course Mth 158 So, when I do multiple q_a assignments I can do them all and send them in one form. I am getting the hang of this I think.
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20:19:52 **** 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) consists of all numbers x for which 0< or = x<1 using the half-open interval [a,b), a
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20:20:22 ** 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 --> Okay! self critique assessment: 2
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20:22:03 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 confidence assessment: 2
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20:22:27 ** 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 --> yep self critique assessment: 2
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20:23:34 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 multiply both sides by -4 confidence assessment: 2
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20:24:20 **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 --> ooops...I didn't reverse but now I know. self critique assessment: 2
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20:29:17 1.5.58 (was 1.5.48). Explain how you solved the inquality 2x + 5 >= 1.
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RESPONSE --> 2x+5>=1 2x+5-5>=1-5 subtract both sides by 5 2x>= -4 2x/2>=-4/2 divide both sides by 2 x>= -2 {x|x>= -2} [-2,infinity) confidence assessment: 2
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20:29:45 ** 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 --> whooah! so I am getting the hang of this. self critique assessment: 2
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20:33:41 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 8 - 8 - 4x<=2x -4x<=2x -4x-2x<=2x-2x -6x<=0 -6x/-6<=0/-6 x= - 0/6 confidence assessment: 2
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20:40:29 1.5.76 (was 1.5.66). Explain how you solved the inquality 0 < 1 - 1/3 x < 1.
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RESPONSE --> ooops. I worked that a little wrong should have changed my - to a plus. 0<1 - 1/3x<1 (3)0<(1-1/3x)3<1(3) multiply by 3 0<3-x<3 0-3<3-x-3<3-3 subtract by 3 -3<-x<0 divide by -1 3>x>0 {x|3>x>0} or (0,3) confidence assessment: 2
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20:41:11 ** 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 --> Okay so I got it right but didn't flip the expression. self critique assessment: 2
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20:45:22 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
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20:48:05 ** 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 --> So to get the 6>1-2x>-6, you need to multiply by 2?? I will work on this kinda of problem. self critique assessment: 2
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20:51:38 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 really have a hard time with problem solving. confidence assessment: 0
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20:58:21 ** 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 --> Ok to get this i need to find an x out of the problem, like in this one x = owner cost, then I need to think of what I am working on like in this one inequalities, given 70
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21:01:22 1.5.112. Why does the inequality x^2 + 1 < -5 have no solution?
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RESPONSE --> Be cause you can't sqrt a negative number. confidence assessment: 2
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?H|I?}???????h???P?????? assignment #014 014. `query 14 College Algebra 02-21-2007
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21:04:41 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 --> ok thats what I got but it kicked me out of the program. who knows! self critique assessment: 2
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21:22:16 1.6.12 (was 1.6.6). Explain how you found the real solutions of the equation | 1 - 2 z | + 6 = 0.
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RESPONSE --> |1-2z|+6=0 |1-2z|+6-6=0-6 |1-2z|= - 6 1-2z= - 6 or 1-2z=6 1-12z= - 6-1 1-1-2z=6-1 -2z= - 7 -2z=5 -2x/-2= - 7/ - 2 -2z/-2=5/-2 x=7/2 or z= - 5/2 {x|x= - 5/2, 7/2} confidence assessment: 2
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21:23:39 ** Starting with | 1-2z| +6 = 9 we add -6 to both sides to get | 1 - 2z| = 3. We then use the fact that | a | = b means that a = b or a = -b: 1-2z=3 or 1-2z= -3 Solving both of these equations: -2z = 2 or -2z = -4 we get z= -1 or z = 2 We express our solution set as {-2/3,2} **
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RESPONSE --> I thought that was equal to 0 so I solved for 0. hmmmmmmm.... self critique assessment: 2
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??·????·??N????h??? assignment #015 015. `query 15 College Algebra 02-21-2007
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21:25:24 1.6.12 (was 1.6.6). Explain how you found the real solutions of the equation | 1 - 2 z | + 6 = 0.
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RESPONSE --> ok so it is equal to 0, so how did you get |1-2z|+6=9??? that confused me. confidence assessment: 3
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21:25:54 ** Starting with | 1-2z| +6 = 9 we add -6 to both sides to get | 1 - 2z| = 3. We then use the fact that | a | = b means that a = b or a = -b: 1-2z=3 or 1-2z= -3 Solving both of these equations: -2z = 2 or -2z = -4 we get z= -1 or z = 2 We express our solution set as {-2/3,2} **
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RESPONSE --> lol Okay! self critique assessment: 2
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21:32:55 1.6.30 (was 1.6.24). Explain how you found the real solutions of the equation | x^2 +3x - 2 | = 2
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RESPONSE --> |x^2+3x-2|=2 |u|=a u=a or u= -a x^2+3x-2=2 or x^2+3x-2= -2 x^2+3x-2-2=2-2 x^2+3x-2+2= -2+2 x^2+3x-4=0 x^2+3x-0=0 (x+4)(x-1)=0 x(x+3)=0 x= -4 or x=1 or x=0 or x= -3 confidence assessment: 2
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21:33:28 ** My note here might be incorrect. If the equation is | x^2 +3x -2 | = 2 then we have x^2 + 3x - 2 = 2 or x^2 + 3x - 2 = -2. In the first case we get x^2 + 3x - 4 = 0, which factors into (x-1)(x+4) = 0 with solutions x = 1 and x = -4. In the second case we have x^2 + 3x = 0, which factors into x(x+3) = 0, with solutions x = 0 and x = -3. **
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RESPONSE --> YES! self critique assessment: 2
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21:38:05 1.6.36 (was 1.6.30). Explain how you found the real solutions of the inequality | x + 4 | + 3 < 5.
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RESPONSE --> |x+4|+3<5 |u| |x+4|+3-3<5-3 |x+4|<2 -2
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21:38:29 STUDENT SOLUTION: | x+4| +3 < 5 | x+4 | < 2 -2 < x+4 < 2 -6 < x < -2
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RESPONSE --> yes yes yes I'm so proud of myself!
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21:45:42 1.6.48 (was 1.6.42). Explain how you found the real solutions of the inequality | -x - 2 | >= 1.
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RESPONSE --> |-x-2|>=1 |u|>=a u<= - a or u>=a -x-2>=1 or -x-2<= - 1 -x-2+2>=1+2 -x-2+2<= - 1+2 -x>=3 -x<=1 x>= -3 or x<= - 1 {x|x>= - 3 or x<= - 1} confidence assessment: 2
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21:47:34 **Correct solution: | -x -2 | >= 1 Since | a | > b means a > b or a < -b (note the word 'or') we have -x-2 >= 1 or -x -2 <= -1. These inequalities are easily solved to get -x >= 3 or -x <= 1 or x <= -3 or x >= -1. So our solution is {-infinity, -3} U {-1, infinity}. **
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RESPONSE --> Ok.....I understand the solution but I think I didn't reverse...again.... self critique assessment: 2
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