07pm1-solution

07pm1-solution - Math 34A Practice Midterm I Solutions Fall...

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Unformatted text preview: Math 34A Practice Midterm I Solutions Fall 2007 Problem 1 Solve for x and y the equations 3 x + 2 y = 3 a x + y = 2 a- b Since the coefficients of x and y are both 1 in the second equation, we can easily isolate one of these. Subtract x from both sides of the second equation to get y = 2 a- b- x Now, we can substitute this expression into the first equation; replace the 2 y with 2(2 a- b- x ): 3 x + 2(2 a- b- x ) = 3 a Distribute the 2 to get 3 x + 4 a- 2 b- 2 x = 3 a x + 4 a- 2 b = 3 a x + 4 a = 3 a + 2 b x = 2 b- a And now that we have x , we can find y : y = 2 a- b- x = 2 a- b- (2 b- a ) = 2 a- b- 2 b + a = 3 a- 3 b So x = 2 b- a and y = 3 a- 3 b . 1 Problem 2 Let: h be the height of the box, l be the length and width of the bottom, V be the volume of the box, A be the surface area of the box. The formula for the surface area and volume are V = l 2 h A = 4 hl + 2 l 2 . We are given that V = 20 m 3 so 20 m 3 = l 2 h h = 20 m 3 l 2 and therefore the total surface area is A = 4 hl + 2 l...
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07pm1-solution - Math 34A Practice Midterm I Solutions Fall...

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