course Mth 158 ²ƒxÏW·Çù³“{ÀkŸx豴zassignment #010 010. `query 10 College Algebra 06-17-2008
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20:35:06 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 -6= -18-y. First you subtract -6 from each side. 5y +6 -6 = 18 -y -6. That leaves you with 5y = -24 - y. Add + y to each side. You get 6y= -2x Divide each side by 6. The answer is y= -4
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20:36:03 ** 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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20:39:04 query 1.1.38 \ 44 (was 1.1.30). Explain, step by step, how you solved the equation (2x+1) / 3 + 16 = 3x
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RESPONSE --> Multiply each side by 3. leaving 2x + 1+ 16 = 9x. Then substract - 2x from each side 17=7x. Divide each side by 7. The answer is x= 17/7
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20:41:22 ** 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 --> I didn't multiply my 16 by 3 which would give me 48 plus the one would be 49. Then divide by 7 would be 7.
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20:50:11 query was 1.1.44 \ 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-6 = X^ 2 +6x + 9 Add +6 to each side, x^2 - x -6 + 6 = x^2 + 6x +9 =6. x^2 -x = x ^ 2 + 6x + 15. Substract x^ 2 from each side. -x = 6x + 15. Substract -15 from each side. -x -15 = 6x Substract -x from each side. -15 = 5x x= -3
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20:53:40 ** 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 --> I messed up. When entering in the data from my paper I was switching something around. Actually on my homework paper I have -9 from each side leavin -15 = 7x. And then I have divided by 7 leaving -15/7
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21:02:57 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 --> Factor x^2 -9 to get (x+ 3) (x-3) X/ (x+3) (x-3) 4 / x+3 = 3/(x=3) (x-3) Multiply each side by (x=3) (x-3) x + 4 (x-3) = 3 x+ 4x -12= 3 Add +12 to each side x+4x= 15 5x =15 Divide each side by 5 X=3
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21:12:05 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 --> Multiply each side by (10 w -7) (5w + 7) (8w + 5) (5w + 7) = (4w -3) (10 w -7) 40 w^2 + 81 w + 35= 40 w^2 -58w -21 Substract 40 w^2 from each side 81w +35 = 58 w -56 Substract -35 from each side 81 w= 58 w -56 Substract - 58 w from each side 23 w= -56 Divide each side by 23 w= -2 Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.
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21:19:00 ** 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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RESPONSE --> Instead of having +21 I had -21.
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21:22:07 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- ax = b Substract -1 from both sides. -ax= b-1 Divide each side by -a x= b-1/-a
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21:26:43 ** 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 --> I needed to multiply both side by -1 to get the( -b+1)/a
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21:54:59 query extra problem (was 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 --> Factor: x(x^2 + 6x -7)=0 I don't know.
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21:58:05 ** 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 --> After factoring out the x on the left side you must factor the trinomial.
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22:08:46 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 --> The test score is worth twice the regular scores so I used 86 + 80 + 84 + 90 + x +x = 80 ( the desired average) after working this out he would need 70 on his final exam. Use sme formula only let it equal 90 for an answer of 100 if he want an A.
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22:13:28 ** 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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RESPONSE --> I see how you would need an average firts and then let your equation reflect the weight amount. In this case 2/3.
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22:17:58 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= -gt + v0 Add +gt to both sides. v + gt = v0 Substract v from each side. gt= v0 Divide each side by g. t=v0 - v/g
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22:21:15 ** 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 so that t = -(v - v0) / g = (-v + v0) / g.
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RESPONSE --> I should have started by adding -v0 to each side.
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