill in the line from x = 0 to x = 1. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): not exactly the same but close ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * 1.5.40 (was 1.5.30). How did you fill in the blank for 'if x < -4 then x + 4 ____ 0'? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If x < -4 then x + 4 < 0 because we already know that x is < - 4 confidence rating #$&*: 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): I see what you are saying when you said algebraically. ------------------------------------------------ Self-critique Rating:3 ********************************************* 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 we multiply both sides by -4 -4x < 8 because we multiply by a negative number the sign changes confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * if x> -2 then if we multiply both sides by -4 we get -4x <8. Recall that the inequality sign has to reverse if you multiply or divide by a negative quantity. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * 1.5.58 (was 1.5.48). Explain how you solved the inequality 2x + 5 >= 1. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2x + 5 >= 1 we are going to have to solve for x first subtract 5 from both sides 2x >= -4 now we will divide by 2 x >= -2 confidence rating #$&*: 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): ok, same solution ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 1.5.64 (was 1.5.54). Explain how you solved the inequality 8 - 4(2-x) <= 2x. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 8 - 4(2-x) <= 2x first we will have to us the distributive property to get rid of the () 8 - 8 + 4x <= 2x now we will combine like terms by subtracting 2x from both sides 2x <= 0 x <= 0 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 8- 4(2-x)<= 2x. Using the distributive law: 8-8+4x<= 2x . Simplifying: 4x<=2x . Subtracting 2x from both sides: 2x<=0. Multiplying both sides by 1/2 we get x<=-0 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok, same answer ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 1.5.76 (was 1.5.66). Explain how you solved the inequality 0 < 1 - 1/3 x < 1. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 0 < 1 - 1/3 x < 1 first we need to make two inequalities and solve them individually 0 < 1 - 1/3 x and 1 > 1 - 1/3 x well solve the latter first 1 > 1 - 1/3 x first subtract 1 from both sides 0 > -1/3 x now we divide by -1/3 0 < x because the sign changes when dividing by a negative 0 < 1 - 1/3 x first we subtract 1 from both sides -1 < -1/3 x now we can get rid of the fraction by multiplying both sides by - 3 3 > x the sign changed because we multiplied both sides by a negative 0 < x < 3 this is our final solution confidence rating #$&*: 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, I didn’t have a problem with this one ------------------------------------------------ Self-critique Rating:3 ********************************************* 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: I believe we are trying to find the original inequality -3 < x < 3 first we multiply by -2 6 > - 2x > -6 now we can add 1 to both sides 7 > 1 - 2x > -5 now we rewrite -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): same answer I just started at a more basic level ------------------------------------------------ Self-critique Rating:3 ********************************************* 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 $300 over owner cost. What is range of commission on a sale? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 70 < x < 300. 25 + .40 * excess over owner cost 25 + .40 * 70 < 25 + .40 x < 25 + .40 * 300 follow order of operations 25 + 28 < 25 + .40 x < 25 + 120 combine like terms 53 < 25 + .40 x < 145 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 $25 + 40% of owner cost is in the range 25 + .40 * 70 < 25 + .40 x < 25 + .40 * 300 or 25 + 28 < 25 + .40 x < 25 + 120 or 53 < 25 + .40 x < 145. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): same answer just not so many steps ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * 1.5.123 \ 112. Why does the inequality x^2 + 1 < -5 have no solution? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x^2 + 1 < -5 we will try to solve for x x^2 < -4 there is no square root of -4 so there is no solution confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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:3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!