IAT SOLUTIONS - C_7.pdf

Chapter 7, continuedCopyright © by McDougal Littell, a division of Houghton Mifflin Company.25661. When h 5 200: h(t) 5 }1 1 13e 20.65t256200 5 }1 1 13e 20.65t1 1 13e 20.65t 5 256 }20013e 20.65t 5 0.28e 20.65t 5 0.28 }13ln e 20.65t ø ln 1 0.28 }13 220.65t 5 ln 1 0.28 }13 220.65t 5 23.84t ø 5.9It takes about 6 weeks for the seedling to reach a heightof 200 centimeters.62. 3x 2 y 5 7 3x 2 y 5 7x 1 2y 5 14 3 23 23x 2 6y 5 2423x 2 5 5 7 → x 5 4The solution is (4, 5).27y 5 235y 5 563. 5x 2 y 5 7 3 5 25x 2 5y 5 352x 1 5y 5 23 2x 1 5y 5 2351 }32 272 2 y 5 7 → y 5 2 }29 27The solution is 1 }32 27 , 2 }29 272 .27x 5 32x 5 32 }2764. x 1 4y 5 26 3 2 2x 1 8y 5 21222x 1 y 5 12 22x 1 y 5 12x 1 4(0) 5 26 → x 5 26The solution is (26, 0).9y 5 0y 5 065. f(x) 5 x 3 2 2x 2 1 5; 2 or 0 positive real zerosf(2x) 5 2x 3 2 2x 2 1 5; 1 negative real zeroPositivereal zerosNegativereal zerosImaginaryzeros2 1 00 1 266. f(x) 5 x 4 1 6x 3 2 x 2 1 7x 2 8;3 or 1 positive real zerosf(2x) 5 x 4 2 6x 3 2 x 2 2 7x 2 8; 1 negative real zeroPositivereal zerosNegativereal zerosImaginaryzeros3 1 01 1 267. f(x) 5 x 5 2 3x 3 1 7x 2 1 6x 1 9;2 or 0 positve real zerosf(2x) 5 2x 5 1 x 3 1 7x 2 2 6x 1 9;3 or 1 negative real zerosPositivereal zerosNegativereal zerosImaginaryzeros2 3 02 1 20 3 20 1 468. f(x) 5 x 7 1 10x 6 2 5x 4 1 12x 3 2 17;3 or 1 positive real zerosf(2x) 5 2x 7 1 10x 6 2 5x 4 2 12x 3 2 17;2 or 0 negative real zerosPositivereal zerosNegativereal zerosImaginaryzeros3 2 23 0 41 2 41 0 669. f(1) f(2) f(3) f(4) f(5) f(6)19 28 27 16 25 2369 21 211 221 231210 210 210 210f(x) 5 ax 2 1 bx 1 c(1, 19): a 1 b 1 c 5 19(2, 28): 4a 1 2b 1 c 5 28(3, 27): 9a 1 3b 1 c 5 27Using a calculator, the solution is a 5 25, b 5 24,and c 5 0. So, a function that fits the data isf(x) 5 25x 2 1 24x.Algebra 2Worked-Out Solution Key447