In order to find the number of chips that would result in the minimum cost, we take the first derivative of the given equation. Note that the derivative refers to the slope of the graph at a given point. We can utilize this concept knowing that at the minimum or maximum point of a graph, the slope is zero.
Taking the derivative of the given equation and equating it to zero, we have:
y' = (0.000015)(2)x - (0.03)x° + 0
0 = (0.00003)x - 0.03
Solving for x or the number of chips produced, we have x = 1000. We then substitute this value in the given equation, such that,
y = (0.000015)(1000)² - (0.03)(1000) + 35
The minimized cost, y, to produce 1000 chips is then calculated to be $20.
3(x + 2) > x
3x + 6 > x
3x - 3x + 6 > x - 3x
6 > -2x
6/-2 < -2x/-2
-3< x
(When dividing by a negative, reverse the direction of the inequality side, reverse it EVERY time you divide by a negative)
ANSWER: -3 < x
Answer:
400/ 81
Step-by-step explanation:
16/25 × (5/3)^4
16/25 × (625/ 81)
10000/ 2025
400/81
Answer:
9 weeks
Step-by-step explanation:
Knowing that the "q" is the total number of quizzes, you can set the equation equal to 12, and solve for w.
12 = w + 3 Subtract 3 both sides
9 = w
