See the next attachment, not this one thanks
enigma_1 ~~~~ Miss Cricket ~~~~ Joined Feb 27, 2013 Messages 4,282 Location Lords Gender Female HSC 2014 Dec 19, 2013 #1 See the next attachment, not this one thanks Attachments question.png 45.2 KB Views: 55 Last edited: Dec 19, 2013
A alficio New Member Joined May 9, 2013 Messages 15 Gender Undisclosed HSC N/A Dec 19, 2013 #2 Integrate both sides to get..... dp/dt =-30(t^-4) Integrating both sides p= (-30t^-3)/-3 + c p=10t^-3 + c subs in t=2, p=5.25 therefore c=4 Just subs in t=5 for five days Look at the 10t^-3 for the last part Just wondering what textbook is this from?
Integrate both sides to get..... dp/dt =-30(t^-4) Integrating both sides p= (-30t^-3)/-3 + c p=10t^-3 + c subs in t=2, p=5.25 therefore c=4 Just subs in t=5 for five days Look at the 10t^-3 for the last part Just wondering what textbook is this from?
