#$&* course Mth 158 Mr. smith, i was wondering if you had received my chapter R test. i had taken it last Wednesday and turned it in to the attendant working in the vhcc learning lab. i am messaging you not to inquire about the grade i received, rather i am messaging you just to be sure that you have it. i have also emailed you per your request. just touching bases, thanks, Shane Watson.
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Given Solution: * * STUDENT SOLUTION WITH INSTRUCTOR COMMENT: I factored this and came up with (z + 2)(z - 3) = 0 Which broke down to z + 2 = 0 and z - 3 = 0 This gave me the set {-2, 3} -2 however, doesn't check out, but only 3 does, so the solution is: z = 3 INSTRUCTOR COMMENT: It's good that you're checking out the solutions, because sometimes we get extraneous roots. But note that -2 also checks: (-2)^2 - (-2) - 6 = 4 + 2 - 6 = 0. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: * 1.2.14 (was 1.3.6). Explain how you solved the equation v^2+7v+6=0 by factoring. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To solve this problem we use the same method used in the prior question. I broke this down into two terms which gave v^2+7v+6=0= (v+1)(v+6)=0. When we distribute this back out we get our original problem. Now we can set these terms equal to zero. So (v+1)=0 and (v+6)=0 which give v=-1 and v=-6. So our solutions are (-1, -6). confidence rating #$&*:ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 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} &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: * 1.2.20 (was 1.3.12). Explain how you solved the equation x(x+4)=12 by factoring. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To solve this we must first distribute x giving us x(x+4)=12= x^2+4x=12, now we must set it to zero by subtracting 12 giving x^2+4x-12=0. Now we will break this down into two terms giving x^2+4x-12=0=(x+6)(x-2)=0. Now that we have two terms we set them equal to zero and isolate x. so (x+6)=0 becomes x=-6 and (x-2)=0 becomes x=2. So our solutions are (-6, 2). confidence rating #$&*:ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Starting with x(x+4)=12 apply the Distributive Law to the left-hand side: x^2 + 4x = 12 add -12 to both sides: x^2 + 4x -12 = 0 factor: (x - 2)(x + 6) = 0 apply the zero property: (x - 2) = 0 or (x + 6) = 0 so that x = {2 , -6} ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: * 1.2.26 \ 38 (was 1.3.18). Explain how you solved the equation x + 12/x = 7 by factoring. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To solve this we eliminate the fraction of 12/x by multiplying everything by the denominator so it becomes x^2+12=7x. then we subtract 7x from both sides giving us x^2-7x+12=0. Now that we have arrived at this form of the equation we can factor this into two terms which would be (x-3)(x-4)=0. Now that we have two terms we set them both equal to zero and isolate the x variable. This gives (x-3)=0 which gives us x-3 and (x-4)=0 which gives us x=4. So our solutions are (3, 4). confidence rating #$&*:ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 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} ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: * 1.2.32 (was 1.3.24). Explain how you solved the equation (x+2)^2 = 1 by the square root method. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To solve this using the sqrt method we take the square root of both sides of the equation giving us x+2=1 or x+2=-1. So we add 1 to both sides and we subtract 1 from both sides giving us x=-1 or x=-3. So our solutions are (-1, -3). confidence rating #$&*:ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * (x + 2)^2 = 1 so that x + 2 = ± sqrt(1) giving us x + 2 = 1 or x + 2 = -1 so that x = {-1, -3} ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: * 1.2.38 (was 1.3.36). Explain how you solved the equation x^2 + 2/3 x - 1/3 = 0 by completing the square. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First to eliminate fractions we multiply by the denominator 3. Making the problem 3x^2+2x-1=0. Now we factor this into two terms giving us (3x^2-1)(x+1)=0. Now we set these terms equal to zero and isolate the x variable giving 3x^2-1=0 and x+1=0 which gives us x=1/3 and x=-1. confidence rating #$&*:ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 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. STUDENT QUESTION: The only thing that confuses me is the 1/3. Is that because of the 3x? INSTRUCTOR RESPONSE: You got the equation (3x - 1) ( x + 1) = 0. The product of two numbers can be zero only if one of the numbers is zero. So (3x - 1) ( x + 1) = 0 means that 3x - 1 = 0 or x + 1 = 0. You left out this step in your solution. x + 1 = 0 is an equation with solution x = -1 Thus the solution to our original equation is x = 1/3 or x = -1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: * 1.2.44 \ 52 (was 1.3.42). Explain how you solved the equation x^2 + 6x + 1 = 0 using the quadratic formula. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We know that the quadratic formula is -b(+)(-)sqrtb^2-4*a*c/2*a so we simply plug in the variables in the formula. This gives us -6(+)(-)sqrt 6^2-4*1*1/2*1 and when we perform operations becomes -6(+)(-)sqrt36-4/2 and then -6(+)(-)sqrt32/2. Now 32 has no sqrt but it does contain factors that do so -6(+)(-)sqrt16*2/2=-6(+)(-)4sqrt2/2. Which can be further simplified by dividing by 2 giving the solutions( -3+2sqrt2) or (-3-2sqrt2). confidence rating #$&*:ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Starting with x^2 + 6x + 1 = 0 we identify our equation as a quadratic equation, having the form a x^2 + b x + c = 0 with a = 1, b = 6 and c = 1. We plug values into quadratic formula to get x = [-6 ± sqrt(6^2 - 4 * 1 * 1) ] / 2 *1 x = [ -6 ± sqrt(36 - 4) / 2 x = { -6 ± sqrt (32) ] / 2 36 - 4 = 32, so x has 2 real solutions, x = [-6 + sqrt(32) ] / 2 and x = [-6 - sqrt(32) ] / 2 Our solution set is therefore { [-6 + sqrt(32) ] / 2 , [-6 - sqrt(32) ] / 2 } Now sqrt(32) simplifies to sqrt(16) * sqrt(2) = 4 sqrt(2) so this solution set can be written { [-6 + 4 sqrt(2) ] / 2 , [-6 - 4 sqrt(2) ] / 2 }, and this can be simplified by dividing numerators by 2: { -3 + 2 sqrt(2), -3 - 2 sqrt(2) }. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: * 1.2.69 \ 1.2.78 \ 72 (was 1.3.66). Explain how you solved the equation pi x^2 + 15 sqrt(2) x + 20 = 0 using the quadratic formula and your calculator. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To solve this we simply insert the proper variables into the quadratic formula. Giving -15sqrt(2)(+)(-)sqrt (15sqrt(2))^2-4*pix^2*20/2*pix^2, once we solve this equation we get -5.62 and -1.13. so our solutions are (-5.62, -1.13). confidence rating #$&*:ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 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 ). The discriminant (-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** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: * 1.2.106 \ 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We solve this by interpreting the information given and making an equation, given by 1(x-2)(2x-2)=4 where 1 is the hight (x-2) is the width and (2x-2) is the length and 4 is the desired volume. Now we solve this by first distributing so the problem then becomes 2x^2-6x+4=4. Now we divide by 2 on both sides giving x^2-3x+2=2. When we reach this point we factor giving (x-2)(x-1)=2 and now to set it equal to zero we subtract 2 from both sides giving x^2-3x then factoring gives x(x-3)=0 meaning that the solutions are x=0 and x=3. And seeing as you can’t have a measure of distance being zero x=3 is our solution. confidence rating #$&*:ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 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. We solve by factoring. x^2 - 3x + 2 = (x - 2) ( x - 1) so we have (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 does. • So our solution to the equation is x = 3. x stands for the shorter side of the rectangle, which is therefore 3. The longer side is double the shorter, or 6. Thus to make the box: We take our 3 x 6 rectangle, cut out 1 ft corners and fold it up, giving us a box with dimensions 1 ft x 1 ft x 4 ft. This box has volume 4 cubic feet, confirming our solution to the problem. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: * 1/2/100 / 1.2.108 \ 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To solve this we use the formula given, s=-4.9t^2+20t and insert our measurement of distance traveled. So this becomes -4.9t^2+20t=15. So we subtract 15 and we get -4.9t^2+20t-15=0. Now we use the quadratic formula giving-20(+)(-)sqrt20^2-4*(-4.9)*(-15)/2(-4.9) and when we perform this formula we get t=3.09 and t=.99. And since all projectiles travel in an arch it passes the 15 m mark twice, once a t=.99 and again at t=3.09. Next we use the same method to determine when the object will cease to travel meaning when it hits the ground. We set the equation equal to zero and use the quadratic formula using the same steps giving t=0 and t=4.1 meaning the object is not in motion at t=0 and strikes the ground at t=4.1. now to discover when the object will reach 100 m in height we use the quadratic formula once again setting the prior formula equal to 100 and about evaluating the formula it is discovered that there is no solution meaning that the objects path of travel will not pass the 100 meter mark. confidence rating #$&*:ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 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. Interpretation: • 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. Interpretation: 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. Interpretation: We conclude that this object will not rise 100 ft. ** STUDENT QUESTION I was lost on this question and even reading the solution, Im still confused about it. Do you have any suggestions on how to look at it in a different way??? INSTRUCTOR RESPONSE s = -4.9 t^2 + 20 t means that if you plug in a value of t, you get the height, which is represented by the variable s. Also if you want to find the value of t that gives you a certain height, you plug in that height for s and solve the equation for t. The first question asks you to find when the height is 15 meters. So you plug in 15 for s. What equation do you get? You get the equation 15 = -4.9 t^2 + 20 t. Now you solve the equation for t. How do you do that? The equation is quadratic, since it contains both t^2 and t. The standard form for a quadratic equation is a t^2 + b t + c = 0 In this form you can try to factor the left-hand side. If this is possible you can then apply the zero property, as you've done in some of the preceding problems. If you can't figure out how to factor the equation (and in real-world problems you usually can't), you can use the quadratic formula. In this case you rearrange the equation 15 = -4.9 t^2 + 20 t to the form -4.9 t^2 + 20 t - 15 = 0 and pretty quickly realize that you won't be able to factor it. So you use the quadratic formula, as shown in the given solution. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: * Add comments on any surprises or insights you experienced as a result of this assignment. When first seeing the word problems it gives a sense of panic but once to start to relate the information to formula things become much simpler. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: * Add comments on any surprises or insights you experienced as a result of this assignment. When first seeing the word problems it gives a sense of panic but once to start to relate the information to formula things become much simpler. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!