A) zeroes
P(n) = -250 n^2 + 2500n - 5250
Extract common factor:
P(n)= -250 (n^2 - 10n + 21)
Factor (find two numbers that sum -10 and its product is 21)
P(n) = -250(n - 3)(n - 7)
Zeroes ==> n - 3 = 0 or n -7 = 0
Then n = 3 and n = 7 are the zeros.
They rerpesent that if the promoter sells tickets at 3 or 7 dollars the profit is zero.
B) Maximum profit
Completion of squares
n^2 - 10n + 21 = n^2 - 10n + 25 - 4 = (n^2 - 10n+ 25) - 4 = (n - 5)^2 - 4
P(n) = - 250[(n-5)^2 -4] = -250(n-5)^2 + 1000
Maximum ==> - 250 (n - 5)^2 = 0 ==> n = 5 and P(5) = 1000
Maximum profit =1000 at n = 5
C) Axis of symmetry
Vertex = (h,k) when the equation is in the form A(n-h)^2 + k
Comparing A(n-h)^2 + k with - 250(n - 5)^2 + 1000
Vertex = (5, 1000) and the symmetry axis is n = 5.
Answer:
none of the above
Step-by-step explanation:
The grocer's revenue will be the product of the number of loaves sold (30-2x) and their price (2.50+0.50x).
Revenue will be positive for values of x between those that make these factors be zero. The number of loaves sold will be zero when ...
... 30 -2x = 0
... 15 -x = 0 . . . . . divide by 2
... x = 15 . . . . . . . add x
The price of each loaf will be zero when ...
... 2.50 +0.50x = 0
... 5 + x = 0 . . . . . . . multiply by 2
... x = -5 . . . . . . . . . . subtract 5
Revenue will be positive for any number of increases greater than -5 and less than 15.
_____
D is the best of the offered choices, but it is incorrect in detail. -5 is a number less than 15, but will give zero revenue.
(13 + 20) + 15 = 13 + (20 + 15)
Answer:
6c^5+4c−301, assuming that you meant 6c^5
Answer:
The quotient is 3x-8
Step-by-step explanation:
(6x^2-x-40)÷ (5 + 2x)
The above equation can be written as
(6x^2-x-40)÷ (2x + 5)
The division is shown in the figure below.
The quotient is 3x-8