course Mth 158 ?????????w?????assignment #010v?????n???Z????
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22:05:09 query 1.1.20 (was 1.1.12). Explain, step by step, how you solved the equation 5y + 6 = -18 - y
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RESPONSE --> 5y+6=-18-y 5y+y=-18-6 (6y)/(6)=(-24)/(6) y=-4
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22:06:12 ** STUDENT SOLUTION WITH INSTRUCTOR COMMENT: 5y + 6 = 18 - y Subtract 6 from both sides, giving us 5y = 12 - y Add y to both sides, 5y + y = 12 or 6y = 12 divide both sides by 6 y = 2 INSTRUCTOR COMMENT: This is correct for equation 5y + 6 = 18 - y but the equation as I note it is 5y + 6 = -18 - y. If that's the correct equation the solution is found by practically the same steps but you end up with y = -4. **
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RESPONSE --> ok
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22:13:28 query 1.1.44 (was 1.1.30). Explain, step by step, how you solved the equation (2x+1) / 3 + 16 = 3x
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RESPONSE --> (x+2)(x-3)=(x+3)^2 (x+2)(x-3)=(x+3)(x+3) * using the foil method you get x^2-3x+2x-6=x^2+3x+3x+9 *add and subtract like terms resulting in..x^2-x-6=x^2+6x+9 continuing and trying to get x on one side you get.. x^2-x^2-x-6x=9+6 again add and subtract like terms, -7x=15..divide both sides by 7.. (-7x)/(-7)=(15)/(-7) x=(15)/(-7)
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22:15:08 ** STUDENT SOLUTION: (2x + 1) / 3 + 16 = 3x First, multiply both sides of the equation by 3 2x +1 + 48 =9x or 2x + 49 = 9x subtract 2x from both sides. 49 = 7x Divide both sides by 7 7 = x **
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RESPONSE --> ok
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22:20:42 query was 1.1.36. Explain, step by step, how you solved the equation (x+2)(x-3) = (x+3)^2
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RESPONSE --> (x+2) (x-3)=(x+3) (x+3) use the foil method getting.. x^2-3x+2x-6=x^2+3x+3x+9...add and subtract like terms getting.. x^2-x-6=x^2+6x+9..again adding and subtracting like terms and getting x on one side results in.. x^2-x^2-x-6x=9+6.. add and subtract like terms.. 7x=15.. divide both sides by 7.. (-7x)/(-7)=(15)/(-7) x=(15)/(-7)
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22:21:32 ** STUDENT SOLUTION: (x+2)(x+3) = (x+3)^2 First, we use the distributive property to remove the parenthesis and get x^2 - x - 6 = x^2 + 6x + 9 subtract x^2 from both sides, -x - 6 = 6x + 9 Subtract 9 from both sides - x - 6 - 9 = 6x or -x - 15 = 6x add x to both sides -15 = 7x Divide both sides by 7 -15/7 = x **
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RESPONSE --> ok
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22:28:05 query 1.1.52 (was 1.1.48). Explain, step by step, how you solved the equation x / (x^2-9) + 4 / (x+3) = 3 / (x^2-9)/
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RESPONSE --> multiply both sides by a common denominator.. (x^2-9) x / (x^2-9) + 4 / (x+3)= (x^2-9) 3 / x^2-9 x+4x-12=3 add 12 to both sides.. x+4x-12+12=3+12 5x=15 divide both sides by 5 x=15 / 5 x=3
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22:31:23 ** Starting with x / (x^2 -9) + 4 / (x+3) = 3 / (x^2 -9), first factor x^2 - 9 to get x / ( (x-3)(x+3) ) + 4 / (x+3) = 3 / ( (x-3)(x+3) ). Multiply both sides by the common denominator ( (x-3)(x+3) ): ( (x-3)(x+3) ) * x / ( (x-3)(x+3) ) + ( (x-3)(x+3) ) * 4 / (x+3) = ( (x-3)(x+3) ) * 3 / ( (x-3)(x+3) ). Simplify: x + 4(x-3) = 3. Simplify x + 4x - 12 = 3 5x = 15 x = 3. If there is a solution to the original equation it is x = 3. However x = 3 results in denominator 0 when substituted into the original equation, and division by 0 is undefined. So there is no solution to the equation. When you multiplied both sides by x-3, if x = 3 you were multiplying by zero, which invalidated your solution **
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RESPONSE --> ok
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22:37:27 query 1.1.58 (was 1.1.54). Explain, step by step, how you solved the equation (8w + 5) / (10w - 7) = (4w - 3) / (5w + 7)
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RESPONSE --> (8w + 5) / (10w - 7) = (4w - 3) / (5w + 7) multiply both sides by the common denominator.. (10w-7) (5w+7) (8w + 5) / (10w - 7)= (4w - 3) / (5w + 7)(10w-7) (5w+7) ..simplify.. (5w+7) (8w+5)= (4w-3) (10w-7).. simplify.. 40w^2+25w+56w+35=40w^2-28w^2-30w+21..simplify.. 40w^2-40w^2+81w+58w= 21-35 139w = -14.. divide both sides by 139.. w= -14 / 139
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22:37:59 06-27-2006 22:37:59 ** STUDENT SOLUTION: 1) clear the equation of fractions by multiplying both sides by the LCM (10w - 7)(5W + 7) After cancellation the left side reads: (5w+7)(8w + 5) After cancellation the right side reads: (10w - 7)(4w - 3) multiply the factors on each side using the DISTRIBUTIVE LAW Left side becomes: (40w^2) + 81w + 35 Right side becomes: (40w^2) - 58w + 21 3) subtract 40w^2 from both sides add 58w to both sides subtract 35 from both sides Rewrite: 139w = - 14 Now divide both sides by 139 to get w = - (14 / 139) DER**
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22:40:25 query 1.1.70 (was 1.1.78). Explain, step by step, how you solved the equation 1 - a x = b, a <> 0.
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RESPONSE --> 1 - a x = b, a <> 0. subtract 1 from both sides..1-1-ax= b-1 divide both sides by -a.. (-ax)/(-a)=(b-1)/(-a) x= (-b+1) / (a)
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22:41:17 06-27-2006 22:41:17 ** Start with 1 -ax = b, a <> 0. Adding -1 to both sides we get 1 - ax - 1 = b - 1, which we simplify to get -ax = b - 1. Divide both sides by -a, which gives you x = (b - 1) / (-a). Multiply the right-hand side by -1 / -1 to get x = (-b + 1) / a or x = (1 - b) / a. **
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22:41:18 ** Start with 1 -ax = b, a <> 0. Adding -1 to both sides we get 1 - ax - 1 = b - 1, which we simplify to get -ax = b - 1. Divide both sides by -a, which gives you x = (b - 1) / (-a). Multiply the right-hand side by -1 / -1 to get x = (-b + 1) / a or x = (1 - b) / a. **
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RESPONSE -->
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22:48:55 query 1.1.72. Explain, step by step, how you solved the equation x^3 + 6 x^2 - 7 x = 0 using factoring.
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RESPONSE --> x is a common term and the rest is factored out.. (x) (x^2+6x-7)=0.. factor out (x^2+6x-7)=0.. (x) (x+7) (x-1)
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22:49:46 ** Starting with x^3 + 6 x^2 - 7 x = 0 factor x out of the left-hand side: x(x^2 + 6x - 7) = 0. Factor the trinomial: x ( x+7) ( x - 1) = 0. Then x = 0 or x + 7 = 0 or x - 1 = 0 so x = 0 or x = -7 or x = 1. **
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RESPONSE --> I factored it out but I did not solve for x, which I could have easily done.
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23:11:39 query 1.1.90 (was 1.2.18). 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) scores 86, 80, 84, 90, scores to ave B (80) and A (90).
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RESPONSE --> I first added his grades of 86, 80, 84, 90 and got 340. I then took 340 and divided it by 4 (.. the number of scores) and got his grade average.. 85. I then took his average thus far, 85.. and multiplyed it by 1/ 3rd.. getting 28.05. (I multiplyed 85 by 1/3rd because that is the part of the grade that does not include the exam). I then took 80, the score needed in order to recieve a B and subtracted 28.05.. (80-28.05=51.95) I then took 51.95 (the 1/3rd of the grade needed in order to recieve an 80) and divided it by 2/3rds.. the amount the exam is worth towards the grade (51.95 / .66 = 79 79 is the grade needed in order to recieve an 80. In order to earn an A, which requires an average of 90, I basically did the same steps.. I took his average thus far, 85.. and multiplyed it by 1/ 3rd.. getting 28.05. I then took 90, the score needed in order to recieve a A and subtracted 28.05.. (80-28.05=61.95) I then took 61.95 (the 1/3rd of the grade needed in order to recieve a 90) and divided it by 2/3rds.. the amount the exam is worth towards the grade (61.95 / .66 = 94.. 94 is the grade needed in order to recieve an A
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23:13:33 06-27-2006 23:13:33 ** This can be solved by trial and error but the only acceptable method for this course, in which we are learning to solve problems by means of equations, is by an equation. Let x be the score you make on the exam. The average of the four tests is easy to find: 4-test average = ( 86 + 80 + 84 + 90 ) / 4 = 340 / 4 = 85. The final grade can be thought of as being made up of 3 parts, 1 part being the test average and 2 parts being the exam grade. We would therefore have final average = (1 * test average + 2 * exam grade) / 3. This gives us the equation final ave = (85 + 2 * x) / 3. If the ave score is to be 80 then we solve (85 + 2 * x) / 3 = 80. Multiplying both sides by 3 we get 85 + 2x = 240. Subtracting 85 from both sides we have 2 x = 240 - 85 = 155 so that x = 155 / 2 = 77.5. We can solve (340 + x) / 5 = 90 in a similar manner. We obtain x = 92.5. Alternative solution: If we add 1/3 of the test average to 2/3 of the final exam grade we get the final average. So (using the fact that the test ave is 85%, as calculated above) our equation would be 1/3 * 85 + 2/3 * x = final ave. For final ave = 80 we get 1/3 * 85 + 2/3 * x = 80. Multiplying both sides by 3 we have 85 + 2 * x = 240. The rest of the solution goes as before and we end up with x = 77.5. Solving 1/3 * 85 + 2/3 * x = 90 we get x = 92.5, as before. **
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23:17:55 query 1.1.82 (was 1.1.90). Explain, step by step, how you solved the equation v = g t + v0 for t.
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RESPONSE --> v = g t + v0 for t. subtracted v0 from both sides.. v-v0=-gt+v0-v0 results in.. v-v0= -gt divide -g from both sides.. (v-v0) /( -g) = (-gt) / (-g) (-v+v0) / (g)= t
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23:18:44 ** NOTE: v0 stands for v with subscript 0; the whole expression v0 stands for the name of a variable. It doesn't mean v * 0. Starting with v = g t + v0, add -v0 to both sides to get v - v0 = gt. Divide both sides by g to get (v - v0) / g = t }or t = (v - v0) / g. **
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RESPONSE --> ok
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23:19:04 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> no comments
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