course Mth 174 You asked me to retry to get the correct and answers - I have been doing it the way you said but it still didn't give me the answers. I'm confused as to what I'm doing wrong and it's making it hard to find my mistakes. Is there any other way to get the answers? úÕúŸèñ‹äÛÂÝÔvÌÀö¸Û±y·±assignment #003 ËWž‘ª˜ä¨ô¿“ÉJ€ìИZ¸è°Ç{ÿ`zdÕî’á Physics II 06-25-2007
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15:05:25 how can you symbolically represent the give statement?
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RESPONSE --> t = s/[(vf+v0)/2] dropped from rest v = a * t s = .5 a t^2
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15:06:33 How can we show that the statement is true?
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RESPONSE --> a ball dropped from rest will have velocity or v = a*t and position s = .5 a t^2
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15:07:21 How can we use a graph to show that the statement is true?
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RESPONSE --> Numbers can be plugged into the equations and make sure the graph follows the intended trend.
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15:08:09 what did you get for the integral and how did you reason out your result?
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RESPONSE --> u = cos(3t) du = -3sin(3t)dt -(1/3)du = sin(3t)dt int['sqrt(cos(3t) ) * sin(3t)] -(1/3)int['sqrt(u)]du -(1/3)*(2/3) u^(3/2) + c -(2/9) (cos(3t))^(3/2) + c
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15:08:24 query problem 7.1.20 (3d edition #21) antiderivative of x^2 e^(x^3+1)
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RESPONSE --> int[(x^2)(e^(x^(3)+1))dx u = x^3 + 1 du = 3x^2 dx (1/3)du = x^2 dx (1/3) int [e^u] du (1/3) e^u + c (1/3) e^(x^3 + 1) + c
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15:08:33 what is the antiderivative?
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RESPONSE --> int[(x^2)(e^(x^(3)+1))dx u = x^3 + 1 du = 3x^2 dx (1/3)du = x^2 dx (1/3) int [e^u] du (1/3) e^u + c (1/3) e^(x^3 + 1) + c
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15:08:42 What substitution would you use to find this antiderivative?
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RESPONSE --> u = x^3 + 1 du = 3x^2 dx (1/3)du = x^2 dx (1/3) int [e^u] du (1/3) e^u + c (1/3) e^(x^3 + 1) + c
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15:09:23 what is the antiderivative?
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RESPONSE --> I'm not sure what to use as u. I tried t+1 and t^2. when I used u= t^2 du = 2tdt (1/2)tdu = dt (1/2)t int[((t+1)^2)/u) du 1/2 t + ln (t^2) I got stuck here. This isn't right, but I can't get the correct answer.
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15:09:34 What substitution would you use to find this antiderivative?
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RESPONSE --> That's where i'm confused.
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15:09:48 query 7.1.64 (3d edition #60). int(1/(t+7)^2, t, 1, 3)
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RESPONSE -->
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15:09:54 What did you get for the definite integral?
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RESPONSE --> u = t + 7 du = dt int[u^-2]du 1/u 1/(t+7) (1, 3) 1/10 - 1/8 = -1/40
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15:10:01 What antiderivative did you use?
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RESPONSE --> int[u^-2]du 1/u 1/(t+7) (1, 3) 1/10 - 1/8 = -1/40
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15:10:16 What is the value of your antiderivative at t = 1 and at t = 3?
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RESPONSE --> 1/(t+7) (1, 3) 1/10 - 1/8 = -1/40
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15:10:55 What were the populations in 1990 and 2000?
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RESPONSE --> 1990 population is 5.3billion 2000 population is 6.1 billion
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15:11:02 What is the average population between during the 1990's and how did you find it?
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RESPONSE --> int 5.3e^(0.014t) (0, 10) int 5.3e^(0.014*0) = 5.3 billion int 5.3e^(0.014*10) = 6.1 billion 6.1 billion - 5.3 billion = .8 billion
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15:11:12 What is the value of your antiderivative at t = 1 and at t = 3?
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RESPONSE --> 0.8 billion work on last problem.
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15:13:14 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I reran this program as you asked me to and after entering my response I clicked ""enter response"" then ""next question/answer"" and I did not recieve the answers yet again. I haven't gotten them and i have been doing it this way and have not been getting answers. It's making it difficult to know how to get the right answers when I don't. do you have any other suggestions. I""m going to do the next assignment and see if it works there.
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º{…‘x‚ò‡f↔²®ÏÑy€‡¤ÔÖß ˆÏ𩬠assignment #004 ËWž‘ª˜ä¨ô¿“ÉJ€ìИZ¸è°Ç{ÿ`zdÕî’á Physics II 06-25-2007
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15:22:20 what is the requested antiderivative?
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RESPONSE --> antiderivative of sin^2 x u = sin x v = -cos x u' = cos x v' = sin x -sinx cosx + int(cos^2 x) -sinx cosx + int(1-sin^2 x) -sinx cosx + int(1) - int (sin^2 x) 2 int (sin^2 x) dx = -sin x cos x + int(1) dx -1/2 sin x cos x + 1/2 x + C
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15:23:16 What substitution, breakdown into parts and/or other tricks did you use to obtain the antiderivative?
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RESPONSE --> antiderivative of sin^2 x I used u & v... u = sin x v = -cos x u' = cos x v' = sin x Then, cos^2 x = 1 - sin^2 x, so I plugged that in for the cos^2 x... -sinx cosx + int(cos^2 x) -sinx cosx + int(1-sin^2 x) -sinx cosx + int(1) - int (sin^2 x) 2 int (sin^2 x) dx = -sin x cos x + int(1) dx -1/2 sin x cos x + 1/2 x + C
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15:32:35 what is the requested antiderivative?
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RESPONSE --> (t+2) `sqrt(2+3t) let u = t +1 and v = 3/6 (2+3t)^(3/2) u'= 1 v'= (2+3t)^(1/2) int [(t+2) `sqrt(2+3t)] = 3/6 (t+1) (2+3t)^(3/2) - int[1* 3/6 (2+3t)^(3/2)] 3/6 (t+1) (2+3t)^(3/2) - (3/6) int[(2+3t)^(3/2)] add one to 3/2 = 5/2 divide by 5/2 giving 3/6 * 2/5 = 6/30 = 1/5 3/6 (t+1) (2+3t)^(3/2) - 1/5 (2+3t)^(5/2)
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15:32:45 What substitution, breakdown into parts and/or other tricks did you use to obtain the antiderivative?
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RESPONSE --> (t+2) `sqrt(2+3t) let u = t +1 and v = 3/6 (2+3t)^(3/2) u'= 1 v'= (2+3t)^(1/2) int [(t+2) `sqrt(2+3t)] = 3/6 (t+1) (2+3t)^(3/2) - int[1* 3/6 (2+3t)^(3/2)] 3/6 (t+1) (2+3t)^(3/2) - (3/6) int[(2+3t)^(3/2)] add one to 3/2 = 5/2 divide by 5/2 giving 3/6 * 2/5 = 6/30 = 1/5 3/6 (t+1) (2+3t)^(3/2) - 1/5 (2+3t)^(5/2)
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15:32:49 What substitution, breakdown into parts and/or other tricks did you use to obtain the antiderivative?
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RESPONSE -->
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15:35:17 query problem 7.2.27 antiderivative of x^5 cos(x^3)
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RESPONSE -->
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