Solve the equation and check your solution.
+ 
= 
A. a = - 
B. a = 8
C. a = -8
D. a = -4
Answer: C
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Solve the following quadratic application using the given graph. An arrow is fired into the air with an initial velocity of 160 feet per second. The formula 
models the arrow's height above the ground, y, in feet, x seconds after it was shot into the air. 
(i) When does the arrow reach its maximum height? (ii) What is the maximum height? (iii) Write the vertex as an ordered pair. (iii) Write the vertex as an ordered pair. (iv) Write the x-intercepts as ordered pairs. What are the real-world meanings of the intercepts?
A. (i) The arrow reaches its maximum height 6 seconds after it is fired into the air. (ii) The maximum height of the arrow is 720 feet. (iii) Vertex: (6, 720) (iv) x-intercepts: (0, 0) and (0, -20); The arrow is on the ground 0 seconds after being fired and -20 seconds after being fired. B. (i) The arrow reaches its maximum height 5 seconds after it is fired into the air. (ii) The maximum height of the arrow is 400 feet. (iii) Vertex: (5, 400) (iv) x-intercepts: (0, 0) and (0, 10); The arrow is on the ground 0 seconds after being fired and 10 seconds after being fired. C. (i) The arrow reaches its maximum height 6 seconds after it is fired into the air. (ii) The maximum height of the arrow is 1200 feet. (iii) Vertex: (6, 1200) (iv) x-intercepts: (0, 0) and (0, 9.69); The arrow is on the ground 0 seconds after being fired and 9.69 seconds after being fired. D. (i) The arrow reaches its maximum height 5 seconds after it is fired into the air. (ii) The maximum height of the arrow is 80 feet. (iii) Vertex: (5, 80) (iv) x-intercepts: (0, 0) and (0, -10); The arrow is on the ground 0 seconds after being fired and -10 seconds after being fired.
Solve the rational equation.
= 
A. -3 B. 3 C. 1 D. -1
Simplify.(3a)4
A. 12a4 B. 81a C. 81a4 D. 12a
Find the opposite of the number.-1
A. 0 B. does not exist C. -1 D. 1