tag:blogger.com,1999:blog-1000693476051168456.post8014414001187705870..comments2019-06-03T03:59:46.148-05:00Comments on JabbaMatheez 40S (Winter 2008): WORKING WITH A(N) HYPERBOLA - CONCLUSION [beware shameless advertising.]Darren Kuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-1000693476051168456.post-25420312067925554512008-05-20T17:30:00.000-05:002008-05-20T17:30:00.000-05:00thank you for noticing.... was really rushing and ...thank you for noticing.... was really rushing and i'm lazy to change it LOL but it's always good to learn.:. J + ME .:.https://www.blogger.com/profile/09320523442838223212noreply@blogger.comtag:blogger.com,1999:blog-1000693476051168456.post-17232320119750415712008-05-19T23:53:00.000-05:002008-05-19T23:53:00.000-05:00I need to agree with that anonymous. Hyperbolas do...I need to agree with that anonymous. Hyperbolas don't have horizontal asymptotes. Good job for having a keen eye and keep up the good work.<BR/><BR/>-m@rkm@rkhttps://www.blogger.com/profile/00509544958003356512noreply@blogger.comtag:blogger.com,1999:blog-1000693476051168456.post-59096540804048366852008-05-18T18:32:00.000-05:002008-05-18T18:32:00.000-05:00On slide 7, shouldn't the equations of the asympto...On slide 7, shouldn't the equations of the asymptotes be:<BR/><BR/>y = ±5x/3<BR/><BR/>(Remember the y = mx + b form? In this case b = 0.)<BR/><BR/>not...<BR/><BR/>y = ±5/3 <-- because this would result in a horizontal line.<BR/><BR/><BR/>&&Vote ZEPH for V.P. =)Anonymousnoreply@blogger.com