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 College Algebra, Trigonometry, and Precalculus Comments, Corrections and Clarifications It is recommended that the student work the 30-35 odd-numbered problems per section up to the "real-life" problems. From that point the student would choose two or three from the odd-numbered problems. The exceptions are those sections in which the major emphasis in the section is word problems, in which case more of the word problems should be worked. At this level of math the author of the solutions guide will often show fewer steps in the solution than would have been the case in the earlier grades. Do not take that as a suggestion to skip steps. You might even regard problems in the solutions guide as outlines for finding solutions rather than exactly what the student should know. Also realize that, at this level, there are often many correct ways to work problems so it is very possible that a solution that is different from the one illustrated in the solutions guide is also just as correct. College Algebra Corrections and Clarifications None Trigonometry, 7th edition(textbook copyright 2007) Since August 2008 withChalk Dust Company Corrections and Clarifications P.8 Transformations of Functions. The Trig 6th edition textbook and DVD #1 reflect P.8 as "Shifting, Reflecting, and Stretching Graphs" and this is identical to "Transformations and Functions." Trigonometry, 6th edition Corrections and Clarifications Section 1.2, #57. The problem in the solutions guide is from a different problem that appeared in the previous edition of the textbook. Please omit the problem. Section 1.4, problem 95. The solution in the solutions guide does not correspond with the problem in the textbook. To solve the problem, replace t with -0.7 in the formula and evaluate the expression on the right side to find the current to be I=0.794 amperes. [Be sure the calculator is in radian mode]. Section 1.4, problem 96. The solution in the solutions guide does not correspond with the problem in the textbook. To solve the problem... In the right triangle illustration, notice that the length of the opposite side is given (6 miles) and the hypotenuse is the distance we want to calculate. The trig ratio involving the opposite side and the hypotenuse is sine. sin theta=opposite/hypotenuse sin theta=6/d Since all of the questions ask for the value of d, the distance to the airplane, it might be helpful to solve for d... d=6/sin theta Plug in the given values of theta to find the distances. (a) When theta=30... [Make sure the calculator is in degree mode]. d=6/sin 30 degrees d=12 miles Section 1.6, problem 9. In the solutions guide, the graph is incorrect. The period should be pi/3 rather than pi.