course mth 163 x????y??o??????assignment #014014.
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19:10:20 `q001. Note that this assignment has 5 questions If a(n) = a(n-1) + 2^n, with a(0) = 3, then substitute in turn the values 1, 2, 3 and 4 into the equation to obtain the values a(1), a(2), a(3) and a(4).
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RESPONSE --> First of all we substitute 1 in for n: a(1) = a(1-1) + 2^1 a(1) = a(0) + 2 **We know that a(0) = 3, so sub** a(1) = 3 + 2 a(1) = 5 Continue to substitute: a(2) = a(2-1) + 2^2 a(2) = a(1) + 4 a(2) = 5 + 4 a(2) = 9 a(3) = a(3-1) + 2^3 a(3) = a(2) + 8 a(3) = 9 + 8 a(3) = 17 a(4) = a(4-1) + 2^4 a(4) = a(3) + 16 a(4) = 17 + 16 a(1) = 23 confidence assessment: 2
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19:10:47 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 we are given 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 we are given 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 we are given a(3) = 16 we now have a(4) = 17 + 16 = 33.
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RESPONSE --> self critique assessment: 3
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19:38:11 `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)?
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RESPONSE --> This problem is done the same way: Substitute a(1) = 2 * a(1-1) + 1 a(1) = 2 * a(0) + 1 a(1) = 2 * 3 + 1 a(1) = 6 + 1 a(1) = 7 a(2) = 2 * a(2-1) + 2 a(2) = 2 * a(1) + 2 a(2) = 2 * 7 + 2 a(2) = 14 + 2 a(2) = 16 a(3) = 2 * a(3-1) + 3 a(3) = 2 * a(2) + 3 a(3) = 2 * 16 + 3 a(3) = 32 + 3 a(3) = 35 a(4) = 2 * a(4-1) + 4 a(4) = 2 * a(3) + 4 a(4) = 2 * 35 + 4 a(4) = 70 + 4 a(4) = 74 confidence assessment: 3
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19:38:34 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.
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RESPONSE --> self critique assessment: 3
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20:04:39 `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?
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RESPONSE --> First solve for y for each x coordinate: x=1- (1, 0) x=3- (3, 4) x=5- )5, 28) x=7- (7, 54) x=9- (9, 88) Slopes (y2-y1)/(x2-x1): x=1 to x=3 (4-0)/(3-1) = 4/2 = 2 x=3 to x=5 (28-4)/(5-3) = 24/2 = 12 x=5 to x=7 (54-28)/(7-5) = 26/2 = 13 x=7 to x=9 (88-54)/9-7) = 34/2 = 17 Not sure what pattern you are looking for.... confidence assessment: 2
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20:07:39 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 = 6. 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.
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RESPONSE --> I miss calculated and the y value for x=3 should have been 10. The slope between 1 and 3 should have been 5, and hte slope between 3 and 5 should be 9 self critique assessment:
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20:12:33 `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?
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RESPONSE --> I used a proportional equation : 4 inches/3 lbs = 24 inches/ x lbs **cross mult. 4x = 72 **Divide x = 18 lbs confidence assessment: 2
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20:14:11 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 = 748.
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RESPONSE --> I understand the use of this formula. I just wasn't thinking of it at the time. self critique assessment: 2
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20:30:25 `q005. Two boxes are each constructed of a single layer of cardboard. The first box is 12 inches by 18 inches by 24 inches and weighs 22 ounces; the second is 36 inches by 54 inches by 72 inches. Using proportionality determine the weight of the second box.
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RESPONSE --> I know I should use the w = kx^3, where w = weight and x = the volume, but I got a strange number: to get the first box volume: 12*18*24 = 5184 using w = kx^3 22 = k(5184)^3 22 = 139314069504k .000000000158 = k Then subsititue: w = k(36*54*72)^3 w = (.000000000158)(139968)^3 w = (.000000000158)(2742118830047232) w = 433254.78 confidence assessment: 1
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20:31:04 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.
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RESPONSE --> I should have used the surface area formula. self critique assessment: 1
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b??J?W?????????? assignment #015 015. Precalculus I 03-05-2007
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21:25:57 `q001. Note that this assignment has 12 questions If you are given $1000 and invest it at 10% annual interest, compounded annually, then how much money will you have after the first year, how much after the second, and how much after the third? Is the the change in the amount of money the same every year, does the change increase year by year, does the change decrease year by year or does it sometimes increase and sometimes decrease?
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RESPONSE --> Using compounded interest formula P(1 +.1): First Year: 1000(1.1) = 1100 in order to find the next few years raise the formula by the year you want to find: Second Year: 1000(1.1)^2 = 1000*1.21= 1210 Third Year: 1000(1.1)^3 = 1000*1.331 = 1331 The change in money each year is not the same. It increases year by year. confidence assessment: 3
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21:26:13 During the first year the interest will be 10% of $1000, or $100. This makes the total at the end of the first year $1100. During the second year the interest will be 10% of $1100, or $110. At the end of the second year the total will therefore be $1100 + $110 = $1210. During the third year the interest will be 10% of $1210, or $121. At the end of the sphere year the total will therefore be $1210 + $121 = $1331. The yearly changes are $100, $110, and $121. These changes increase year by year.
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RESPONSE --> self critique assessment: 3
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21:33:00 `q002. In the preceding problem you obtained amounts $1100, $1210 and $1331. What number would you multiply by $1000 to get $1100? What number we do multiply by $1100 to get $1210? What number would we multiply by $1210 to get $1331? What is the significance of this number and how could we have found it from the original information that the amount increases by 10 percent each year?
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RESPONSE --> I used the formla which multiplies 1000 by 1.1 to get 1100 You also, multiply1100 by 1.1 to get 1210. Same to get 1331. See formula that I have used in these problems. confidence assessment: 3
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21:33:08 To get $1100 you have to multiply $1000 by 1100 / 1000 = 1.1. To get $1210 you have to multiply $1210 by 1210 / 1100 = 1.1. To get $1331 you have to multiply $1331 by 1331 / 1210 = 1.1. If the amount increases by 10 percent, then you end up with 110 percent of what you start with. 110% is the same as 1.1.
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RESPONSE --> self critique assessment: 3
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21:41:19 `q003. Given the recurrence relation P(n) = 1.10 * P(n-1) with P(0) = 1000, substitute n = 1, 2, and 3 in turn to determine P(1), P(2) and P(3). How is this equation related to the situation of the preceding two problems?
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RESPONSE --> You get the same answers as the previous questions. confidence assessment: 3
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21:41:26 Substituting 1 into P(n) = 1.10 * P(n-1) we obtain P(1) = 1.10 * P(1-1), or P(1) = 1.10 * P(0). Since P(0) = 1000 we get P(1) = 1.1 * 1000 = 1100. Substituting 2 into P(n) = 1.10 * P(n-1) we obtain P(2) = 1.10 * P(1). Since P(1) = 1100 we get P(1) = 1.1 * 1100 = 1210. Substituting 3 into P(n) = 1.10 * P(n-1) we obtain P(3) = 1.10 * P(2). Since P(2) = 1000 we get P(3) = 1.1 * 1210 = 1331.
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RESPONSE --> self critique assessment: 3
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21:45:18 `q004. If you are given $5000 and invest it at 8% annual interest, compounded annually, what number would you multiply by $5000 to get the amount at the end of the first year? Using the same multiplier, find the results that the end of the second and third years.
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RESPONSE --> Using the P(1 + interest): you would end up multiplying by 1.08 First Year: 5000(1.08) = 5400 Second Year: 5400(1.08) = 5832 Third Year: 5832(1.08) = 6298.56 confidence assessment: 3
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21:45:47 If your money increases by 8% in a year, then it the end of the year you will have 108% as much as at the beginning. Since 108% is the same as 1.08, our yearly multiplier will be 1.08. If we multiply $5000 by 1.08, we obtain $5000 * 1.08 = $5400, which is the amount the end of the first your. At the end of the second year the amount will be $5400 * 1.08 = $5832. At the end of the third year the amount will be $5832 * 1.08 = $6298.56.
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RESPONSE --> self critique assessment: 3
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21:48:37 `q005. How would you write the recurrence relation for a $5000 investment at 8 percent annual interest?
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RESPONSE --> 5000(1.08) confidence assessment: 3
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21:49:44 Just as the recurrence relation for 10 percent annual interest, as seen in the problem before the last, was P(n) = 1.10 * P(n-1), the recurrence relation for 8 percent annual interest is P(n) = 1.08 * P(n-1).
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RESPONSE --> My equation equals out to the one given. self critique assessment: 3
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21:51:11 `q006. If you are given amount $5000 and invest it at annual rate 8% or .08, then after n years how much money do you have? What does a graph of amount of money vs. number of years look like?
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RESPONSE --> 5000(1.08)^n The graph will increase slowly and the begin to grow quickly or exponentially confidence assessment: 2
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21:51:30 After 1 year the amount it $5000 * 1.08. Multiplying this by 1.08 we obtain for the amount at the end of the second year ($5000 * 1.08) * 1.08 = $5000 * 1.08^2. Multiplying this by 1.08 we obtain for the amount at the end of the third year ($5000 * 1.08^2) * 1.08 = $5000 * 1.08^3. Continuing to multiply by 1.08 we obtain $5000 * 1.08^3 at the end of year 3, $5000 * 1.08^4 at the end of year 4, etc.. It should be clear that we can express the amount at the end of the nth year as $5000 * 1.08^n. If we evaluate $5000 * 1.08^n for n = 0, 1, 2, ..., 10 we get $5000, $5400, 5832, 6298.56, 6802.45, 7346.64, 7934.37, 8569.12, 9254.65, 9995.02, 10,794.62. It is clear that the amount increases by more and more with every successive year. This result in a graph which passes through the vertical axis at (0, 5000) and increases at an increasing rate.
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RESPONSE --> self critique assessment: 3
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21:56:43 `q007. With a $5000 investment at 8 percent annual interest, how many years will it take to double the investment?
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RESPONSE --> If you keep multiplying by 1.08...after 10 years you will have just over $10,000 confidence assessment: 3
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21:57:02 Multiplying $5000 successively by 1.08 we obtain amounts $5400, 5832, 6298.56, 6802.45, 7346.64, 7934.37, 8569.12, 9254.65, 9995.02, 10,794.62 at the end of years 1 thru 10. We see that the doubling to $10,000 occurs very shortly after the end of the ninth year. We can make a closer estimate. If we calculate $5000 * 1.08^x for x = 9 and x = 9.1 we get about $10,072. So at x = 9 and at x = 9.1 the amounts are $9995 and $10072. The first $5 of the $77 increase will occur at about 5/77 of the .1 year time interval. Since 5/77 * .1 = .0065, a good estimate would be that the doubling time is 9.0065 years. If we evaluate $5000 * 1.08^9.0065 we get $10,000.02.
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RESPONSE --> self critique assessment: 3
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22:00:52 `q008. If you are given amount P0 and invest it at annual rate r (e.g., for the preceding example r would be 8%, which in numerical form is .08), then after n years how much money do you have?
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RESPONSE --> If you have P0 inverst at rate r in this case .08 after n years you get the formula: P0 * (1+.08)^n confidence assessment: 3
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22:01:15 If the annual interest rate is .08 then each year we would multiply the amount by 1.08, the amount after n years would be P0 * 1.08^n. If the rate is represented by r then each year then each year we multiply by 1 + r, and after n years we have P0 * (1 + r)^n.
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RESPONSE --> I just did not put in the variable for .08 as r. self critique assessment: 3
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22:15:10 `q009. If after an injection of 800 mg an antibiotic your body removes 10% every hour, then how much antibiotic remains after each of the first 3 hours? How long does it take your body to remove half of the antibiotic?
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RESPONSE --> Using Q(t) - Q0* (1-r)^t 800(1-.1)^3 800(.9)^3 800(.729) 583.2 You use trial and error and will find that between 6.5 and 6.75 hours half the antibiotic will be removed. confidence assessment: 3
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22:17:19 If 10 percent of the antibiotic is removed each hour, then at the end of the hour the amount left will be 90 percent of what was present at the beginning of the hour. Thus after 1 hour we have .90 * 800 mg, after a second hour we have .90 of this, or .90^2 * 800 mg, and after a third hour we have .90 of this, or .90^3 * 800 mg. The numbers are 800 mg * .90 = 720 mg, then .90 of this or 648 mg, then .90 of this or 583.2 mg.
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RESPONSE --> self critique assessment: 3
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22:19:25 `q010. In the preceding problem, what function Q(t) represents the amount of antibiotic present after t hours? What does a graph of Q(t) vs. t look like?
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RESPONSE --> Q(t) = Q0 (1-r)^t The graph will be decreasing at a decreasing rate. confidence assessment: 3
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22:19:58 After t hours we will have 800 * .9^t mg left. So Q(t) = 800 * .9^t. The amounts for the first several years are 800, 720, 648, 583.2, etc.. These amounts decrease by less and less each time. As a result the graph, which passes through the vertical axes at (0,800), decreases at a decreasing rate. We note that no matter how many times we multiply by .9 our result will always be greater than 0, so the graph will keep decreasing at a decreasing rate, approaching the horizontal axis but never touching it.
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RESPONSE --> self critique assessment: 3
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22:24:21 `q011. Suppose that we know that the population of fish in a pond, during the year after the pond is stocked, should be an exponential function of the form P = P0 * b^t, where t stands for the number of months after stocking. If we know that the population is 300 at the end of 2 months and 500 at the end of six months, then what system of 2 simultaneous linear equations do we get by substituting this information into the form of the function?
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RESPONSE --> Substituting: 300 = P0(b^2) 500 = P0(b^6) confidence assessment: 1
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22:24:34 We substitute the populations 300 then 500 for P and substitute 2 months and 6 months for t to obtain the equations 300 = P0 * b^2 and 500 = P0 * b^6.
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RESPONSE --> I actually had it right
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22:27:30 `q012. We obtain the system 300 = P0 * b^2 500 = P0 * b^6 in the situation of the preceding problem. If we divide the second equation by the first, what equation do we obtain? What do we get when we solve this equation for b? If we substitute this value of b into the first equation, what equation do we get? If we solve this equation for P0 what do we get? What therefore is our specific P = P0 * b^t function for this problem?
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RESPONSE --> I have gotten the equation down to 5/3 = b^4 but not sure where to go from there. confidence assessment: 1
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22:28:19 Dividing the second equation by the first the left-hand side will be left-hand side: 500/300, which reduces to 5/3, and the right-hand side will be right-hand side: (P0 * b^6) / (P0 * b^2), which we rearrange to get (P0 / P0) * (b^6 / b^2) = 1 * b^(6-2) = b^4. Our equation is therefore b^4 = 5/3. To solve this equation for b we take the 1/4 power of both sides to obtain (b^4)^(1/4) = (5/3)^(1/4), or b = 1.136, to four significant figures. Substituting this value back into the first equation we obtain 300 = P0 * 1.136^2. Solving this equation for P0 we divide both sides by 1.136^2 to obtain P0 = 300 / (1.136^2) = 232.4, again accurate to 4 significant figures. Substituting our values of P0 and b into the original form P = P0 * b^t we obtain our function P = 232.4 * 1.136^t.
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RESPONSE --> I remember that you have to raise it to the inverse now. Had a duh moment. self critique assessment: 3
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