Given the function \(T(z) = z - 6\), find \(T(-4)\).
A. 10
B. -10
C. 2
D. -2
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What is the range of the function: \(\{(1, 2), (2, 4), (3, 6), (4, 8)\}\)?
A. \(\{2, 4, 6, 8\}\)
B. \(\{1, 2, 3, 4\}\)
C. \(\{6, 8\}\)
D. \(\{1, 2, 3, 4, 6, 8\}\)
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What is the domain of the function: \(\{(1, 3), (3, 5), (5, 7), (7, 9)\}\)?
A. \(\{3, 5, 7, 9\}\)
B. \(\{1, 3, 5, 7\}\)
C. \(\{1, 9\}\)
D. \(\{1, 3, 5, 7, 9\}\)
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Suppose \(p\) varies directly as \(d\), and \(p = 2\) when \(d = 7\). What is the value of \(d\) when \(p = 10\)?
A. \(d = \frac{20}{7}\)
B. \(d = 15\)
C. \(d = \frac{7}{5}\)
D. \(d = 35\)
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The number of calories burned, \(C\), varies directly with the time spent exercising, \(t\). When Lila bikes for 3 hours, she burns 900 calories. Which of the following equations shows this direct linear variation?
A. \(C = 300t\)
B. \(C = t\)
C. \(C = 3t\)
D. \(C = 900t\)
Answer (2)
( 1 ) T ( z ) = z − 6 ⇒ T ( − 4 ) = − 4 − 6 = − 10 ⇒ \ A n s . B . ( 2 ) r an g e : Y = { 2 ; 4 ; 6 ; 8 ; } ⇒ A n s . A . ( 3 ) d o main : D = { 1 ; 3 ; 5 ; 7 } ⇒ A n s . B . ( 4 ) d p = co n s t an t 7 2 = d 10 ⇔ 2 d = 7 ⋅ 10 ⇔ d = 2 7 ⋅ 2 ⋅ 5 = 35 ⇒ \ A n s . D .
( 5 ) 900 c a l or i es → 3 h o u rs x → 1 h o u r x = 3 900 c a l or i es = 300 c a l or i es C = 300 ⋅ t ⇒ A n s . A .
The answers to the questions are: 1. B. -10, 2. A. {2, 4, 6, 8}, 3. B. {1, 3, 5, 7}, 4. D. d = 35, 5. A. C = 300t.
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