Aphrodite1228idn
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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

Sagot :

Kate200468idn
[tex](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.[/tex]

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