course Mth 158 hvȆcע~HIθassignment #013
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22:13:12 **** 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 < or = to x <1
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22:14:13 ** 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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RESPONSE --> I believe I said the same answer but did not describe the graph
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22:16:13 **** query 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 --> 'if x < -4 then x + 4 ____ 0'? 'if x < -4 then x + 4 < 0
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22:16:26 07-03-2006 22:16:26 ** 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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22:18:51 **** query 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 --> if x > -2 then -4x ____ 8'? x > -2 then -4x > 8'?
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22:21:55 **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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RESPONSE --> I forgot the reversing of the sign when multiplying or divideing by a negative quantity.
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22:25:50 **** query 1.5.58 (was 1.5.48). Explain how you solved the inquality 2x + 5 >= 1.
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RESPONSE --> 2x + 5 >= 1 subtract 5 from both sides.. 2x>-4 divide by 2.. x>-2 {x/x>-2} or (-2,infint.)
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22:26:02 ** 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
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22:28:31 **** query 1.5.64 (was 1.5.54). Explain how you solved the inquality 8 - 4(2-x) <= 2x.
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RESPONSE --> 8 - 4(2-x) <= 2x. simplify.. 8-8-4x<=2x add 4x to both sides.. 0<=6x {x/x<=6}
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22:29:16 ** 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 --> I messed up in the problem but I DO understand completely
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22:16:50 **** query 1.5.76 (was 1.5.66). Explain how you solved the inquality 0 < 1 - 1/3 x < 1.
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RESPONSE --> 0 < 1 - 1/3 x < 1 subtract 1 from each side to isolate the term with x...... -1<-1/3x<0 multiply each part by -3 to get rid of the denominator..... 3>x>0 reverse order so numbers get larger as read from left to right.. 0
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22:19:16 ** 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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RESPONSE --> i did the problem differently.. i did not seperate the inequalties.. but i did get the same answer.
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22:29:49 **** query 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 --> if -3 < x < 3 then a < 1 - 2x < b. change the middle part of the inequality to -2x... -3< -2x <3 multiply each part by -2.. 6 > -2x > -6 add 1 to each part.. 7> 1 -2x > -5 a= 7, b= -5
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22:32:14 ** 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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RESPONSE --> i did not write in the traditional ""left toright order"".. other than that.. i did the problem correctly.. however forgetting that part made my a and b answers backwards..
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22:49:02 **** query 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 --> 70<25+40%x<300 (I converted 40% to the fraction 2/5).. 70 < 25 + (2)/(5)x < 300 subtract -25 from each side.. 45< (2)/(5)x < 275 multiply each side by 5 to get rid of the denominator.. 225 < 2x < 1500 divide each side by 2 112.50 < x < 750
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22:52:57 ** 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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RESPONSE --> i did the problem wrong.. but i understand to have multiplyed the 40% with the 70 and 300 and then to add in the owner cost range of 25+40%.. i thought I would have to get rid of the owner cost range completely..
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22:59:47 **** query 1.5.112. Why does the inequality x^2 + 1 < -5 have no solution?
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RESPONSE --> x^2 + 1 < -5 have no solution? x^2 < -5-1 x^2 < -6 ^2*(sqrt of x^2 )< sqrt of -6 no solution because there is no sqrt of -6?
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23:02:18 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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RESPONSE --> i got the sense of the problem and did figure out why there was no solution, however I came up with -6 because I wanted to subtract 1 instead of add 1
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