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Design and development The Fokker CX was originally designed for the Royal Dutch East Indies Army, in order to replace the Fokker CVLike all Fokker aircraft of that time, it was of mixed construction, with wooden wing structures and a welded steel tube frame covered with aluminium plates at the front of the aircraft and with fabric at the rear. Solution Any diagonal n n matrix looks like 0 BBB BBB BBB B@ a1 an 1 CCC CCC CCC CA = a1E11 anEnn where Eii is the matrix with entries all 0 except a 1 at the i’th diagonal entry This tells us that (E11;. La E , I a a U O a E T a F a , R N A ca Ma F , W V a U Sa a H a E.
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See also the proof of e u du = e u PROOF 2 You need not memorize this theorem Derive it each time you use it Consider this example if you have the integral 2 x dx There is no need to memorize the formula We will get this integral into the easier form, e u du Recall that e ln(2) = 2 2 x dx = ( e ln (2)) x dx = e ln (2) x dx set u = ln(2) x. DYaåchgr cV gXdä \^cr ^ g`V\h X ^ibac^^ «U chdh madX`, `V`^b Wqa fVcrn U cVim^agå kdZ^hr X @ik, shd edacdghrä ^ cVXgYZV ^bc^ad bdä \^cr!» $ 9 E8 GHED 6 7 М ое мД ух гл б ина др ст , чтобы Духом Моим ты мог приобретать знание, говорит Дух Благодати. # F « ª e O \ K % 1 !.

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For positive m and nonnegative kThe theorem is Every nonnegative integer is the sum of m2 polygonal numbers of order m2 When m=0, u 0 (k)=kAnd with nonnegative k we trivially have f(NxN)=N when f(x,y)=xyEven simpler is f(N)=N when f(x)=x But the open question is for inputs both positive and negative. Â 5 Ê ³ % é W ÿ ó ê ° ¥ v 4 ´ k g 0 › é 6 W < § * ‘ z € m P Ý « G Ä š › ä < A ` Ñ v ˜ ç “ Ê ¡ Q !. Oh, so what they want you to do is as follows a For f(g(x) you take the f(x) equation, except, instead of the "x"s, you put in the "g(x)" equation.

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Solution Any diagonal n n matrix looks like 0 BBB BBB BBB B@ a1 an 1 CCC CCC CCC CA = a1E11 anEnn where Eii is the matrix with entries all 0 except a 1 at the i’th diagonal entry This tells us that (E11;. SOLUTION 13 Begin with x 2 xy y 2 = 1 Differentiate both sides of the equation, getting D ( x 2 xy y 2) = D ( 1 ) , 2x ( xy' (1)y) 2 y y' = 0 , so that (Now solve for y' ) xy' 2 y y' = 2x y, (Factor out y' ) y' x 2y = 2 x y, and the first derivative as a function of x and y is (Equation 1). ( 6 ) u ~tjfs q f¶6 g f ;.

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