2. State the domain and range of the relation. Is it a function?
x -1 -2 -3 -4 -5; Domain:{-1, -2, -3, -4, -5} and
y 9 7 3 -1 -8; Range:{-8, -1, 3, 7, 9}
it is one one relation and therefore a function.
4.
Evaluate for f(-2). Show your work.
f(x) = -5x + 4 or f(-2) = -5*-2 + 4 = 14
5.
Evaluate for h(1/3). Show your work.
h(x) = 2x2 - x or h(1/3) = 2*(1/3)^2 -(1/3) = (2/9) - (1/3) = -1/90
6.
Evaluate for g(1.5). Show your work.
g(x) = | x - 3.5| or g(1.5) = | 1.5 - 3.5| = 2.0
7. On graph paper, graph the function below. In words describe what you see. Use complete sentences.
f(x) = - |6x| + 2
For x<0 y = 6x+2 is a straight line with slope +6 and intercept 2; and
For x>0 y = -6x+2 is a straight line with slope -6 and intercept 2
Intercept 2 is the value of the function at x = 0
The graph is consisting of two rays may be called branches
i) one branch: ray with initial point (0,2) and the other point {(-1/3), 0} and,
ii) other branch: with initial point (0,2) and the other point {(1/3), 0}.
8.
The stretch of a rubber band varies directly with the weight it supports. If the weight of a 3 oz. stretches a certain rubber band 5 in., what weight will stretch a rubber band 15 in.? Show all work.
Required weight = 3*(15/5) = 9 oz
9.
The area of a square varies directly with the square's side length. If the area of a square with side length of 6.4 in. is 40.96 in.2, what is the area of a square with a side length of 7.96 in.?
A. Write a quadratic variation equation.
A = ks^2. we have 40.96^2 = k*6.4^2 we have = k = 40.96^2/{6.4)^2} = 1
or required area = 7.96^2 = 63.3616 in^2
B. Solve the equation. Make sure your final answer has a label.
10.
The tire pressure varies inversely with the volume of air in the tire. If the pressure is 40 psi when the volume is 105 in.3, find the pressure when the volume is 120 in.3.
Required Volume = 40*(105/120) = 35 psi