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that is, if .3: = —2.
12. ”6+1D'H2 = (13+1m  {5+tﬁ) (by hint)
=“ﬁF+HﬂF+2W4m[mmkﬁmMHﬁhmwﬂ
E Hi?“2 + ”ﬁll” + 2HﬁTHltﬁll {by CauchyHSCth} = (llﬁll + “one, so that
“5+ ﬁll? 5 [Iran + Illa")? Taking square roots of both sides, we ﬁnd that ”1? + 13“ 5 “if“ + tﬁt as claimed. 14. The horizontal oompouents of 131 and 1'32 are —[F‘.1siuﬁ and EF2 sing. reapeotively {the
horizontal component of F3 is zero}. Since the system is at rest, the horizontal compononts must add up to I], so that —  F1  sin 13+
t ~_ 4 . ﬂ _
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.g—é 5,5 ﬁgﬁ, so that. Leonardo was mistaken.
2 §‘ = 0.20 + 0315 + ll5c : 76,
o lilinhm‘ze quantit at"  at“ = 2 3
{swarm} tells us that (55. 3})? 5311531]ng o + b + c2. The CauchySc] 1 ’ i « 
o2 + .52 .f. C2 3, m? [[2, or T62 5 (a? Warz meliuélhty ___ ' + 62 + 03(1),? + I132 2
033‘ Quantity :12 + 62 + c2 is minimal when a2 + b” + 2 — $20.5 J {’1'
is the case who 1’ E 0.2%  c H m This
13 a, :.. : 0.3%; for some ' ' I
_ poﬁttl‘b'e constant t: It ' ' I
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. . it ~.:
: The Student mus { J 76, 90 that k = 200. Thus the first exam, 60 hours for 34. rref['A) = [ [.‘r 1 2 A basis of ker(A) is 51 = 10—1—2 3 DHMH 2 1 _ 1 W1! — $5
6* ﬁrme72 £1 _ 1 ‘ ifwzii — Ire—tapas M — m ...
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 Spring '08
 lee
 Math

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