course mth 158
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22:41:43 **** query 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 - 2 z | + 6 = 0 l 1-2z l= -6 1 - 2 z =-6 or 1 - 2 z = 6 -2z=-7 or -2z=5 z= 7/2 or z=-5/2 {-5/2, 7/2}
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22:45:07 ** 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 think somebody is wrong here. i added -6 to both sides as well but the problem that was listed here showed | 1-2z| +6 = 0 and not 9 so i believe i am correct ** 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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23:07:45 **** query 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 | x^2 +3x - 2 | = 2 or | x^2 +3x - 2 | = -2 x^2 +3x - 2-2 = 2-2 or x^2 +3x - 2 = -2+2 x^2 +3x - 4= 0 or x^2 +3x +0 = 0 (x+4)(x-1)=0 or (x+3)(x+0)=0 -4,1 or -3,0 {-4,-3,0,1}
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23:08:16 ** 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 --> ok
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23:21:12 **** query 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 | x + 4 | + 3 -3< 5-3 | x + 4 | < 2 -2 < x + 4 < 2 -2 -4 < x + 4 -4 < 2-4 -6 < x < -2 {x I -6 < x < -2} (-6,-2)
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23:21:22 STUDENT SOLUTION: | x+4| +3 < 5 | x+4 | < 2 -2 < x+4 < 2 -6 < x < -2
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RESPONSE --> ok
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23:42:47 **** query 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 -x - 2 >= 1 or -x - 2 <= -1 -x - 2 +2 >= 1+2 or -x - 2 +2 <= -1+2 -x >=3 or -x <=1 divide by-1 x<=-3 or x>=-1 (-infinity, -3) or (1,infinity)
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23:43:13 **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 --> | -x - 2 | >= 1 -x - 2 >= 1 or -x - 2 <= -1 -x - 2 +2 >= 1+2 or -x - 2 +2 <= -1+2 -x >=3 or -x <=1 divide by-1 x<=-3 or x>=-1 (-infinity, -3) or (-1,infinity)
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ɓ燝ˤɜԽ~ assignment #015 ؾ{]Uӂ͛ξ College Algebra 10-18-2005
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00:31:00 **** query 1.7.20 (was 1.2.30). Explain how you set up and solved an equation for the problem. Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text) million to lend at 19% or 16%, max lent at 16% to average 18%.
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RESPONSE --> i made a chart & filled it in with the info given then made an equation (1000000 x .18=180000 this is how much she needs to make) .16x+.19(1000000-x)= 180,000 -.03x=333,333 how much to lend at 16% with interest of 53333.28 1,000,000(ttl of amt to lend for year)-333,333(amt at 16%)=666,667(amt at 19%to be lent) She can lend $333,333 at 16% .
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00:31:09 ** Good. The details: If x is the amount lent at 16%, then the amount lent at 19% is 1,000,000 - x. Interest on x at 16% is .16 x, and interest on 1,000,000 - x at 19% is .19 (1,000,000 - x). This is to be equivalent to a single rate of 18%. 18% of 1,000,000 is 180,000 so the total interest is 1,000,000. So the total interest is .16 x + .19(1,000,000 - x), and also 180,000. Setting the two equal gives us the equation .16 x + .19(1,000,000 - x) = 180,000. Multiplying both sides by 100 to avoid decimal-place errors we have 16 x + 19 ( 1,000,000 - x) = 18,000,000. Using the distributive law on the right-hand side we get 16 x + 19,000,000 - 19 x = 18,000,000. Combining the x terms and subtracting 19,000,000 from both sides we have -3 x = 18,000,00 - 19,000,000 so that -3 x = -1,000,000 and x = -1,000,000 / (-3) = 333,333 1/3. **
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RESPONSE --> ok
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12:29:12 **** query 1.7.36 (was 1.2.36). Explain how you set up and solved an equation for the problem. Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text) 3 mph current, upstream takes 5 hr, downstream 2.5 hr. Speed of boat?
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RESPONSE --> let x-3 be the speed of the boat going upstream and x+3 be the speed of the boat going down stream so x-3=5 and x+3=2.5 we then solve each equation x-3+3 =5+3 we add 3 to both sides x=8 the speed of the boat x+3=2.5 x+3-3=2.5-3 subtract 3 from both sides x=-.5 this cannot work so we throw it out the speed of the boat is 8 miles per hour
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12:33:23 STUDENT SOLUTION: Speed of the boat is 9 mph, I used the equation 5(x - 3) = 2.5(x + 3) Reasoning is that it took 5 hours for the boat to travel against the 3mph current, and then traveled the same distance with the 3mph current in 2.5 hours. INSTRUCTOR COMMENT: Good. The details: If we let x be the water speed of the boat then its actual speed upstream is x - 3, and downstream is x + 3. Traveling for 5 hours upstream, at speed x - 3, we travel distance 5 ( x - 3). Traveling for 2.5 hours downstream, at speed x + 3, we travel distance 2.5 ( x + 3). The two distance must be the same so we get 5 ( x - 3) = 2.5 ( x + 3) or 5 x - 15 = 2.5 x + 7.5. Adding -2.5 x + 15 to both sides we get 2.5 x = 22.5 so that x = 22.5 / 2.5 = 9. So the water speed is 9 mph. **
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RESPONSE --> I see where I was wrong I understand the process that was used
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13:33:27 **** query 1.7.32 (was 1.2.42). Explain how you set up and solved an equation for the problem. Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text) pool enclosed by deck 3 ft wide; fence around deck 100 ft. Pond dimensions if pond square, if rectangular 3/1 ratio l/w, circular; which pond has most area?
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RESPONSE --> for circle= pi (d)=100 d=100/pi=31.831-6=25.831 for width of deck you -6 circumfrence of pond 25.831 12.92is radius a=pi r^2 = pi*12.92^2=524.05is area for sq 100/4=25 length of each side of sq fence -3foot from each side for width of deck25-3=22 22 length of each side of sq a=lw a=484 for rect. 3w for length w= width using the equation (2)length + width=perrimeter substitute 2(3w+w)=100 6w+2w=100 8w=100 100/8=w=12.5 is width 3(12.5)=37.5 is length -6 for deck is 37.5-6=31.5 and the width is 6.5 a=lw =31.5*6.5=204.75is area the circle pond has the most area
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13:35:54 10-19-2005 13:35:54 ** If the deck is circular then its circumference is C = 2 pi R and its radius is r = C / (2 pi). C is the 100 ft length of the fence so we have R = 100 ft / ( 2 pi ) = 50 ft / pi. The radius of the circle is 3 ft less, due to the width of the deck. So the pool radius is r = 50 ft / pi - 3 ft. This gives us pool area A = pi r^2 = pi ( 50 / pi - 3)^2 = pi ( 2500 / pi^2 - 300 / pi + 9) = 524, approx.. If the pool is square then the dimensions around the deck are 25 x 25. The dimensions of the pool will be 6 ft less on each edge, since each edge spans two widths of the deck. So the area would be A = 19 * 19 = 361. The perimeter of the rectangular pool spans four deck widths, or 12 ft. The perimeter of a rectangular pool is therefore 12 ft less than that of the fence, or 100 ft - 12 ft = 88 ft. If the pool is rectangular with length 3 times width then we first have for the 2 l + 2 w = 88 or 2 (3 w) + 2 w = 88 or 8 w = 88, giving us w = 11. The width of the pool will be 11 and the length 3 times this, or 33. The area of the pool is therefore 11 * 33 = 363. The circular pool has the greatest area, the rectangular pool the least. **
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NOTES -------> i did get the end correct answer for the circle pool and the correct area for the circle and i do see where i messed up on the sq and rect portion and am keeping this for notes If the deck is circular then its circumference is C = 2 pi R and its radius is r = C / (2 pi). C is the 100 ft length of the fence so we have R = 100 ft / ( 2 pi ) = 50 ft / pi. The radius of the circle is 3 ft less, due to the width of the deck. So the pool radius is r = 50 ft / pi - 3 ft. This gives us pool area A = pi r^2 = pi ( 50 / pi - 3)^2 = pi ( 2500 / pi^2 - 300 / pi + 9) = 524, approx.. If the pool is square then the dimensions around the deck are 25 x 25. The dimensions of the pool will be 6 ft less on each edge, since each edge spans two widths of the deck. So the area would be A = 19 * 19 = 361. The perimeter of the rectangular pool spans four deck widths, or 12 ft. The perimeter of a rectangular pool is therefore 12 ft less than that of the fence, or 100 ft - 12 ft = 88 ft. If the pool is rectangular with length 3 times width then we first have for the 2 l + 2 w = 88 or 2 (3 w) + 2 w = 88 or 8 w = 88, giving us w = 11. The width of the pool will be 11 and the length 3 times this, or 33. The area of the pool is therefore 11 * 33 = 363. The circular pool has the greatest area, the rectangular pool the least. **
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16:02:17 **** query 1.7.44 (was 1.2.54). Explain how you set up and solved an equation for the problem. Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text) 20 lb bag 25% cement 75% sand; how much cement to produce 40% concentration?
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RESPONSE -->
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16:02:18 **** query 1.7.44 (was 1.2.54). Explain how you set up and solved an equation for the problem. Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text) 20 lb bag 25% cement 75% sand; how much cement to produce 40% concentration?
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RESPONSE -->
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16:04:12 ** If x stands for the amount of cement added then we have the following: Original amount of cement in bag is 25% of 20 lb, or 5 lb. Original amount of sand in bag is 75% of 20 lb, or 15 lb. The final amount of cement will therefore be 5 lb + x, the final amount of sand will be 15 lb and the final weight of the mixture will be 20 lb + x. The mix has to be 40%, so (amt of cement) / (total amt of mixture) = .40. This gives us the equation (5 + x) / (20 + x) = .40. Multiplying both sides by 20 + x we have 5 + x = .40 ( 20 + x ). After the distributive law we have 5 + x = 80 + .40 x. Multiplying by 100 we get 500 + 100 x = 800 + 40 x. Adding -40 x - 500 to both sides we have 60 x = 300 so that x = 300 / 60 = 5. We should add 5 lbs of cement to the bag. **
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RESPONSE --> i tried for so long to come up and check that answer and i actually came up with 5 but was so confused by the time i finished i couldn't get it to check out properly If x stands for the amount of cement added then we have the following: Original amount of cement in bag is 25% of 20 lb, or 5 lb. Original amount of sand in bag is 75% of 20 lb, or 15 lb. The final amount of cement will therefore be 5 lb + x, the final amount of sand will be 15 lb and the final weight of the mixture will be 20 lb + x. The mix has to be 40%, so (amt of cement) / (total amt of mixture) = .40. This gives us the equation (5 + x) / (20 + x) = .40. Multiplying both sides by 20 + x we have 5 + x = .40 ( 20 + x ). After the distributive law we have 5 + x = 80 + .40 x. Multiplying by 100 we get 500 + 100 x = 800 + 40 x. Adding -40 x - 500 to both sides we have 60 x = 300 so that x = 300 / 60 = 5. We should add 5 lbs of cement to the bag. **
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16:06:01 **** query 1.7.52 (was 1.2.60). Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text): without solving what's wrong with prob how many liters 48% soln added to 20 liters of 25% soln to get 58% soln?
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RESPONSE -->
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16:06:46 ** Solution from Previous Student and Instructor Comment: It's not possible, adding a 25% solution to a 48% solution is only going to dilute it, I don't really know how to prove that algebraically, but logically that's what I think. (This is much like the last problem, that I don't really understand). INSTRUCTOR COMMENT: Right but the 48% solution is being added to the 25% solution. Correct statement, mostly in your words Adding a 48% solution to a 25% solution will never give you a 58% solution. Both concentrations are less than the desired concentration. **
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RESPONSE --> INSTRUCTOR COMMENT: Right but the 48% solution is being added to the 25% solution. Correct statement, mostly in your words Adding a 48% solution to a 25% solution will never give you a 58% solution. Both concentrations are less than the desired concentration. **
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