course Mth 158 Ϛ兰`ՆŲassignment #010
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10:56:38 1.1.20 (was 1.1.12). Explain, step by step, how you solved the equation 5y + 6 = -18 - y
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11:03:59 ** 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 --> 5y+6 (-6)=18-y(-6) 5y=12-y 5y (+y) = 12-y (+y) 6y=12 divide both sides by 6 6y/6=12/6 y=2 ------------------------------------------------ Self-critique Rating:ent: 0
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11:08:09 1.1.38 (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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11:13:19 ** 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 --> 2x+1 + 16 = 3x 2x+1 + 16 (*3)=3x * 3 2x +1 +48 = 9x 2x+49 (-2x) = 9x (-2x) 49 = 7x divide both sides by 7 49/7=7x/7 x=7 ------------------------------------------------ Self-critique Rating:ent: 1
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11:17:23 ** 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 --> x^2 - x - 6= x^2 +6x+9 take x^2 from both sides -x-6=6x+9 -x - 6 (-9)=6x + 9 (-9) -x-15=6x -x-15(+x) = 6x+x -15=7x divide both by 7 x=2.14 ------------------------------------------------ Self-critique Rating:ent: 3
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諭친x|էwO assignment #011 011. `query 11 College Algebra 07-22-2009
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11:20:44 1.2.5. Explain, step by step, how you solved the equation z^2 - z - 6 = 0 using factoring.
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11:20:50 ** 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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11:21:33 1.2.14 (was 1.3.6). Explain how you solved the equation by factoring.
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11:21:37 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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11:21:45 1.2.20 (was 1.3.12). Explain how you solved the equation x(x+4)=12 by factoring.
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11:21:48 ** 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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11:22:01 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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11:22:03 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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11:22:06 ** 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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11:22:14 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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11:22:17 ** (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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11:22:24 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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11:22:27 ** 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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11:22:35 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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11:22:37 ** 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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11:22:46 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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11:22:50 ** 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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11:22:59 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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11:23:02 ** 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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11:23:09 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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11:23:13 ** 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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