5 Steps to a 5 AP English Literature (5 Steps to a 5 on the by Barbara Murphy, Estelle Rankin

By Barbara Murphy, Estelle Rankin

Note: CD-ROM/DVD and different supplementary fabrics usually are not included.

A excellent PLAN for the appropriate SCORE

STEP 1
Set up your learn plan with 3 personalized learn schedules

STEP 2
Determine your readiness with an AP-style diagnostic exam

STEP 3
Develop the concepts that would provide the side on try day

STEP 4
Develop the information you want to ranking High

STEP 5
Build your self assurance with full-length perform exams

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Extra resources for 5 Steps to a 5 AP English Literature (5 Steps to a 5 on the Advanced Placement Examinations, 2014-2015 Edition, 5th Edition)

Sample text

LI. 12. L3. 14. L5. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 50. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. i . [i - ( i . (i 51. 1 INTRODUCTION If 2 is substituted in the equation 3x- 2 = 4 the two sides become equal. We say that x = 2 is a solution of the equation. The following operations are helpful for finding the solu­ tions of an equation: I. II. The same quantity may be added to or subtracted from both sides of an equation. Both sides of an equation may be multiplied (or divided) by the same non-zero quantity.

To explain the meaning of x , where a is an irrational number, is beyond the scope of this text. But it is worth noting that it is possible to use arbitrary real numbers as exponents, while pre­ serving all the laws of exponents. a EXAMPLE 4 16 8 1 / 2 2/3 =Vl6 = 4 . ( V5 V5 = 5 3 3 ^ , 2 . 2 5 = 5 1 / 2 1 / 3 2 = 4 9 o r g 2/3 . V? = V5I25 EXAMPLE 5 (2 xV) 2 1 / 2 -2 2 / V / 2 y 6 / 2 - 2x y 2 3 28. 1/4 16 does not equal 4. Beware: EXERCISE SET In fact, 16 1/4 2,3 Simplify each of the following: 1.

Thus, x=0 or x+ 7 = 0 , which yields the solutions x=0 (b) or x = -7 or x - 3 = 0, or x=3 . x - 5x+6 = 0 2 (x- 2)(x- 3) = 0. Thus, x- 2 = 0 which yields the solutions x=2 (c) x - 2x+l = 0 2 (x- l)(x- 1) = 0. Thus, x- 1 = 0 , which yields the solution x= 1 . EXAMPLE 3 S O L V E : 2y = 30- 4y 2 62. SOLUTION W e begin by bringing all terms to the left side 2 y + 4 y - 30 = 0. 2 Dividing by 2, we have y +2y-15 = 0 2 (y+5)(y-3) = 0. Thus, y+ 5 = 0 or y- 3 = 0 , or y=3. which yields the solutions y = -5 The solutions of a quadratic equation 2 ax + bx + c = 0 can always be obtained by the quadratic formula: (3) This formula is not "magic", but has a logical derivation from still another method of solving a quadratic equation called "completing the square".

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