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

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern