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| rev | Admin | (Created page with "'''Solution: E''' <math>100=a(0)=M .110=a(1)=K+L+M</math> so <math>K+L=10.136=a(2)=4 K+2 L+M</math> so <math>18=2 K+L</math> so <math>K=8</math> and <math>L=2</math>. Thus <math>a(t)=8 t^2+2 t+100</math>. So <math>\delta(t)=\frac{a^{\prime}(t)}{a(t)}=\frac{16 t+2}{8 t^2+2 t+100}</math> so <math>\delta(.5)=\frac{10}{103}=.097</math>. '''References''' {{cite web |url=https://web2.uwindsor.ca/math/hlynka/392oldtests.html |last=Hlynka |first=Myron |website=web2.uwindsor.c...") | Nov 26'23 at 17:43 | +578 |