On a given workday, the rate, in tons per hour, at which unprocessed ore arrives at a refinery is modeled by O(t) = 20 + 10 cos (t^2/25), where t is measured in hours and 0 ≤ t ≤ 10. At the beginning of the workday, t = 0, the refinery has 100 tons of unprocessed ore. During the hours of operation, the refinery processes ore at a constant rate of 25 tons per hour. Determine the time at which the total rate of ore at the refinery is neither increasing nor decreasing? Justify your answer.

{x = -1/2 ,   y = -1

step-by-step explanation:

solve the following system:

{2 x + y = -2

{10 x - 3 y = -2

in the first equation, look to solve for y:

{2 x + y = -2

{10 x - 3 y = -2

subtract 2 x from both sides:

{y = -2 x - 2

{10 x - 3 y = -2

substitute y = -2 x - 2 into the second equation:

{y = -2 x - 2

{10 x - 3 (-2 x - 2) = -2

10 x - 3 (-2 x - 2) = (6 x + 6) + 10 x = 16 x + 6:

{y = -2 x - 2

{16 x + 6 = -2

in the second equation, look to solve for x:

{y = -2 x - 2

{16 x + 6 = -2

subtract 6 from both sides:

{y = -2 x - 2

{16 x = -8

divide both sides by 16:

{y = -2 x - 2

{x = -1/2

substitute x = -1/2 into the first equation:

{y = -1

{x = -1/2

collect results in alphabetical order:

answer:   {x = -1/2 ,   y = -1

move it to the other side

step-by-step explanation:

p

step-by-step explanation: