Answer:
we conclude that the intervals in which the function P(t) is decreasing are:
Step-by-step explanation:
At x< 1, the function P(t) is decreasing
- A function p is an increasing function on an open interval if f(y) > f(x) for any two input values x and y in the given interval where y>x
- A function p is a decreasing function on an open interval if f(y) > f(x) for any two input values x and y in the given interval where y>x
From the figure, it is clear that the function seems to be increasing
from (1, 3) and then 4 to onwards.
But it is clear that the function seems to be decreasing from the x < 1 and from the interval (3, 4).
Therefore, we conclude that the intervals in which the function P(t) is decreasing are:
Answer: 3,295.5
Step-by-step explanation:
Answer:
-35
Step-by-step explanation:
- 4ac = (1)(1) - 4(-3)(3) = 1 - 36 = -35
I FOUND YOUR COMPLETE QUESTION IN OTHER SOURCES.
PLEASE SEE ATTACHED IMAGE.
For this case the area is given by:
A = (22 + x) * (28 + x) = 722
Rewriting we have:
616 + 22x + 28x + x ^ 2 = 722
x ^ 2 + 50x + 616 - 722 = 0
x ^ 2 + 50x - 106 = 0
Solving the polynomial we have:
x1 = 2.04
x2 = -52.04
We take the positive root:
x = 2.04 inches:
Answer:
The width of the border to the nearest inch is:
x = 2inches