9514 1404 393
Answer:
split the number into equal pieces
Step-by-step explanation:
Assuming "splitting any number" means identifying parts that have the number as their sum, the maximum product of the parts will be found where the parts all have equal values.
We have to assume that the number being split is positive and all of the parts are positive.
<h3>2 parts</h3>
If we divide number n into parts x and (n -x), their product is the quadratic function x(n -x). The graph of this function opens downward and has zeros at x=0 and x=n. The vertex (maximum product) is halfway between the zeros, at x = (0 + n)/2 = n/2.
<h3>3 parts</h3>
Similarly, we can look at how to divide a (positive) number into 3 parts that have the largest product. Let's assume that one part is x. Then the other two parts will have a maximum product when they are equal. Their values will be (n-x)/2, and their product will be ((n -x)/2)^2. Then the product of the three numbers is ...
p = x(x^2 -2nx +n^2)/4 = (x^3 -2nx^2 +xn^2)/4
This will be maximized where its derivative is zero:
p' = (1/4)(3x^2 -4nx +n^2) = 0
(3x -n)(x -n) = 0 . . . . . . . . . . . . . factor
x = n/3 or n
We know that x=n will give a minimum product (0), so the maximum product is obtained when x = n/3.
<h3>more parts</h3>
A similar development can prove by induction that the parts must all be equal.
Answer:
13
Step-by-step explanation:
Given the equation h(x) = 2x²
To solve for h(-2) + 5 then, we need to break the problem into two parts
First, we will solve for the function h(x) at x = -2
h(x) = 2x² so
h(-2) = 2(-2)²
h(-2) = 8
Knowing this, we can plug 8 into our original expression for h(-2)
8 + 5
This gives us 13
Answer:
<h2>
<em><u>Answer</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>1</u></em><em><u>.</u></em><em><u>1</u></em><em><u>.</u></em><em><u> </u></em></h2>
<em><u>Hope</u></em><em><u> </u></em><em><u>this</u></em><em><u> </u></em><em><u>helps</u></em><em><u> </u></em><em><u>you</u></em>
Answer:
The following are the solution to the given points:
Step-by-step explanation:
for point A:


The set A is not part of the subspace 
for point B:


The set B is part of the subspace
for point C:

In this, the scalar multiplication can't behold

∉ C
this inequality is not hold
The set C is not a part of the subspace
for point D:

The scalar multiplication s is not to hold
∉ D
this is an inequality, which is not hold
The set D is not part of the subspace 
For point E:

The
is the arbitrary, in which
is arbitrary

The set E is the part of the subspace
For point F:

The
arbitrary so, they have
as the arbitrary 
The set F is the subspace of 