Perform the indicated operation.60.5 m - 2.11 dam (in meters)
A. 58.39 m
B. 60.289 m
C. 602.89 m
D. 39.4 m
Answer: D
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Find all the first order partial derivatives for the following function.f(x, y) = sin2 (4xy2 - y)
A. fx(x, y) = 2 sin (4xy2 - y) cos (4xy2 - y); fy(x, y) = 2 sin(4xy2 - y) cos(4xy2 - y) B. fx(x, y) = 2 sin (4xy2 - y) cos (4xy2 - y); fy(x, y) = (16x - 2) sin (4xy2 - y) cos (4xy2 - y) C. fx(x, y) = 8y2 sin (4xy2 - y) cos (4xy2 - y); fy(x, y) = 2 sin (4xy2 - y) cos (4xy2 - y) D. fx(x, y) = 8y2 sin (4xy2 - y) cos (4xy2 - y); fy(x, y) = (16xy - 2) sin (4xy2 - y) cos (4xy2 - y)
Find log 3.5 × 10-2.
A. -1.46 B. -0.54 C. 0.54 D. 1.46
Solve the following quadratic application using the given graph. The profit that the vendor makes per day by selling x pretzels is given by the function 
(i) Find the number of pretzels that must be sold to maximize profit. (ii) What is the maximum profit? (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) 800 pretzels must be sold to maximize profit. (ii) The maximum profit is $80 million. (iii) Vertex: (800, 80) (iv) x-intercepts: (0, 155.05) and (0, 644.95). There is no profit when 155.05 and 644.95 pretzels are sold. B. (i) 80 pretzels must be sold to maximize profit. (ii) The maximum profit is $800 million. (iii) Vertex: (80, 800) (iv) x-intercepts: (0, 310.1) and (0, -644.95). There is no profit when 310.1 and -644.95 pretzels are sold. C. (i) 240 pretzels must be sold to maximize profit. (ii) The maximum profit is $400 million. (iii) Vertex: (240, 400) (iv) x-intercepts: (0, 310.1) and (0, -644.95). There is no profit when 310.1 and -644.95 pretzels are sold. D. (i) 400 pretzels must be sold to maximize profit. (ii) The maximum profit is $240 million. (iii) Vertex: (400, 240) (iv) x-intercepts: (0, 155.05) and (0, 644.95). There is no profit when 155.05 and 644.95 pretzels are sold.
Find the difference.-10.3 - (-6.9)
A. 3.4 B. 17.2 C. -3.4 D. -17.2