course Mth 163 ̧䚽ӕݪڧЮassignment #016
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23:51:27 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 --> a^n*b^m 2^3 * 2^2 = (2*2*2)*(2*2) = 8 * 4 = 32 (ab)^(n*m) (2*2)^(3*2) = 4^6 = 4096 32 is not equal to 4096, so we find, a^n*a^m = a^(n+m) not a^(n*m)
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23:51:29 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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23:52:14 Why is the follow erroneous: a^(-n) = - a^n
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RESPONSE --> 2^(-3) = 1/(2*2*2) = 1/8 (-2)^3 = (-2*-2*-2) = -8 so, 1/8 is not equal to -8
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23:52:26 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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23:53:36 Why is the following erroneous: a^n + a^m = a^(n+m)
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RESPONSE --> a^n + a^m = a^(n+m) 2^3 + 2^2 = (2*2*2) + (2*2) = 8+4 = 12 2^(3+2) = 2^5 = 32 12 is not equal to 32
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23:53:40 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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23:55:08 STUDENT RESPONSE: Why is the following erroneous: a^0 = 0
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RESPONSE --> 4^0 = 1 if you had 4^3 * 4^0, according to the laws of exponents you could change that to 2(3+0) = 2^3, which is equal to the first amount paired with 4^0 and in order to get the same amount out as you put in, the second amount must be equal to 1, therefore, 4^0 is equal to one.
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23:55:11 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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23:56:45 Why is the following erroneous: a^n * a^m = a^(n*m).
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RESPONSE --> a^n * a^m = a^(n*m) 2^3*2^2 = (2*2*2)*(2*2) = 8*4 = 32 2^(3*2) = 2^6 = 64 32 is not equal to 64, a^n*a^m = a^(n+m)
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23:56:51 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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23:57:45 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 --> (0,1200) (1,1304.082) negative x axis is asymptote ratio is 1304.082/1200 = 1.086735
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23:57:49 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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23:58:25 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 --> (0,400) (1,428) negative x axis asymptote ratio is 428/400 = 1.07
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23:58:28 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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23:58:49 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 asymptote ratio is 220/250 = .88
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23:58:52 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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23:59:29 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) (1,0.032) positive x axis asymptote ratio is .032/.04 = .8
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23:59:31 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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23:59:35 problem 3. y = f(x) = 5 (1.27^x).
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00:00:37 What is the ratio between the y values at x = 0 and at x = 1?
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RESPONSE --> f(0) = 5*1.27^0 = 5 f(1) = 5*1.27^1 = 6.35 ratio 6.35/5 = 1.27
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00:00:39 ** f(1) / f(0) = 5 * 1.27^1 / ( 5 * 1.27^0) = 1.27 **
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00:01:31 What is the ratio between the y values at to x = 3.4 and x = 4.4?
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RESPONSE --> f(3.4) = 5(1.27^3.4) = 11.2694483 f(4.4) = 5(1.27^4.4) = 14.31219934 ratio 14.31219934/11.2694483 = 1.27
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00:01:33 ** f(4.4) / f(3.4) = 5 * 1.27^4.4 / ( 5 * 1.27 ^3.4) = 1.27. **
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00:02:58 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 --> f(5) = 5(1.27^5) = 16.5191847 f(6) = 5(1.27^4) = 20.97936457 ratio 20.97936457/16.5191847 = 1.27 once again
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00:03:01 ** STUDENT RESPONSE: My points were 4.5 and 5.5, and the y ratio was again 1.27 **
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00:04:54 What is the ratio of y values when x values are separated by two units?
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RESPONSE --> f(2) = 5(1.27^2) = 8.0645 f(4) = 5(1.27^4) = 13.00723205 ratio is 13.00723205/8.0645 = 1.613
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00:04:57 ** 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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00:06:32 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 of y values at x = x1 and x = x1+1 is b, which is derived from Ab^(x1+1)/Ab^x1 = b^(x1+1-x1) = b^1 = b. Since the ratio is simply b, there is no dependence on the x1 value.
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00:06:36 ** 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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00:08:25 problem 5. y = 3 (2 ^ (.3 x) ). What is the ratio of the two basic-point y values?
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RESPONSE --> y =f(x)= 3 (2 ^ (.3 x) ) f(0) = 3(2^(.3*0)) = 3(2^0) = 3*1 = 3 f(1) = 3(2^(.3*1)) = 3(2^.3) = 3.69343324 ratio is 3.69343324/3 = 1.231 approx
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00:08:29 ** 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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00:08:43 What is the y = A b^x form of this function?
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RESPONSE --> y=3(1.231^x)
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00:08:46 ** 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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00:09:54 What does the value of 2 ^ .3 have to do with this situation?
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RESPONSE --> the value of 2^.3 ends up being the b value out of the form y=Ab^x
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00:09:58 ** CORRECT STUDENT RESPONSE: this is the b value in the form y = A b^x. **
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00:14:59 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 --> P(n+1) = (1+r) P(n), with r = .1 and P(0) = $1000 P(0+1) = (1+.1)P(0) so p(1) = (1.1)(1000) = $1100 P(1+1) = (1+.1)P(1) so p(2) = (1.1)(1100) = $1210 P(2+1) = (1+.1)P(2) so p(3) = (1.1)(1210) = $1331 P(3+1) = (1+.1)P(3) so p(4) = (1.1)(1331) = $1464.10 P(4+1) = (1+.1)P(4) so p(5) = (1.1)(1464.10) = $1610.51
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00:15:03 ** 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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00:17:54 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 --> Q(n+1) = .85 Q(n), Q(0) = 400 Q(0+1) = .85Q(0), so Q(1) = .85(400) = 340 Q(1+1) = .85Q(1), so Q(2) = .85(340) = 289 Q(2+1) = .85Q(2), so Q(3) = .85(289) = 245.65 Q(3+1) = .85Q(3), so Q(4) = .85(245.65) = 208.8025
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00:17:56 ** 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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00:18:27 What is the growth rate for this equation?
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RESPONSE --> the growth rate is -.15, which is obtained by .85-1 (the growth factor is 1+growth rate)
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00:18:29 ** The growth factor is .85, which is 1 + growth rate. It follows that the growth rate is .85 - 1 = .15 **
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00:19:33 problem 9. interest rate 12%, initial principle $2000. What is your difference equation?
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RESPONSE --> you need the growth factor to make a difference equation, which is found by 1+.12 = 1.12 so P(n+1) = (1.12)P(n), with P(0) = 2000
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00:19:35 ** 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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00:22:00 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 --> P(0+1)=(1.12)2000 P(0+1)=(1.12)2000, so P(1) = 2240 P(1+1)=(1.12)2240, so P(2) = 2508.80 P(2+1)=(1.12)2508.80, so P(3) = 2809.86 P(3+1)=(1.12)2809.86, so P(4) = 3147.04
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00:22:04 ** 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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00:22:40 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 --> To fall 1/8 of the original value would take about 68.26 min.
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00:22:47 ** 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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00:24:21 What are your ratios of temperature excess to average rate, and are they nearly constant?
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RESPONSE --> first segment 50/23.35 = 2.14 second segment 23.35/10.90 = 2.14 third segment 10.90/5.09 = 2.14 fourth segment 5.09/2.38 = 2.14 these are the y values divided.
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00:25:48 ** 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 --> Since the difference in t values change, the resulting ratio would be different, but as long as it is the same throughout, or have I gone about the wrong process of evaluating the problem.
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00:27:16 What are your estimates of the times required to fall to half of the three values?
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RESPONSE --> 25 = 50(.97^t) => about 22.76 min 12.5 = 50(.97^t) => about 45.51 min 6.25 = 50(.97^t) => about 68.26 min Time is approx. 22.757 min constantly.
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00:27:20 ** STUDENT RESPONSE: The temperature falls to 50/2 = 25 at t = 22.75. The temperature falls to 25/2 = 12.5 at t = 45.51 The temperature falls to 12.5/2 = 6.25 at t = 68.26. The time interval required for each subsequent fall is very close to 22.75, demonstrating that the half-life is constant. **
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00:27:52 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 --> 25=50(.97^t); divide both sides by 50 .5=.97^t
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00:27:55 ** 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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00:28:12 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 --> temp(t) = 50(.97^t)+25
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00:28:17 ** 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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00:29:00 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)+25 A = 50 b = .97 c = 25
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00:29:03 ** 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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00:31:00 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 --> (200,40) y=kx 40=200k k=40/200 k=.2, so y= .2 x y=.2*70=14mg/hr.
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00:31:06 ** 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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00:31:10 Add comments on any surprises or insights you experienced as a result of this assignment.
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