Victory for the ACT Student Text 15e
V ICTORY P RACTICE T EST A NSWERS AND E XPLANATIONS • 455 The shaded circle at ͷ indicates or , without the equal sign (the symbol indicates 15. (E) (p. 381) Mathematics/Number and Quantity/Basic Arithmetic Manipulations . symbol indicates “greater than”). The graph indicates x ͷ # or x 3 ! . The answers are connected by “or,” meaning the union of those two sets of numbers. Therefore, the answer is x ͷ # or x 3 ! . Note that x ͷ # can be rewritten as x ͷ $ . 16. (H) (p. 381) Mathematics/Number and Quantity/Basic Arithmetic Manipulations . First, convert the mixed numbers to improper fractions: 2 5 2 1 8 7 5 12 8 15 − = − ͶͲǡ ǣ 5 12 8 15 40 96 40 75 40 21 − = − = either $ “less than,” and the ! or # , and the unshaded circle at 3 indicates either !
Use the equation for circumference of a circle:
17. (D) (p. 382) Mathematics/Geometry/Circles .
r
r
C
2
circle
r
2 (6)
12 r Therefore, 3 r 360 90 c c . 18. (J) (p. 382) Mathematics/Algebra and Functions/Solving Algebraic Equations or Inequalities with One Variable/ Equations Involving Rational Expressions and Equations Involving Radical Expressions . Solve the equation for x : 2 3 2 is 12 3 4 1 r r of the circumference, which, in degrees, is ( ) 4 1
x
x
x x x x 3 2
2 ] ^ g
^ h h
2
3 2 3 2
]
g
2 9 2 18
2
9
x
3 ! Therefore, the positive value of x is 3. 19. (B) (p. 382) Mathematics/Algebra and Functions/Expressing and Evaluating Algebraic Functions/Function Notation . To determine the value of ( ( )) g f 2 ǡ ϐ ʹ x in ( ) f x and evaluate: ( )
( ) ( ) ( ) f x x x f 2 2 2 2 2 4 4 0 2 2 = − = − = − = Now, substitute this value for x ( ) 4 0 0 4 4 = − = − = − ( ) g x x g
in ( ) g x
and evaluate:
20. (F) (p. 382) Mathematics/Algebra and Functions/Solving Algebraic Equations or Inequalities with One Variable/ Equations Involving Absolute Value . Since x 5 2 7 − = , we know that x 5 2 7 − = or ( ) x 5 2 7 − − = . Solve each for possible values of x : x x x 5 2 7 2 2 1 − = − = = −
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