problemset2s - MAT 137Y 2008-2009 Winter Session, Solutions...

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MAT 137Y 2008-2009 Winter Session, Solutions to Problem Set 2 1 (a) (SHE 1.6 #78) Solving the system of equations, the lines intersect at the point ( - 2 23 , 38 23 ) . Since m 1 = 4 = tan θ 1 and m 2 = - 19 - tan θ 2 , we have tan α = - 23 1 - 76 = 23 75 . (c) (SHE 1.7 #52, #54) If f and g are odd, then ( f · g )( - x ) = f ( - x ) · g ( - x ) = - f ( x ) · ( - g ( x )) = ( f · g )( x ) , so fg must be even. If f is even and g are odd, then ( f · g )( - x ) = f ( - x ) · g ( - x ) = f ( x ) · ( - g ( x )) = - ( f · g )( x ) , so fg must be odd. (d) If f and g are odd, then ( f g )( - x ) = f ( g ( - x )) = f ( - g ( x )) = - f ( g ( x )) = - ( f g )( x ) , so ( f g ) is odd. If f is even and g is odd, then ( f g )( - x ) = f ( g ( - x )) = f ( - g ( x )) = f ( g ( x )) = ( f g )( x ) , so ( f g ) is even. 2 (i) Let a and b be the lengths of the sides and h be the length of the hypoteneuse. By the Pythagorean Theorem, it follows a 2 + b 2 = h 2 . Also, the area of the triangle can be computed two ways; giving
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problemset2s - MAT 137Y 2008-2009 Winter Session, Solutions...

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