A function
is periodic if there is some constant
such that
for all
in the domain of
. Then
is the "period" of
.
Example:
If
, then we have
, and so
is periodic with period
.
It gets a bit more complicated for a function like yours. We're looking for
such that
Expanding on the left, you have
and
It follows that the following must be satisfied:
The first two equations are satisfied whenever
, or more generally, when
and
(i.e. any multiple of 4).
The second two are satisfied whenever
, and more generally when
with
(any multiple of 10/7).
It then follows that all four equations will be satisfied whenever the two sets above intersect. This happens when
is any common multiple of 4 and 10/7. The least positive one would be 20, which means the period for your function is 20.
Let's verify:
More generally, it can be shown that
is periodic with period
.
Answer:
1
Step-by-step explanation:
Remember that anything to the 0th power is going to be 1 in terms of exponent rules!
Answer:
$2.71
Step-by-step explanation:
Hey!
I multiplied $2.71 by 2 and that gave me $5.42
Answer:
(1, 10)
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
<u>Algebra I</u>
- Solving systems of equations using substitution/elimination
Step-by-step explanation:
<u>Step 1: Define Systems</u>
-5x + 6y = 55
4x + 3y = 34
<u>Step 2: Rewrite Systems</u>
4x + 3y = 34
- Multiply everything by -2: -8x - 6y = -68
<u>Step 3: Redefine Systems</u>
-5x + 6y = 55
-8x - 6y = -68
<u>Step 4: Solve for </u><em><u>x</u></em>
<em>Elimination</em>
- Combine equations: -13x = -13
- Divide -13 on both sides: x = 1
<u>Step 5: Solve for </u><em><u>y</u></em>
- Define equation: 4x + 3y = 34
- Substitute in <em>x</em>: 4(1) + 3y = 34
- Multiply: 4 + 3y = 34
- Isolate <em>y</em> term: 3y = 30
- Isolate <em>y</em>: y = 10