#$&* course Mth 163 10/28 930 014.
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Given Solution: If we substitute n = 1 into a(n) = a(n-1) + 2^n we get a(1) = a(1-1) + 2^1 or, since 1-1 = 0 and 2^1 = 2 a(1) = a(0) + 2. Since we are given a(0) = 3 we now have a(1) = 3 + 2 = 5. If we substitute n = 2 into a(n) = a(n-1) + 2^n we get a(2) = a(2-1) + 2^2 or, since 2-1 = 1 and 2^2 = 4 a(2) = a(1) + 4. Since in the previous step we found that a(1) = 5 we now have a(2) = 5 + 4 = 9. If we substitute n = 3 into a(n) = a(n-1) + 2^n we get a(3) = a(3-1) + 2^3 or, since 3-1 = 2 and 2^3 = 8 a(3) = a(2) + 8. Since in the previous step we found that a(2) = 9 we now have a(3) = 9 + 8 = 17. If we substitute n = 4 into a(n) = a(n-1) + 2^n we get a(4) = a(4-1) + 2^4 or, since 4-1 = 3 and 2^4 = 16 a(4) = a(3) + 16. Since in the previous step we found that a(3) = 17 we now have a(4) = 17 + 16 = 33. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ok ********************************************* Question: `q002. If a(n) = 2 * a(n-1) + n with a(0) = 3, then what are the values of a(1), a(2), a(3) and a(4)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: a(1) = 2 * a(1-1) + 1 = 2 * a0 + 1 = 2 * 3 + 1 = 7 a(2) = 2 * a(2-1) + 2 = 2 * a1 + 2 = 2 * 7 + 2 = 16 a(3) = 2 * a (3-1) + 3 = 2 * a2 + 3 = 2 * 16 + 3 = 35 a(4) = 2 * a(4-1) + 4 = 2 * a3 + 4 = 2 * 35 + 4 = 74 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If we substitute n = 1 into a(n) = 2 * a(n-1) + n we get a(1) = 2 * a(1-1) + 1 or since 1-1 = 0 a(1) = 2 * a(0) + 1. Since we know that a(0) = 3 we have a(1) = 2 * 3 + 1 = 7. If we substitute n = 1 into a(n) = 2 * a(n-1) + n we get a(2) = 2 * a(2-1) + 2 or since 2-1 = 1 a(2) = 2 * a(1) + 2. Since we know that a(0) = 3 we have a(2) = 2 * 7 + 2 = 16. If we substitute n = 1 into a(n) = 2 * a(n-1) + n we get a(3) = 2 * a(3-1) + 3 or since 3-1 = 2 a(3) = 2 * a(2) + 3. Since we know that a(0) = 3 we have a(3) = 2 * 16 + 3 = 35. If we substitute n = 1 into a(n) = 2 * a(n-1) + n we get a(4) = 2 * a(4-1) + 4 or since 4-1 = 3 a(4) = 2 * a(3) + 4. Since we know that a(0) = 3 we have a(4) = 2 * 35 + 4 = 74. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ok ********************************************* Question: `q003. What are the average slopes of the graph of y = x^2 + x - 2 between the x = 1 and x= 3 points, between the x = 3 and x = 5 points, between the x = 5 and x = 7 points, and between the x = 7 and x = 9 points? What is the pattern of this sequence of slopes? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = 1^2 + 1 - 2 = 0 y = 3^2 + 3 - 2 = 10 slope = 5 y = 3^2 + 3 - 2 = 10 y = 5^2 + 5 - 2 = 28 slope = 9 y = 28 and y = 54 slope = 13 y = 54 and y = 88 slope = 17 The pattern of the sequences of the slopes is a difference of 4 each time and all are divided by 2 when figuring the slope. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: At x = 1, 3, 5 , 7 and 9 we find by substituting that y = 0, 10, 28, 54 and 88. The x = 1, 3, 5, 7 and 9 points are therefore (1,0), (3,10), (5,28), (7,54) and (9,88). The run from one point to the next is always 2. The rises are respectively 10, 18, 26 and 34. The slopes are therefore slope between x = 1 and x = 3: slope = rise / run = 10 / 2 = 5. slope between x = 3 and x = 5: slope = rise / run = 18 / 2 = 9. slope between x = 5 and x = 7: slope = rise / run = 26 / 2 = 13. slope between x = 7 and x = 9: slope = rise / run = 34 / 2 = 17. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ok ********************************************* Question: `q004. If a solid stone sphere 4 inches in diameter weighs 3 pounds, then what would be the weight of a solid stone sphere 2 feet in diameter? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: These are the ones I’m having trouble with…looking below..sorry! confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The volume of a sphere is proportional to the cube of its diameters, and weight is directly proportional to volume so we have the proportionality w = k d^3, where w and d stand for weight and diameter and k is the proportionality constant. Substituting the known weight and diameter we get 3 = k * 4^3, where we understand that the weight is in pounds and the diameter in inches. This gives us 3 = 64 k so that k = 3 / 64. Our proportionality equation is now w = 3/64 * d^3. So when the diameter is 2 feet, we first recall that diameter must be in inches and say that d = 24, which we then substitute to obtain w = 3/64 * 24^3. A simple calculation gives us the final weight w = 648. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Hallelujah I totally understand this now!! YES!!!! I have taken good notes, and I am finally getting this!! Thank you!! ------------------------------------------------ Self-critique rating: OK!!!!
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Given Solution: The ratios 36/12, 54/18 and 72/24 of the corresponding sides are all the same and all equal to 3, so the dimensions of the sides of the second box are 3 times those of the first. Since the thickness of the cardboard is the same on both boxes, only the dimensions of the rectangular sides change. The only thing that matters, therefore, is the surface area of the box. The proportionality is therefore of the form w = k x^2, where w is the weight of the box and x stands any linear dimension. It follows that w2 / w1 = (x2 / x1)^2. Since as we just saw x2 / x1 = 3, we see that w2 / w1 = 3^2 = 9. Since w1 = 22 oz, we write this as w2 / 22 oz = 9. Multiplying both sides by 22 oz we see that w2 = 22 oz * 9 = 198 oz. STUDENT QUESTION I reduced the sides of each box by dividing each side measurement by 6. Then since we were looking for weight, I used the y = kx^3 formula to calculate the weight of the second box. I used y as the weight and x as the volume (l*w*h). Obviously my calculation was way off. Why wouldn’t this work? INSTRUCTOR COMMENT Presumably the cardboard is of the same thickness for both boxes. So the amount of cardboard is determined by the surface area of the the box, not the volume. If the box was filled with cardboard as well as being constructed of cardboard, then the proportionality with volume would be appropriate and your solution would be correct. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I know what I did wrong. Could I use that formula to find the average of the box widths, which I got 3, and so did you, but I want to make sure that is an ok way, or should I use your way? I have corrected my notes to your way just in case. I see what I messed up as well. I did not plug the numbers into the formula correctly. I have fixed my notes and is understood ********************************************* Question: `q006. Between t = 1 and t = 3 the function y = .5 t^2 - 5 t + 9 has average slope -3; between t = 3 and t = 5 the average slope is -1; and between t = 5 and t = 7 the average slope is 1. • What do you conjecture will be the average slope between t = 7 and t = 9? • What linear function do you think gives the slope as a function of t? • What do you think will be the slope at t = 14? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The average slope between t = 7 and t = 9 would be 3. Each slope goes up positive 2 each time. Linear function y = mx + b would be t = -3x + 2 ??? The slope at t = 14 : t = 9 and 11 would be 5 t = 11 and 13 would be 7 t = 13 and 15 would be 9 so t = 14 would be 8. confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties. ********************************************* Question: `q007. A sphere of radius .5 meters has mass 1100 kg and requires 11 ounces of paint to cover its surface. What would be the mass of a sphere of the same material whose radius is 1.3 meters, and how much paint would it take to cover its surface with a coat whose thickness was the same as that used to cover the first? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1.3 / .5 = 2.6 M2/m1 = x2/x1^3 = 2.6^3 = 17.576 M1 = 1100 Mass = M2/1100 = 17.576 M2 = 17.576 * 1100 = 19333.6 Paint required = y = kx^2 = 19333.6^1/2 = 26.839 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique rating: ????? " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ????? " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!