course MTH 158 {?D?????`???assignment #011 011. `query 11 College Algebra 06-19-2008
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13:35:03 1.2.5. Explain, step by step, how you solved the equation z^2 - z - 6 = 0 using factoring.
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RESPONSE --> z^2 - z - 6 = 0 factors into (z-3)(z+2) = 0 so z = {3, -2} confidence assessment: 2
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13:35:46 ** STUDENT SOLUTION WITH INSTRUCTOR COMMENT: I factored this and came up with (z + 2)(z - 3) = 0 Which broke down to z + 2 = 0 and z - 3 = 0 This gave me the set {-2, 3} -2 however, doesn't check out, but only 3 does, so the solution is: z = 3 INSTRUCTOR COMMENT: It's good that you're checking out the solutions, but note that -2 also checks: (-2)^2 - (-2) - 6 = 4 + 2 - 6 = 0. **
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RESPONSE --> ok self critique assessment: 2
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13:39:29 1.2.14 (was 1.3.6). Explain how you solved the equation by factoring.
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RESPONSE --> v^2 + 7v + 6 = 0 factors into (v+6)(v+1) = 0 so v = {-6, -1} but when I checked with -1, it did not come out to be 0, so the answer is v = -6 confidence assessment: 3
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13:39:52 STUDENT SOLUTION: v^2+7v+6=0. This factors into (v + 1) (v + 6) = 0, which has solutions v + 1 = 0 and v + 6 = 0, giving us v = {-1, -6}
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RESPONSE --> ok self critique assessment: 3
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13:43:20 1.2.20 (was 1.3.12). Explain how you solved the equation x(x+4)=12 by factoring.
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RESPONSE --> x(x+4)=12 -12 -12 = x(x+4) - 12 = 0 x^2 + 4x -12 = 0 (x + 6)(x - 2) = 0 x = {-6, 2} confidence assessment: 3
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13:43:32 ** Starting with x(x+4)=12 apply the Distributive Law to the left-hand side: x^2 + 4x = 12 add -12 to both sides: x^2 + 4x -12 = 0 factor: (x - 2)(x + 6) = 0 apply the zero property: (x - 2) = 0 or (x + 6) = 0 so that x = {2 , -6} **
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RESPONSE --> ok self critique assessment: 3
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13:49:07 1.2.26 (was 1.2.38) (was 1.3.18). Explain how you solved the equation x + 12/x = 7 by factoring.
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RESPONSE --> Multiply each side by x (x) x + 12/x = 7(x) x^2 + 12 = 7x -7x -7x = x^2 - 7x + 12 = 0 which factors to (x-4)(x-3) = 0 x = {4, 3} confidence assessment: 3
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13:49:27 ** Starting with x + 12/x = 7 multiply both sides by the denominator x: x^2 + 12 = 7 x add -7x to both sides: x^2 -7x + 12 = 0 factor: (x - 3)(x - 4) = 0 apply the zero property x-3 = 0 or x-4 = 0 so that x = {3 , 4} **
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RESPONSE --> ok self critique assessment: 3
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13:51:51 1.2.32 (was 1.3.24). Explain how you solved the equation (x+2)^2 = 1 by the square root method.
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RESPONSE --> Take the sqrt of both sides out which equals x+2 = 1 then subtract 1 from each side x+2 = 1 -1 -1 = x + 1 = 0 x = -1 confidence assessment: 3
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13:53:01 ** (x + 2)^2 = 1 so that x + 2 = ?sqrt(1) giving us x + 2 = 1 or x + 2 = -1 so that x = {-1, -3} **
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RESPONSE --> understood self critique assessment: 2
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23:19:18 1.2.44 (was 1.3.36). Explain how you solved the equation x^2 + 2/3 x - 1/3 = 0 by completing the square.
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RESPONSE --> (3) x^2 + 2/3 x - 1/3 = 0 (3) 3x^2 + 2x - 1 = 0 Using the quadratic equation x = -b +/- sqrt(b^2 - 4ac) / 2*a , we get x = -2 +/- sqrt (2^2 - 4 * 3 * 1) / 2*3 x = -2 +/- sqrt (4 - 12) / 6 x = -2 +/- sqrt (-8) / 6 x = -2 -2 / 6 or x = 2 -2 / 6 since 0 cannot be the numerator, the answer is x = -2/3 confidence assessment: 2
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23:20:44 ** x^2 + 2/3x - 1/3 = 0. Multiply both sides by the common denominator 3 to get 3 x^2 + 2 x - 1 = 0. Factor to get (3x - 1) ( x + 1) = 0. Apply the zero property to get 3x - 1 = 0 or x + 1 = 0 so that x = 1/3 or x = -1. DER**
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RESPONSE --> I don't get how 3x - 1 turns into 1/3 self critique assessment: 1
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23:28:21 1.2.50 (was 1.2.52) (was 1.3.42). Explain how you solved the equation x^2 + 6x + 1 = 0 using the quadratic formula.
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RESPONSE --> x = -6 +/- sqrt (6^2 - 4*1*1) / 2*1 x = -6 +/- sqrt (36 - 4) / 2 x = -6 +/- sqrt (32) / 2 x = {-3 -sqrt 32, -3 +sqrt 32} confidence assessment: 2
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23:29:06 ** Starting with x^2 + 6x + 1 = 0 we identify this as having the form a x^2 + b x + c = 0 with a = 1, b = 6 and c = 1. We plug values into quadratic formula to get x = [-6 ?sqrt(6^2 - 4 * 1 * 1) ] / 2 *1 x = [ -6 ?sqrt(36 - 4) / 2 x = { -6 ?sqrt (32) ] / 2 36 - 4 = 32, so x has 2 real solutions, x = { [-6 + sqrt(32) ] / 2 , [-6 - sqrt(32) ] / 2 } Note that sqrt(32) simplifies to sqrt(16) * sqrt(2) = 4 sqrt(2) so this solution set can be written { [-6 + 4 sqrt(2) ] / 2 , [-6 - 4 sqrt(2) ] / 2 }, and this can be simplified by dividing numerators by 2: { -3 + 2 sqrt(2), -3 - 2 sqrt(2) }. **
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RESPONSE --> understood self critique assessment: 2
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11:07:29 1.2.74 (was 1.2.72) (was 1.3.66). Explain how you solved the equation pi x^2 + 15 sqrt(2) x + 20 = 0 using the quadratic formula and your calculator.
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RESPONSE --> With quad formula we get x = -15 sqrt(2) +/- [sqrt (15 sqrt2)^2 - 4pi*20] / 2pi x = -15 sqrt(2) +/- [sqrt (450 - 80pi)] / 2pi x = -15 sqrt(2) +/- 31.61972439 x = 10.41 or -52.83 {10.41, -52.83} confidence assessment: 2
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11:10:34 ** Applying the quadratic formula with a = pi, b = 15 sqrt(2) and c = 20 we get x = [ (-15sqrt(2)) ?sqrt ((-15sqrt(2))^2 -4(pi)(20)) ] / ( 2 pi ). (-15sqrt(2))^2 - 4(pi)(20) = 225 * 2 - 80 pi = 450 - 80 pi = 198.68 approx., so there are 2 unequal real solutions. Our expression is therefore x = [ (-15sqrt(2)) ?198.68] / ( 2 pi ). Evaluating with a calculator we get x = { 5.62, 1.13 }. DER**
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RESPONSE --> understood self critique assessment: 2
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11:37:27 1.2.97 (was 1.2.98) (was 1.3.90). 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) box vol 4 ft^3 by cutting 1 ft sq from corners of rectangle the L/W = 2/1.
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RESPONSE --> Since some of the previous questions on here have been different from the questions in my book, I will go ahead and type up the question as it appears in my book, just in case they are different. 97. An open box is to be constructed from a square piece of sheet metal by removing a square of side 1 foot from each corner and turning up the edges. If the box is to hold 4 cubic feet, what should be the dimensions of the sheet metal? The formula in the book is a(x-2a)^2 = b, a representing the height of the box, x representing the length of a side and b representing the volume. So... 1(x - 2)^2 = 4 x-2 = 2 x=4 confidence assessment: 2
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11:38:13 ** Using x for the length of the shorter side of the rectangle the 2/1 ratio tells us that the length is 2x. If we cut a 1 ft square from each corner and fold the 'tabs' up to make a rectangular box we see that the 'tabs' are each 2 ft shorter than the sides of the sheet. So the sides of the box will have lengths x - 2 and 2x - 2, with measurements in feet. The box so formed will have height 1 so its volume will be volume = ht * width * length = 1(x - 2) ( 2x - 2). If the volume is to be 4 we get the equation 1(x - 2) ( 2x - 2) = 4. Applying the distributive law to the left-hand side we get 2x^2 - 6x + 4 = 4 Divided both sides by 2 we get x^2 - 3x +2 = 2. (x - 2) (x - 1) = 2. Subtract 2 from both sides to get x^2 - 3 x = 0 the factor to get x(x-3) = 0. We conclude that x = 0 or x = 3. We check these results to see if they both make sense and find that x = 0 does not form a box, but x = 3 checks out fine. **
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RESPONSE --> This says that it is a rectangle, and my book said it was a square... self critique assessment: 1
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12:18:00 1.2.100 (was 1.3.96). 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) s = -4.9 t^2 + 20 t; when 15 m high, when strike ground, when is ht 100 m.
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RESPONSE --> I tried using the quadratic equation for (a), and it came out with both negative answers, and there can't exactly be negative seconds, so that didn't work.. and the book doesn't give any direction or hints or anything on how to figure out the other two! confidence assessment: 0
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12:20:01 ** To find the clock time at which the object is 15 m off the ground we set the height s equal to 15 to get the equation -4.9t^2 + 20t = 15 Subtracting 15 from both sides we get -4.9t^2 +20t - 15 = 0 so that t = { -20 ?sqrt [20^2 - 4(-4.9)(-15) ] } / 2(-4.9) Numerically these simplify to t = .99 and t = 3.09. The object passes the 15 m height on the way up at t = 99 and again on the way down at t = 3.09. To find when the object strikes the ground we set s = 0 to get the equation -4.9t^2 + 20t = 0 which we solve to get t = [ -20 ?sqrt [20^2 - 4(-4.9)(0)] ] / 2(-4.9) This simplifies to t = [ -20 +-sqrt(20^2) ] / (2 * -4.9) = [-20 +- 20 ] / (-9.8). The solutions simplify to t = 0 and t = 4.1 approx. The object is at the ground at t = 0, when it starts up, and at t = 4.1 seconds, when it again strikes the ground. To find when the altitude is 100 we set s = 100 to get -4.9t^2 + 20t = 100. Subtracting 100 from both sides we obtain -4.9t^2 +20t - 100 = 0 which we solve using the quadratic formula. We get t = [ -20 ?sqrt (20^2 - 4(-4.9)(-100)) ] / 2(-4.9) The discriminant is 20^2 - 4 * -4.9 * -100, which turns out to be negative so we do not obtain a solution. We conclude that this object will not rise 100 ft. **
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RESPONSE --> How did t = { -20 ?sqrt [20^2 - 4(-4.9)(-15) ] } / 2(-4.9) simplify to .99 and 3.09??? I got -18.95 and -21.05 even when I went over it! self critique assessment: 1
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12:20:47 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> There was alot that the book and/or CD ROMs didn't give information for, which left me quite confused. confidence assessment: 3
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