Given the function T(z) = z – 6, find T(–4).
A. 10
B. –10
C. 2
D. –2

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}

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}

Suppose p varies directly as d, and p = 2 when d = 7. What is the value of d when p = 10?
A. d =20/7
B. d = 15
C. d =7/5
D. d = 35

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

$$(1)\\T(z)=z-6\ \ \ \Rightarrow\ \ \ T(-4)=-4-6=-10\ \ \ \Rightarrow\ \ \ Ans. \ B. \\\\(2)\\range:\ \ \ Y=\{2;\ 4;\ 6;\ 8;\}\ \ \ \Rightarrow\ \ \ Ans. \ A. \\\\(3)\\domain:\ \ \ D=\{1;\ 3;\ 5;\ 7\}\ \ \ \Rightarrow\ \ \ Ans. \ B. \\\\(4)\\ \frac{p}{d} =constant\\\\\frac{2}{7} =\frac{10}{d} \ \ \ \Leftrightarrow\ \ \ 2d=7\cdot10\ \ \ \Leftrightarrow\ \ \ d= \frac{7\cdot2\cdot5}{2} =35\ \ \ \Rightarrow\ \ \ Ans. \ D.$$

$$(5)\\900\ calories\ \rightarrow\ 3\ hours\\x\ \rightarrow\ \ 1\ hour\\\\x= \frac{900}{3} \ calories=300\ calories\\\\C=300\cdot t\ \ \ \Rightarrow\ \ \ Ans. \ A.$$

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