By Francoise,Naber

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4 3 3 3 + 3 3 −1 + 4 , = , 2 2 2 2 = √ 3 2 3, 2 46 66. Chapter 1: Graphs, Functions, and Models √ √ √ − 5+ 5 2+ 7 , 2 2 67. = 0, √ 2+ 7 2 For the side with vertices (7, −6) and (12, 6): 7 + 12 −6 + 6 19 = , ,0 2 2 2 For the side with vertices (12, 6) and (0, 11): 17 12 + 0 6 + 11 , = 6, 2 2 2 For the side with vertices (0, 11) and (−5, −1): 5 0 + (−5) 11 + (−1) , = − ,5 2 2 2 For the quadrilateral whose vertices are the points found 7 above, one pair of opposite sides has endpoints 1, − , 2 19 17 5 , 0 and 6, , − , 5 .

For (−3, 1) and (2, −1): d= (−3 − 5)2 + (4 − 2)2 = d= (6 − + (1 − √ √ = 142 + 62 = 232 √ √ √ Since ( 116)2 + ( 116)2 = ( 232)2 , the points could be the vertices of a right triangle. d= 18 √ 18 The endpoints of the diagonals are (−3, 4), (5, 2) and (2, −1), (0, 7). We find the length of each. For (−4, 5) and (−8, −5): (−8))2 √ (0 − (−3))2 + (7 − 4)2 = d= + (5 − √ (−10)2 + 42 = 116 For (6, 1) and (−8, −5): 50 For (0, 7), (−3, 4): For (−4, 5) and (6, 1): (−4 − (2 − 5)2 + (−1 − 2)2 = d= 53. First we find the distance between each pair of points.

7 1. 6x − 15 = 45 6x = 60 (a + x)3 3 (a + x)2 (a + x)3/2 (a + x)2/3 √ = 4 (a + x)1/4 a+x = a/2 (2a3 b5/4 c1/7 )4 16a12 b5 c4/7 = (54a−2 b2/3 c6/5 )−1/3 54−1/3 a2/3 b−2/9 c−2/5 √ = 16 3 54a34/3 b47/9 c34/35 = (a11/2 )1/3 122. Rationalizing the denominator = x2 y = (xy)1/4 (x2 y)1/3 = (xy)3/12 (x2 y)4/12 = x3 y 3 x8 y 4 121. x2 1 − x2 − √ 2 1 − x2 √ x2 1 − x2 1 − x2 − 2(1 − x2 ) = = (xy)3 (x2 y)4 120. 1 + x2 · 128. √ √ 2 4 8 = 21/2 (23 )1/4 = 21/2 23/4 = 25/4 = √ √ 4 25 = 2 4 2 √ 4 1 1 + x2 √ 1 + x2 1 1 + x2 √ √ + · 2 2 1+x 1+x 1 + x2 √ √ (1 + x2 ) 1 + x2 1 + x2 = + 1 + x2 1 + x2 √ (2 + x2 ) 1 + x2 = 1 + x2 = = (62 23 )1/6 √ = 6 36 · 8 √ = 6 288 118.

### Encyclopedia Of Mathematical Physics. Contents list by subject. Contents by Francoise,Naber

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