I think it would be the number and it’s exponent
Answer:

Step-by-step explanation:
Let the one of the side lengths of the rectangle be
and the other be
.
We can write the following equations, where
will be the side opposite to the wall:

From the first equation, we can isolate
and substitute into the second equation:

Therefore, the parabola
denotes the area of this rectangular enclosure. The maximum area possible will occur at the vertex of this parabola.
The x-coordinate of the vertex of a parabola in standard form
is given by
.
Therefore, the vertex is:

Plug in
to the equation to get the y-coordinate:

Thus the vertex of the parabola is at
. This tells us the following:
- The maximum area occurs when one side (y) of the rectangle is equal to 25
- The maximum area of the enclosure is 1,250 square meters
- The other dimension, from
, must be
And therefore, the desired answers are:

Answer:
I am a person and I'm going to say it is. c.
The surface area of the crate is the summation of all the areas of its six faces. The area of the two of its faces is the product of 50 cm and 70 cm which is 3500 cm^2. The other will have an area of 2000 cm^2. The last two will have the area of 2800 cm^2. Thus, the unit of measurement used to figure out the surface area is the cm^2. The answer is letter A. cm2
Answer:
y has a finite solution for any value y_0 ≠ 0.
Step-by-step explanation:
Given the differential equation
y' + y³ = 0
We can rewrite this as
dy/dx + y³ = 0
Multiplying through by dx
dy + y³dx = 0
Divide through by y³, we have
dy/y³ + dx = 0
dy/y³ = -dx
Integrating both sides
-1/(2y²) = - x + c
Multiplying through by -1, we have
1/(2y²) = x + C (Where C = -c)
Applying the initial condition y(0) = y_0, put x = 0, and y = y_0
1/(2y_0²) = 0 + C
C = 1/(2y_0²)
So
1/(2y²) = x + 1/(2y_0²)
2y² = 1/[x + 1/(2y_0²)]
y² = 1/[2x + 1/(y_0²)]
y = 1/[2x + 1/(y_0²)]½
This is the required solution to the initial value problem.
The interval of the solution depends on the value of y_0. There are infinitely many solutions for y_0 assumes a real number.
For y_0 = 0, the solution has an expression 1/0, which makes the solution infinite.
With this, y has a finite solution for any value y_0 ≠ 0.