Let the two positive real numbers be x and y.
The sum of two numbers is 156, so we can write:
x+ y = 156
or
y = 156 - x
The product of two numbers = xy
Using the value of y in previous equation we can write:
Product = x(156 - x) = - x² + 156x
The above equation is a quadratic equation and results in a parabola. The maximum value of parabola with negative coefficient of x² lies at its vertex.
The x component of vertex will be =

Here b is the coefficient of x term, and a is the coefficient of x² term.
So,
a = -1
b = 156
Using the values in the formula we get the x component of vertex x =

So x = 78
and
y = 156 - x = 156 - 78 = 78
Thus, the two numbers whose sum is 156 and which result in maximum product are 78, 78
Here is a step-by-step method for solving them:
1. Substitute y = uv, and
dy/dx = u
dv/dx + v
du/dx
into
dy/dx + P(x)y = Q(x)
2. Factor the parts involving v
3. Put the v term equal to zero (this gives a differential equation in u and x which can be solved in the next step)
4. Solve using separation of variables to find u
5. Substitute u back into the equation we got at step 2
6. Solve that to find v
7. Finally, substitute u and v into y = uv to get our solution!
I'm so sorry if the explanation is confusing but I hope it's helps you to solve ^ ^
Answer:
80 sq units
Step-by-step explanation:
Starting with the largest rectangle, we see it has a width of 6 and a length of 8, meaning the area of that only is 48 sq units. (area of rect. = l x w, so 6 x 8 is 48)
There are two smaller rectangles with width 6 and length 2. One of those rectangles would be 12 units (6 x 2). Since there is two of them, then 24 sq. units.
Lastly, there are four small triangles. The area formula for a triangle is 1/2(b)(h). The triangle has a base of 2 and a height of 2. (1/2)(2)(2) =2. Since there are four triangles, the total for them only would be 8 (2 x 4).
Adding each of these pieces together will give us the answer:
48+24+8=80 sq units
Answer: 8.0
8.035714286 is rounded to 8
Answer:
Always
Step-by-step explanation:
The Parallel Postulate:
When a line is already given, let's say the line is a, and the point be called A. If there is a point outside the line, only one line can be drawn through the point which doesn't intersect the line i.e. which is parallel to the given line. Hence, this statement is always true ..