Victory for the ACT Student Text 15e

458 • C AMBRIDGE P RACTICE T EST R EINFORCEMENT

26. (F) (p. 384) Mathematics/Statistics and Probability/Sets . .” 27. (B) (p. 384) Mathematics/Algebra and Functions/Manipulating Algebraic Expressions/Basic Algebraic Manipulations . Compare all the answer choices to the original using cross-multiplication. The original is & b a t r at br . For all conditional statements, the original statement and its , then not p contrapositive are equivalent, “if p , then q ” is equivalent to “if not q

A. r a B. t a

t b at br r b ar bt

9

8

C. D.

b a b +

t r t +

=

( t a b b r t at bt br bt at br + = + + = + = ) ( )

9

a b r t at br E. at br

9 9

28. (K) (p. 384) Mathematics/Algebra and Functions/Solving Algebraic Equations or Inequalities with One Variable/ Simple Equations . Simply solve the given equation for x : ( ) x x x x x x x 2 5 2 5 2 2 5 2 3 2 5 3 4 8 4 2 − + = − − = − + = − + = − + = = 29. (E) (p. 385) Mathematics/Geometry/Trigonometry/Trigonometric Relationships . Use the trigonometric relationships to evaluate the given expression. First, multiply the binomials in the numerator and in the denominator using the FOIL (First, Inner, Outer, Last) method: = ( sin ) ( sin ) sin sin sin x x x x x 1 1 1 2 + −

cos cos cos − + − − + −

+

−

x

x

x

x

x

2

( cos ) ( cos ) 1 1

1

= − Using the identity sin cos x x 2 2 + sin cos x x 1 + = 2 2 = − Therefore, by substitution: cot x = = cos sin sin cos x x x x 1 1 2 2 2 2 2 − − cos sin 1 1 2 2 − sin cos x x 1 = − cos sin x x 1 2 2 2 2

x

x

, we have:

=

1