A consultant advises the owners of a phone company that their profit p each month car-be modeled by
p(x) = -50x2 + 3500x - 2500
The variable x is the average cost that a customer is charged. What
range of costs will bring in a profit of at least $40,000?

$15.63 ≤ x ≤ $54.365

Step-by-step explanation:

Profit of the phone company is modeled by the equation,

p(x) = -50x² + 3500x - 2500

For the profit of at least $40000,

-50x² + 3500x - 2500 ≥ 40000

-50x² + 3500x ≥ 40000 + 2500

-50x² + 3500x ≥ 42500

-x² + 70x ≥ 850

x² - 70x + 850 ≤ 0

By quadratic formula,

x - intercept of the inequality will be,

x =

x =

x = 15.635, 54.365

Therefore, $15.635 ≤ x ≤ $54.365 will be the range of cost for which profit will be at least $40000.

b

step-by-step explanation:

hi there!

first we need to find out how much they are adding to each number to make the next number.

ex: 4 + 4 = 8, 8 + 8 = 16, and so on.

so continuing with 32,

32 + 32 = 64

64 + 64 = 128

128 + 128 = 256

256 + 256 = 512

4, 8, 16, 32, 64, 128, 256, 512

the 8th term of the sequence will be 512

hope this ! ~