MATH133_U5_answer - The vertical asymptote is x=-5 X...

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NAME: Jen Harper____________________________ Student Answer form Unit 5; show all work for full credit; trial and error (guess and check) methods are not appropriate; use methods covered in class. 1. A . answer 147.56 when working with this equation and messing with both sides the new formula looks like 0.05t=ln*1600 which then could be t=1600/0.05=147.56 B answer is 5.02 ln(4x)=3 4x=e^3 (e^3)/4=5.02 C answer -4 Log2(8-6x)=5 8-6x=2^5=32 6x=8-32=-24 -24/6=-4 D I could not find a solution for this equation. . 2. A on the graph it looks like it moved to the left about 5 units
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Unformatted text preview: The vertical asymptote is x=-5 X intercept in the form of (x,y) =(-4,0) B .it then reflects around the y axis Vertical asymptote (x=0) Y intercept (-1,0) 3. a.68% was the score s(0)=68-20log(oh)=68% B at 4mths the average score is 54.02 then at 24mths it is 40.04 S(4)=68-20log(4+1)=54.02% S(24)=68-20log(24+1)=40.04% C 6 ½ -7mths 50=68-20log(t+1) Log(t+1)=(68-50)/20=0.9 T+1=10^0.9=7.94 So roughly 7 mths? 4. A .$2938.66 A=2000[1+0.08]^5=2938.66 b. C $2983.66 A=2000*2.7183^(0.08*5)=2983.66...
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This note was uploaded on 08/30/2011 for the course ECON 103 taught by Professor Froyd during the Spring '11 term at American InterContinental University.

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MATH133_U5_answer - The vertical asymptote is x=-5 X...

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