Assignment 11

course Mth 158

I need help with this one. At the bottom of the query I wrote which ones I needed help on. Thanks for your help.

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

⡁L{ڏy College Algebra 03-15-2006

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16:14:34 query 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 factor the problem (z + 2)(z - 3) = 0 make each equal to 0 z + 2 = 0 ; z - 3 = 0 thus giving: z = -2 ; z = 3

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16:15:19 ** 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 --> got it correct, but -2 doesn't check out.

-2 does in fact check out. Check that again (and be very careful about negatives). If you can't get it to check out, send me the details.

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16:21:43 **** query 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 (v + 6)(v + 1) = 0 v + 6 = 0 ; v + 1 = 0 v = -6 and v = -1 Solution set {-6,-1}

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16:21:52 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 --> got it correct

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16:31:57 **** query 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 distribute and subract 12 to other side. x^2 + 4x - 12 = 0 factor (x + 6)(x - 2) = 0 x + 6 = 0 ; x - 2 = 0 x = -6 ; x = 2 {-6 , 2}

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16:32:10 ** 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 --> got it correct

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16:46:54 **** query 1.2.38 (was 1.3.18). Explain how you solved the equation x + 12/x = 7 by factoring.

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RESPONSE --> x + 12/x = 7 x^2 + 5x = 0 x(x + 5) = 0 x = 0 ; x + 5 = 0 x = 0 ; x = -5 {0 ; -5} -5 doesn't factor out so the answer is, {0}

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16:48:36 ** 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 --> opps! I forgot to drop the x off of the 12. I multiplied x into everything. I knew there was something wrong with my problem. Now I know why I had such a hard time.

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17:19:05 **** query 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 --> (x + 2)^2 = 1 x + 2 = + - sqrt (1) x + 2 = + - 1 x + 2 = 1 x + 2 = -1 x = -1 x = -3 solution set : {-1, -3}

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17:19:14 ** (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 --> correct

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18:20:29 **** query 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 --> x^2 + 2/3x -1/3 = 0 x^2 + 2/3x = 1/3 x^2 + 2/3x + 2/3 = 1/3 + 2/3 x^2 + 2/3x +2/3 = 1 (x + 2/3)(x + 1) = 1 x + 2/3 = 1 x + 1 = 1

This only works for equations that are equal to 0. If two quantities multiply to give you 0 then at least one of the quantities must be 0.

If however two quantities multiply to give you 1 then it does NOT follow that at least one of the quantities must be 1.

x = 1/3 x = 0 solution {1/3, 0} I really had a hard time figuring this one out. I have a really hard time with fractions. They get me every time.

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18:21:33 ** 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 --> this one was tough. When I looked in the book, it said to solve it a totally different way than you did.

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18:27:13 **** query 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^2 + 6x + 1 = 0 a = 1 b = 6 c = 1 b^2 - 4ac so.... 6^2 - 4 (1)(1) 36 - 4 32 Solution: {32} because 32 > 0

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18:29:18 ** 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 --> In the book, it said to use b^2 - 4ac

b^2 - 4 a c is the discriminant, and it tells you how many real roots the equation has. But this quantity by itself does not solve the equation as you can tell by plugging your results into the equation.

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19:51:35 **** query 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 --> pi x^2 + 15sqrt(2) + 20 = 0 x =[ -15sqrt(2) +- sqrt((15sqrt2)^2 - 4(pi)(20)) ] / 2(pi) x =[-15sqrt(2) +- sqrt(450 - 251.33)] / 6.28 x = -15sqrt2 +- sqrt198.67 / 6.28 x = -15sqrt2 +- 14.095 / 6.28 -15sqrt2 + 14.095 / 6.28 and -15sqrt2 - 14.095 / 6.28

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19:51:55 ** 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)) sqrt(198.68)] / ( 2 pi ). Evaluating with a calculator we get x = { -5.62, -1.13 }. DER**

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RESPONSE --> I couldn't figure this one out.

You had it. All you needed to do was evaluate

(-15sqrt2 + 14.095) / 6.28 and (-15sqrt2 - 14.095) / 6.28

You will get the same results shown here.

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18:31:38 **** query 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 --> h(2) = 4^3 divide by 2 on both sides h = 4^3 / 2 h = 64 / 2 h = 32

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18:32:03 ** 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 --> I didn't really understand how to work this problem out.

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19:03:11 **** query 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 --> a) It goes 20 meters per second, so I did 20 / 15 to see how fast it goes with 15 meters. I got 3/4 for the answer. So I put 3/4 in for t. s = -4.9(3/4)^2 + 20(3/4) s = -2.76 + 15 s = 12.24 At 15 meters it will be above the ground in 12.24sec. b) I don't know how to do this one. c) Since it takes 1 second for 20 meters, it takes 5 seconds for 100 meters. s = -4.9(5)^2 + 20(5) s = -122.5 + 100 s = -22.5

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19:04:33 ** 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 --> got it wrong I was going to put 15 in for t, but i couldn't figure out how to solve it. So I tried to take another route and it didn't work.

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19:05:35 **** Query Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE --> If you could, please look over 44, 52, and 98. I did something wrong and couldn't figure out what I did wrong. Thank you.

Overall you did fine here.

The place to ask for help is on the problem itself. I need to go through these files sequentially, and it takes longer to go back and find the problems than to read through the work and insert the comments in the first place.

I did look back and there are notes on 44 and 52. Let me know if they don't answer your quesion.

There is a complete solution given for #98, and you are expected to completely analyze that solution and tell me exactly what you do and do not understand about it. After you have done this, I can help you address the things you still do not understand.

You are always welcome to copy the problem, your solution, and the given solution into the form and add your self-critique. However at this stage, not knowing what you do and do not understand, all I could do would be to tell you that your solution isn't correct, and repeat the given solution.

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