R(x) = 60x - 0.2x^2
The revenue is maximum when the derivative of R(x) = 0.
dR(x)/dx = 60 - 0.4x = 0
0.4x = 60
x = 60/0.4 = 150
Therefore, maximum revenue is 60(150) - 0.2(150)^2 = 9000 - 4500 = $4,500
Maximum revenue is $4,500 and the number of units is 150 units
58 * 1,000 = 58,000 its multiplication
Answer:
A. (f + g)(1) = - 9
B. (f - g)(0) = -7
C. (fg)(3) = 0
Step-by-step explanation:
A. (f + g)(1) = f(1) + g(1) f(1) = 1^2 - 9 = 1 - 9 = - 8 g(1) = 1 - 2 = - 1
= -8 -1 = -9
B. (f - g)(0) = f(0) - g(0) f(0) = 0^2 - 9 = 0 - 9 = -9 g(0) = 0 - 2 = -2
= -9 + 2 = -7
C. (fg)(3) = f(3)(g(3) f(3) = 3^2 - 9 = 9 - 9 = 0 g(3) = 3 - 2 = 1
= 0(1) = 0
Use the formula or complete the square.
The zeroes of the quadratic can be real and rational; real and irrational; complex conjugates.
If the quadratic is ax²+bx+c, x=(-b+√b²-4ac)/2a.
If b² > 4ac the solutions are real. If b²-4ac is a perfect square, the solutions are real and rational; otherwise they’re real but irrational.
If b² < 4ac the solutions are complex.
