course mth158
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22:53:27 5.3.38. Explain how you used transformations to graph f(x) = 2^(x+2)
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RESPONSE --> found the domain which is all real numbers the range which is (0,inf) the horizontal asymptote is y=0 there are no x-intercepts the y- intercept is 4 so we would start by graphing the basic line 2^x then shift to the left 2 units.
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22:54:09 The function y = 2 ^ ( x+2) is a transformation of the basic function y = 2^x, with x replaced by x - 2. So the graph moves 2 units in the x direction. The graph of y = 2^x is asymptotic to the negative x axis, increases at a rapidly increasing rate and passes through ( 0, 1 ) and (1, 2). The graph of y = 2^(x-2), being shifted in only the x direction, is also asymptotic to the negative x axis and passes through the points (0 + 2, 1) = (2, 1) and (1 + 2, 2) = (3, 2). The graph also increases at a rapidly increasing rate. All the points of the graph of y = 2^(x-2) lie 2 units to the right of points on the graph of y = 2^x.
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RESPONSE --> o.k.
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23:01:19 5.3.42. Transformations to graph f(x) = 1 3 * 2^x
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RESPONSE --> domain(-inf.inf) range (1,inf) horizantal asymptote y=1 using the basic line 2^x we would then multpliy -1 to reflect the x-axis stretch the gragh verticallyby a factor of 3 and shift up 1 unit.
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23:02:08 1 - 3 * 2^x is obtained from y = 2^x by first vertically stretching the graph by factor -3, then by shifting this graph 1 unit in the vertical direction. The graph of y = 2^x is asymptotic to the negative x axis, increases at a rapidly increasing rate and passes through ( 0, 1 ) and (1, 2). -3 * 2^x will approach zero as 2^x approaches zero, so the graph of y = -3 * 2^x will remain asymptotic to the negative x axis. The points (0, 1) and (1, 2) will be transformed to (0, -3* 1) = (0, -3) and (1, -3 * 2) = (1, -6). So the graph will decrease from its asymptote just below the x axis through the points (0, -3) and (1, -6), decreasing at a rapidly decreasing rate. To get the graph of 1 - 3 * 2^x the graph of -3 * 2^x will be vertically shifted 1 unit. This will raise the horizontal asymptote from the x axis (which is the line y = 0) to the line y = 1, and will also raise every other point by 1 unit. The points (0, -3) and (1, -6) will be transformed to (0, -3 + 1) = (0, -2) and (1, -6 + 5) = (1, -5), decreasing from its asymptote just below the line y = 1 through the points (0, -2) and (1, -5), decreasing at a rapidly decreasing rate.
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RESPONSE --> o.k.
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23:11:17 5.3.60 Solve (1/2)^(1-x) = 4.
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RESPONSE --> we start by getting the same base on both sides of the equation. 1/2^(1-x)=4 (2^-1)^(1-x)=2^2 1-x=2 -x=2-1 -x=1 x=1/-1 x=-1
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23:12:45 (1/2)^(-2) = 1 / (1/2)^2 = 1 / (1/4) = 4, so we know that the exponennt 1 - x must be -2. If 1 - x = -2, it follows that -x = -2 - 1 = -3 and x = 3.
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RESPONSE --> I see where i went wrong.
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23:40:15 5.3.78 A(n) = A0 e^(-.35 n), area of wound after n days. What is the area after 3 days and what is the area after 10 days? Init area is 100 mm^2.
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RESPONSE --> A(3)=100*e^-0.35n A(3)=100*e^-0.35(3)=34.994 A(10)=100*e^-0.35(10)=3.02
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23:41:23 If the initial area is 100 mm^2, then when n = 0 we have A(0) = A0 e^(-.35 * 0) = 100 mm^2. Since e^0 = 1 this tells us that A0 = 100 mm^2. So the function is A(n) = 100 mm^2 * e^(-.35 n). To get the area after 3 days we evaluate the function for n = 3, obtainind A(3) = 100 mm^2 * e^(-.35 * 3) = 100 mm^3 * .35 = 35 mm^2 approx.. After to days we find that the area is A(10) = 100 mm^2 * e^(-.35 * 10) 100 mm63 * .0302 = 3.02 mm^2 approx..
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RESPONSE --> I did exact on 3 days but i guess thats close enought.
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00:15:17 5.3.84. Poisson probability 4^x e^-4 / x!, probability that x people will arrive in the next minute. What is the probability that 5 will arrive in the next minute, and what is the probability that 8 will arrive in the next minute?
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RESPONSE --> 5 people in the next minute= 4^5e^-4/5=connot find a explaination of the x! and how to use it in this chapter.
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00:18:13 he probability that 5 will arrive in the next minute is P(5) = 4^5 * e^-4 / (5 !) = 1024 * .0183 / (5 * 4 * 3 * 2 * 1) = 1024 * .0183 / 120 = .152 approx..
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RESPONSE --> o.k. i see how it is done
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00:18:27 Te probability that 8 will arrive in the next minute is P(8) = 4^8 * e^-8 / (8 !) = .00055 approx..
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RESPONSE --> o.k
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ᕂʇƕ]v`G assignment #037 ؾ{]Uӂ͛ξ College Algebra 12-08-2005
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00:47:05 5.4.14. 2.2^3 = N. Express in logarithmic notation.
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RESPONSE --> 2.2^3 = N is the equivalant to 3=log2.2
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00:49:45 b^x = y is expressed in logarithmic notation as log{base b}(y) = x. In this case b = 2.2, x = 3 and y = N. So we write lob{base b}(y) = x as log{base 2.2}(N) = 3.
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RESPONSE --> I had it in the form that the book shows x=bym iwill do it your way from now on.
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00:56:10 5.4.18.. x^pi = 3. Express in logarithmic notation.
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RESPONSE --> you have x^pi = 3 which would be log(x)3=pi The book has x^pi=e which would be log(x)e=pi
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00:56:36 b^a = y is expressed in logarithmic notation as log{base b}(y) = a. In this case b = x, a = pi and y = 3. So we write lob{base b}(y) = a as log{base x}(3) = pi.
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RESPONSE --> o.k.
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00:59:58 5.4.26.. log{base 2}?? = x. Express in exponential notation.
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RESPONSE --> log{base 2}6 =x in exponet form would be 2^x=6
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01:00:12 log{base b}(y) = a is expressed in exponential notation as b^a = y. In this case b = 2, a = x and y = ?? so the expression b^a = y is written as 2^x = ??.
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RESPONSE --> o.k.
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01:11:41 5.4.36. Exact value of log{base 1/3}(9)
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RESPONSE --> log{base 1/3}(9) y=log(1/3)(9)=(1/3)^y=9 (1/3)^y=3^3 y=3
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01:12:28 log{base b}(y) = a is expressed in exponential notation as b^a = y. In this case b = 1/3, a is what we are trying to find and y = 9 so the expression b^a = y is written as (1/3)^a = 9. Noting that 9 is an integer power of 3 we expect that a will be an integer power of 1/3. Since 9 = 3^2 we might try (1/3)^2 = 9, but this doesn't work since (1/3)^2 = 1/9, not 9. We can correct this by using the -2 power, which is the reciprocal of the +2 power: (1/3)^-2 = 1/ ( 1/3)^2 = 1 / (1/9) = 9. So log base 1/3}(9) = -2.
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RESPONSE --> see where i went wrong
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11:28:01 . What is the domain of G(x) = log{base 1 / 2}(x^2-1)
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RESPONSE --> (-1,inf)
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11:29:35 For any positive value of b the domain of log{base b}(x) consists of all positive real numbers. It follows that log{base 1/2}(x^2-1) consists of all real numbers for which x^2 - 1 > 0. We solve x^2 - 1 > 0 for x by first finding the values of x for which x^2 - 1 = 0. We can factor the expression to get (x-1)(x+1) = 0, which is so if x-1 = 0 or if x + 1 = 0. Our solutions are therefore x = 1 and x = -1. It follows that x^2 - 1 is either positive or negative on each of the intervals (-infinity, -1), (-1, 1) and (1, infinity). We can determine which by substituting any value from each interval into x^2 - 1. On the interval (-1, 1) we can just choose x = 0, which substituted into x^2 - 1 gives us -1. We conclude that x^2 - 1 < 0 on this interval. On the interval (-infinity, -1) we could substitute x = -2, giving us x^2 - 1 = 4 - 1 = 3, which is > 0. So x^2 - 1 > 0 on (-infinity, -1). Substituting x = 2 to test the interval (1, infinity) we obtain the same result so that x^2 - 1 > 0 on (1, infinity). We conclude that the domain of this function is (-infinity, -1) U (1, infinity).
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RESPONSE --> o.k see where i went wrong
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12:55:36 5.4.62. a such that graph of log{base a}(x) contains (1/2, -4).
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RESPONSE --> f(x)=log{base a}(x)contains point (1/2,-4) f(1/2)=log(base a)(1/2)=a^(1/2)=1/2 a=1/2
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12:59:03 log{base a}(x) = y if a^y = x. The point (1/2, -4) will lie on the graph if y = -4 when x = 1/2, so we are looking for a value of a such that a ^ (-4) = 1/2. We easily solve for a by taking the -1/4 power of both sides, obtaining a = (1/2)^-(1/4) = 16.
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RESPONSE --> o.k. thats not what I got and when I raise 1/2^-(1/4) I get the answer 1.189 which is not 16 i do not understand how you arrived at your answer.
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13:05:40 . Transformations to graph h(x) = ln(4-x). Given domain, range, asymptotes.lineCount = lineCount + 1: bLine$(lineCount) = ""
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RESPONSE --> we first graph the natural logarithum function f(x)=In(x) then shift to the left 4 units.
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13:06:29 The graph of y = ln(x) is concave down, has a vertical asymptote at the negative y axilineCount = lineCount + 1: bLine$(lineCount) = s and passes through the points (1, 0) and (e, 1) (the latter is approximately (2.71828, 1) ). The graph of y = ln(x - 4), where x is replaced by x - 4, is shifted +4 units in the x direction so the vertical asymptote shifts 4 units right to the line x = 4. The points (1, 0) and (e, 1) shift to (1+4, 0) = (5, 0) and (e + 4, 1) (the latter being approximately (3.71828, 1). Since x - 4 = - (4 - x), the graph of y = ln(4 - x) is 'flipped' about the y axis relative to ln(x-4), so it the vertical asymptote becomes x = -4 and the graph passes through the points (-5, 0) and (-e-4, 1). The graph will still be concave down.
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RESPONSE --> o.k. i got confussed on which way to shift.
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13:10:03 5.4.102. Solve log{base 6}(36) = 5x + 3.
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RESPONSE --> log{base 6}(36) = 5x + 3. 6^(5x+3)=36 6^(5x+3)=6^2 5x+3=2 5x=-1 x=-1/5
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13:10:16 log{base b}(y) = a is expressed in exponential notation as b^a = y. In this case b = 6, a is what we are trying to find and y = 5x + 3 so the expression b^a = y is written as 6^(5x+3) = 36. We know that 6^2 = 36, so (5x + 3) = 2. We easily solve this equation to get x = -1 / 5.
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RESPONSE --> o.k.
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13:30:47 6.3.102. F(t) = 1 e^(-.15 t) prob of car arriving within t minute of 5:00. How long does it take for the probability to reach 50%? How long to reach 80%?
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RESPONSE --> you mean 5.4.116 that is the only question in this book I could find that was simuliar. f(.5)=1-e^(-.15t) -.5=-e^(-.15t)=>In(.5)=-.15t t=In(.5)/-.15 t=4.62 or aproximately 4 and 1/2 minutes that the probability will be 50% f(.8)=1-e^(-.15t) -.8=-e^(-.15t)=>In(.8)=-.15t t=In(.8)/ -.15 t=1.487 or aproximately 1and 1/2 minutes that the probability will be 80%.
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13:31:01 The probability is F(t), so F(t) = 50% = .5 when 1 - e^(-.15 t) = .5. We subtract 1 from both sides the multiply by -1 to rearrange this equation to the form e^(-.15 t) = .5. Taking the natural log of both sides we get ln( e^(-.15 t) ) = ln(.5), which tells us that -.15 t = ln(.5) so that t = -ln(.5)/.15 = 4.6, approx..
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RESPONSE --> o.k.
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