course mth 163 For some reason Assignment 16 is not showing up in my portfolio. ”ú®£„R{脺†®åø‘ãö°÷ïS§Êò®ì¦éassignment #016
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17:00:42 behavior and source of exponential functions problem 1, perversions of laws of exponents Why is the following erroneous: a^n * b^m = (ab) ^ (n*m)
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RESPONSE --> In order to be able to multiply the powers, you must have the same variable. For example (4^3) * (5^2) is not the same as (4*5)^(3*2). To see this we cna multiply both out: (4*4*4) * (5*5) = 64 * 25 = 1600 the other is: (4*5)^(3*2) = (20)^(6) = 64000000 confidence assessment: 3
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17:01:07 STUDENT RESPONSE: my example was (4^2)(5^3) did not equal 20^6 INSTRUCTOR COMMENT ** more generally a^n * b^n = a^(n+m), which usually does not equal (ab) ^ (n * m) **
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RESPONSE --> self critique assessment: 3
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17:05:02 Why is the follow erroneous: a^(-n) = - a^n
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RESPONSE --> Again a substitution example works well here, but a negative exponent makes a fraction, not the variable negative: 2^(-3) = 1/(2)^3 = 1/8 or .125 - 2 ^ 3 = -8 confidence assessment: 3
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17:05:13 STUDENT RESPONSE: 2^-3 is not equal to -2^3 INSTRUCTOR COMMENT: ** A more general counterexample: a^(-n) = 1 / a^n is positive when a is positive whereas -a^n is negative when a is positive **
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RESPONSE --> self critique assessment: 3
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17:06:53 Why is the following erroneous: a^n + a^m = a^(n+m)
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RESPONSE --> Again, a substitute: 2^2 + 2^4 = 4 + 16 = 20 2^(2+4) = 2^(6) = 64 confidence assessment: 3
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17:07:04 STUDENT RESPONSE: (5^3)+(5^4) is not equal to 5^7 INSTRUCTOR COMMENT: (5^3) * (5^4) = (5^7) since (5*5*5) * (5*5*5*5) = 5*5*5*5*5*5*5 = 5^7. However (5^3) + (5^7) = 5*5*5 + 5*5*5*5*5 = 5*5*5( 1 + 5*5) = 5^3( 1 + 5^2), not 5^7.**
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RESPONSE --> self critique assessment: 3
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17:09:55 STUDENT RESPONSE: Why is the following erroneous: a^0 = 0
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RESPONSE --> I just know that a^0 = 1 I am not sure how to explain this. self critique assessment: 1
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17:10:51 4^0 is not equal to 0 INSTRUCTOR COMMENT: ** a^(-n) * a^n = 0 but neither a^(-n) nor a^n need equal zero **
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RESPONSE --> self critique assessment: 2
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17:13:54 Why is the following erroneous: a^n * a^m = a^(n*m).
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RESPONSE --> 4^2 * 4^4 = 16 * 256 = 4096 4^(2*4) = 4^8 = 65536 They are not equal. confidence assessment: 3
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17:14:04 STUDENT RESPONSE: (4^7)(4^2) is not equal to 4^14 INSTRUCTOR COMMENT: Right. Generally a^n * a^m = a^(n+m), not a^(n*m).
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RESPONSE --> self critique assessment: 3
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17:16:57 problem 2. Graph and describe Give basic points and asymptote; y values corresponding to the basic points, ratio of these y values: y = 1200 (2^(.12 t) )
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RESPONSE --> I was slightly confused....Do we use the -3, -2, -1 , 0, 1, 2, & 3 or am I missing something? confidence assessment: 1
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17:55:13 STUDENT RESPONSE (0,1200),(1,1304) negative x-axis ratio=163/150 INSTRUCTOR COMMENT: the precise ratio is 2^.12, which is probably pretty close to 163/150
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RESPONSE --> I understand where the numbers came from. self critique assessment: 1
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18:04:11 give basic points and asymptote; y values corresponding to the basic points, ratio of these y values: y = 400 ( 1.07 ) ^ t
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RESPONSE --> The points are (0, 400) & (1, 428) negative x-axis ratio: 107/100 confidence assessment: 2
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18:04:23 STUDENT RESPONSE (0,400),(1,428) Neg. x-axis 1.07 or 107/100 is ratio INSTRUCTOR COMMENT: that ratio is correct and is of course equal to 1.07
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RESPONSE --> self critique assessment: 3
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18:09:55 give basic points and asymptote; y values corresponding to the basic points, ratio of these y values: y = 250 ( 1 - .12 ) ^ t
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RESPONSE --> (0, 250) & (1, 220) Positive x axis ratio:22/25 confidence assessment: 3
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18:11:27 STUDENT RESPONSE The basic points are (0,250),(1,220) The positive x-axis is the horizontal asymptote The ratio of y values at the basic points is 220 / 250 = .88. INSTRUCTOR COMMENT: Note also that the ratio .88 is equal to 1-.12; .88 is the growth factor and .12 is the growth rate.
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RESPONSE --> self critique assessment: 3
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18:16:15 give basic points and asymptote; y values corresponding to the basic points, ratio of these y values: y = .04 ( .8 ) ^ t
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RESPONSE --> (0, .04) (1, .032) Positive x-axis Ratio: .8 confidence assessment: 3
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18:16:24 STUDENT RESPONSE (0,.04),(1,.032) are the basic points and the asymptote is the positive x-axis. The ratio is .32 / .4 = .8. The pattern is that the ratio is equal to b, where b is the base for the form y = A b ^ x.
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RESPONSE --> self critique assessment: 3
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18:25:48 problem 3. y = f(x) = 5 (1.27^x).
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RESPONSE --> (0, 5) (1, 6.35) Positive x axis asymtote ratio: 127/100 confidence assessment: 3
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18:29:34 What is the ratio between the y values at x = 0 and at x = 1?
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RESPONSE --> The ratio as I showed in the last answer: 127/100 or 1.27 confidence assessment: 3
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18:29:38 ** f(1) / f(0) = 5 * 1.27^1 / ( 5 * 1.27^0) = 1.27 **
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RESPONSE --> self critique assessment: 3
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18:35:57 What is the ratio between the y values at to x = 3.4 and x = 4.4?
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RESPONSE --> First you find the y values: y = 5(1.27^3.4) y = 5(2.25389) y = 11.27 (3.4, 11.27) y = 5(1.27^4.4) y = 5(2.86244) y = 14.31 (4.4, 14.31) The ratio is the same as before: 14.31/11.27 = 1.27 confidence assessment: 3
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18:36:38 ** f(4.4) / f(3.4) = 5 * 1.27^4.4 / ( 5 * 1.27 ^3.4) = 1.27. **
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RESPONSE --> self critique assessment: 3
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18:42:21 Verify that the ratio of y values is again the same for your own points where x differs by 1 unit.
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RESPONSE --> In the first example with the points (0, 5) (1, 6.35) the ratio was 1.27. In the previous example with points (3.4, 11.27) & (4.4, 14.31) the ratio is 1.27 confidence assessment: 2
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18:42:52 ** STUDENT RESPONSE: My points were 4.5 and 5.5, and the y ratio was again 1.27 **
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RESPONSE --> I didn't realize I was supposed to pick two more points. self critique assessment: 3
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18:45:01 What is the ratio of y values when x values are separated by two units?
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RESPONSE --> I chose to use (0, 5) & (2, 8.0645) the ratio was 1.6 confidence assessment: 2
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18:45:12 ** If x values are separated by 2 units then the ratio is 1.27^(x+2) / 1.27^x = 1.27^(x+2 - x) = 1.27^2 = 1.61 approx. **
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RESPONSE --> self critique assessment: 3
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19:10:41 problem 4. Ratio of y values at x = x1 and x = x1+1 What does your result tell you about how the ratio depends on the x value x1?
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RESPONSE --> the ratio will be the same as long as there is one other unit is higher or lower than the first. confidence assessment: 1
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19:10:51 ** If y = A b^x then the value at x1 is A b^x1 and the value at x1 + 1 is A b ^(x1 + 1). The ratio of these values is A b^(x1+1) / A b^x1 = b^(x1 + 1 - x1) = b^1 = b. The ratio should have been b, where b is the base in the form y = A b^x. This is the same as in all previous examples, which shows that there is no dependence on x1. **
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RESPONSE --> self critique assessment: 3
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19:13:44 problem 5. y = 3 (2 ^ (.3 x) ). What is the ratio of the two basic-point y values?
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RESPONSE --> you get as the basic points: (0, 3) and (1, 3.69) The ratio is 1.23 confidence assessment: 2
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19:13:48 ** The basic points are the x = 0 and x = 1 points. The corresponding y values are 3(2^.(3*0) ) = 3 and 3(2^(.3 *1) ) = 3 * 2^.3 = 3.69 approx. The ratio of these values is 3.69 / 3 = 1.23. **
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RESPONSE --> self critique assessment: 3
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19:15:11 What is the y = A b^x form of this function?
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RESPONSE --> I am not sure. confidence assessment: 0
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19:17:33 ** 3 (2 ^ (.3 x) ) = 3 (2^.3)x = 3 * 1.23^x, approx. This is in the form y = A b^x for A = 3 and b = 1.23. **
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RESPONSE --> I think I understand the explaination. self critique assessment: 1
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19:20:19 What does the value of 2 ^ .3 have to do with this situation?
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RESPONSE --> If you need to find other pointsyou can use it to find b confidence assessment: 1
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19:20:35 ** CORRECT STUDENT RESPONSE: this is the b value in the form y = A b^x. **
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RESPONSE --> My answer was close. self critique assessment: 3
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19:29:07 problem 6 P(n+1) = (1+r) P(n), with r = .1 and P(0) = $1000. What are P(1), P(2), ..., P(5)?
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RESPONSE --> After substituting: P(1) = 1100 P(2) = 1210 P(3) = 1331 P(4) = 1464.10 P(5) = 1610.51 confidence assessment: 2
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19:30:17 ** If n = 0 we get P(0 + 1) = (1 + .1) * P(0), or P(1) = 1.1 * P(0). Since P(0) = $1000 we have P(1) = 1.1 * $1000 = $1100. If n = 1 we get P(1 + 1) = (1 + .1) * P(1), or P(2) = 1.1 * P(1). Since P(1) = $1000 we have P(2) = 1.1 * $1100 = $1210. If n = 2 we get P(2 + 1) = (1 + .1) * P(2), or P(3) = 1.1 * P(2). Since P(2) = $1000 we have P(3) = 1.1 * $1210 = $1331. If n = 3 we get P(3 + 1) = (1 + .1) * P(3), or P(4) = 1.1 * P(3). Since P(3) = $1000 we have P(4) = 1.1 * $1331 = $1464/1. If n = 4 we get P(4 + 1) = (1 + .1) * P(4), or P(5) = 1.1 * P(4). Since P(4) = $1000 we have P(5) = 1.1 * $1464.1 = $1610.51. **
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RESPONSE --> I didn't show everything multiplied out (some of the other problems didn't do this) self critique assessment: 3
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19:42:33 problem 8. Q(n+1) = .85 Q(n), Q(0) = 400. What are Q(n) for n = 1, 2, 3 and 4 /
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RESPONSE --> Substitute n = 0 we get Q(0 + 1) = (.85) * Q(0), or Q(1) = .85 * Q(0). Since Q(0) = 400 we have Q(1) =. 85 * 400 = 340 n = 1: Q(1 + 1) = (.85) * Q(1), or Q(2) = .85 * 340 =289 n = 2: Q(2 + 1) = (.85) * Q(2), or Q(3) = .85 * 289 =245.65 n = 3: Q(3 + 1) = (.85) * Q(3), or Q(4) = .85 * 245.65 = 208.80 confidence assessment: 3
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19:42:45 ** For n = 0 we have Q(0 + 1) = .85 Q(0) so Q(1) = .85 * 400 = 340. For n = 1 we have Q(1 + 1) = .85 Q(1) so Q(2) = .85 * 340 = 289. For n = 2 we have Q(2 + 1) = .85 Q(2) so Q(3) = .85 * 400 = 245.65. For n = 3 we have Q(3 + 1) = .85 Q(3) so Q(4) = .85 * 400 = 208.803. **
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RESPONSE --> self critique assessment: 3
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19:44:35 What is the growth rate for this equation?
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RESPONSE --> .85 confidence assessment: 1
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19:45:40 ** The growth factor is .85, which is 1 + growth rate. It follows that the growth rate is .85 - 1 = .15 **
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RESPONSE --> I got growth rate and growth factor backward. self critique assessment: 2
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19:48:55 problem 9. interest rate 12%, initial principle $2000. What is your difference equation?
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RESPONSE --> y(t) = P(1+r)^t or in this case: y(t) = 2000(1.12)^t confidence assessment: 2
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19:49:26 ** The growth rate is 12% = .12 The growth factor is therefore 1 + .12 and the difference equation is P(n+1)=(1+.12)P(n), P(0)=2000. **
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RESPONSE --> The equation I gave is equal to this one. self critique assessment: 1
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19:54:27 How did you use your difference equation to find the principle after 1, 2, 3 and 4 years and what did you get?
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RESPONSE --> Substituted in : n = 0: P(0 + 1) = (1.12) * Q(0), or Q(1) = 1.12 * 2000 =2240 n = 1: P(1 + 1) = (1.12) * Q(1), or Q(2) = 1.12 * 2240 =2508.80 n = 2: P(2 + 1) = (1.12) * Q(2), or Q(3) = 1.12 * 22508.80 = 2809.86 n = 0: P(3 + 1) = (1.12) * Q(33), or Q(4) = 1.12 * 2809.86 =3147.04 confidence assessment: 3
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19:54:39 ** STUDENT RESPONSE P(0+1)=(1+.12)2000 and so on up to P(4) was found. P1=2240 P2=2508.8 P3=2809.856 P4=3147.03872 **
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RESPONSE --> self critique assessment: 3
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20:04:17 problem 11. Texcess(t) = 50 (.97 ^ t). What is your estimate of the time required to fall to 1/8 of the original value?
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RESPONSE --> I think I know how to do this, but I want to check out the explaination. confidence assessment: 0
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20:07:00 ** The original value takes place at t = 0 and is Texcess(0) = 50 * .97^0 = 50. 1/8 of the original value is therefore 1/8 * 50 = 6.25. You have to find t such that 50 * .97^t = 6.25. Dividing both sides by 50 you get .97^t = 6.25 / 50 or .97^t = .125. Use trial and error to find t: Try t = 10: .97^10 = .74 approx. That's too high. Try t = 100: .97^100 = .04 approx. That's too low. So try a number between 10 and 100, probably closer to 100. Try 70: .97^70 = .118. Lucky guess. That's close to .125 but a little low. {Try 65: .97^65 = .138. Too high. Try a number between 65 and 70, closer to 70 but not too much closer. Try 68: .97^68 = .126. That's good to the nearest whole number. The process could be continued and refined to get more accurate values of t. **
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RESPONSE --> This explaination makes sense to me. It was not what I had, but I understand how to work the problem. self critique assessment: 2
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21:05:21 What are your ratios of temperature excess to average rate, and are they nearly constant?
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RESPONSE --> not sure where to get the numbers. confidence assessment: 0
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21:06:26 ** If the function is 50 (.97^t) then at t = 0, 17, 34, 51, 68 we have function values 50, 29.79130219, 17.75043372, 10.57617070, 6.301557949 and average rates of change -1.18756929, -0.7082863803, -0.4220154719, -0.2514478090, -0.1498191533. Trapezoids have 'average altitudes' 39.89565109, 23.77086796, 14.16330221, 8.438864327. Ratio of temp excess to ave rate, using 'average altitudes' as temp excess, are therefore 39.9 / (-1.19), 23.8 / (-.708), 14.2 / (-.422), 8.44 / (-.251). These quantities vary slightly but all are close to the same value around 33. **
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RESPONSE --> I guess just pick some numbers? Then look at what info they give. self critique assessment: 1
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21:09:10 Give the original and the simplified equation to determine the time required for Texcess to fall to half its original value.
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RESPONSE --> Texcess(t) = (1/2)50(.97^t) Texcess(t) = 25(.97^t) confidence assessment: 1
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21:10:00 ** Texcess has an original value at t = 0, which gives us Texcess(0) = 50 * .97^0 = 50. Half its original value is therefore 25. So our equation is 25 = 50 * .97^t. This equation is simplified by dividing both sides by 50 to get .97^t = 1/2. **
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RESPONSE --> I didn't quite have it right. I should have had 25 on the other side and then divided. self critique assessment: 1
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21:18:00 problem 12. Texcess(t) = 50(.97^t), room temperature {{ 25 if you used Celsius and 75 if you used Farenheit in your observations What function Temp(t) gives temperature as a function of time?
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RESPONSE --> The temperature problems have given me trouble tonight. I will have to look back over it. confidence assessment: 0
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21:24:00 ** Using 75 for room temperature and realizing that temperature is room temperature + temperature excess we obtain the function Temp(t)=50(.97^t)+75.**
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RESPONSE --> The formula makes sense, I just didn't have a clue. self critique assessment: 1
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21:25:23 Identify the values of A, b and c in the generalized form y = A b^x + c.
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RESPONSE --> Temp(t)=50(.97^t)+75 A=50 b = .97 c = 75 confidence assessment: 3
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21:25:29 ** Since the function is Temp(t) = 50(.97^t)+75 , the y = A b^x + c has y = Temp(t), A = 50, b = .97 and c = 75. **
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RESPONSE --> self critique assessment: 3
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21:48:26 problem 14. Antiobiotic removal, 40 mg/hour when there are 200 milligrams present At what rate would antibiotic be removed when there are 70 milligrams present?
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RESPONSE --> Since this is a proportionality problem: 40 = k* 200^3 40 = 8000000k .000005 = k r = (.000005) * (70^3) r = (.000005) * 343000 r = 1.75 confidence assessment: 1
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21:50:05 ** If the rate of removal is directly proportional the quantity present then we have y = k x where y is the rate of removal and x the amount present. Since y = 40 when x = 200 we have 40 = k * 200 so that k = 40/200 = .2. Thus y = .2 x. If x = 70 then we have y = .2 * 70 = 14. When there are 70 mg present the rate of removal is 14 mg/hr. **
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RESPONSE --> I cubed 200 and I shouldn't have. self critique assessment: 2
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