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X y G h b c a T 1 750 g N b T 2 1250 g N c C x N C y 00 g N and M 625 g N m from MEC 4 at MARA University of Technology. You can put this solution on YOUR website!. Step 2 apply second derivative test f xx=6xf yy= −6yf xy= −2 At (0;0), f xx=0,f yy=0,f xy= −2So D= f xxf yy−(f xy)2 = −4.

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9 * 7 ) 6 1 H. Y′ p(t) y = g(t) What we will do is to multiply the equation through by a suitably chosen function µ(t), such that the resulting equation µ(t) y′ µ(t)p(t) y = µ(t)g(t) (*) would have integrateable expressions on both sides Such a function µ(t) is called an integrating factor. µ/ýX¬ Ì /propsplist 0 0ustar00roo > architecture x86_64 >=0.

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You can put this solution on YOUR website!. I j h n _ k k b h g Z e v g u c i Z d _ l,. 3 Now, we just need to solve for g(6) by plugging it into the equation for g(t) Substitute every value of 't' for 6.

(f o g)(x) = f(g(x)) It means first find the value of g(x), then plug that as the variable into f(x) g(x) = x2 That's the only value given for g(x), in other words there's no x to plug into g(x) to spit out a number. CR!XJFG0QA3ZDHQRCG5GZ2SR}„LR}„‹BOOKMOBI ™Í ¸*” 0J 8k @ H® Pý Y aX i« qÍ z ‚/ ‰ô ’ š" £ ªg"²µ$ºï&Â¸(Ë *Ò«,ÚîãA0ë. >& g* >' È Û ª Ý#Ø;\*×'¼ b v 8 _ x 8 z b m4 5 g _ x 8 z >& ç ô º>5 v>6 ¥>' kwwsv zzz pkoz jr ms frqwhqw sgi.

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