BoumaQuiz1Solutions

BoumaQuiz1Solutions - MATHEMATICS 23a, FALL 2007 Practice...

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Unformatted text preview: MATHEMATICS 23a, FALL 2007 Practice Quiz # 1 October 2007 Solutions by Julian Bouma For purposes of this quiz, treat the following approximations as exact: cos 37 o = sin 53 o = 4 5 , cos 53 o = sin 37 o = 3 5 1. The answer is (c). The angle α between vectors ~ v and ~ w is defined as cos α = ~ v · ~ w | ~ v || ~ w | 4 2 · 2- 1 is equal to the dot product 4 · 2 + 2 · (- 1) = 6, and the lengths of the vectors are √ 4 2 + 2 2 and p 2 2 + (- 1) 2 , making the expression for the angle arccos 6 √ 20 · √ 5 = arccos 3 5 = 53 o 2. The answer is (d). From the equations T [1] 3 = [2] 3 [1] 3 T [1] 3 = [1] 3 [1] 3 you can deduce that [T] is equal to [2] 3 [1] 3 [1] 3 [1] 3 . The value of ~ v that T~ v = [1] 3 is [2] 3 [2] 3 3. The answer is (d). All numbers in the set satisfy (2 n 2 > m 2 ) = (2 > m 2 /n 2 ) However, since there are an infinite amount of rational numbers that satisfy this property and thus there is no greatest element, the least upper bound must be a number outside the set but is still greater than any element in the set, namely √ 2. 4. (a) Original: ∃ T ∈ S such that ∀ A ∈ S,TA = A Negation: ∀ T ∈ S, ∃ A ∈ S such that TA 6 = A (b) The original given statement P is true. What it is saying is that there exists a 2 x 2 matrix such that for any 2 x 2 matrix, the...
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This note was uploaded on 05/29/2010 for the course MATH 23 taught by Professor Bamberg during the Spring '10 term at Aarhus Universitet.

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BoumaQuiz1Solutions - MATHEMATICS 23a, FALL 2007 Practice...

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