Answer:
Option 2
Step-by-step explanation:
Let's say that Phalicia opens her savings account one year (12 months) before she goes to college.
With Option 1, she would have saved 300 + 50 * 11 = $850. We do 50 * 11 and not 50 * 12 because she deposits $50 for 11 months, not 12.
With Option 2, she would have saved 5 * 3¹¹ = $885735. Note that we do 3¹¹ and not 3¹² because 5 is being tripled 11 times.
Obviously, she should choose Option 2 because she saves A LOT more money.
Additionally, we can notice that Option 1 is an example of an arithmetic sequence whereas Option 2 is an example of a geometric sequence. Their explicit formulas would be aₙ = 50n + 250 and aₙ = 5 * 3⁽ⁿ⁻¹⁾ respectively.
The period of
is
.
Recall that
and
both have periods of
. This means
![\sin(x + 2\pi) = \sin(x)](https://tex.z-dn.net/?f=%5Csin%28x%20%2B%202%5Cpi%29%20%3D%20%5Csin%28x%29)
![\cos(x + 2\pi) = \cos(x)](https://tex.z-dn.net/?f=%5Ccos%28x%20%2B%202%5Cpi%29%20%3D%20%5Ccos%28x%29)
Replacing
with
, we have
![\sin(3x + 2\pi) = \sin\left(3 \left(x + \dfrac{2\pi}3\right)\right) = \sin(3x)](https://tex.z-dn.net/?f=%5Csin%283x%20%2B%202%5Cpi%29%20%3D%20%5Csin%5Cleft%283%20%5Cleft%28x%20%2B%20%5Cdfrac%7B2%5Cpi%7D3%5Cright%29%5Cright%29%20%3D%20%5Csin%283x%29)
In other words, if we change
by some multiple of
, we end up with the same output. So
has period
.
Similarly,
has a period of
,
![\cos(5x + 2\pi) = \cos\left(5 \left(x + \dfrac{2\pi}5\right)\right) = \cos(5x)](https://tex.z-dn.net/?f=%5Ccos%285x%20%2B%202%5Cpi%29%20%3D%20%5Ccos%5Cleft%285%20%5Cleft%28x%20%2B%20%5Cdfrac%7B2%5Cpi%7D5%5Cright%29%5Cright%29%20%3D%20%5Ccos%285x%29)
We want to find the period
of
, such that
![f(x + p) = f(x)](https://tex.z-dn.net/?f=f%28x%20%2B%20p%29%20%3D%20f%28x%29)
![\implies \sin(3x + p) + \cos(5x + p) = \sin(3x) + \cos(5x)](https://tex.z-dn.net/?f=%20%5Cimplies%20%5Csin%283x%20%2B%20p%29%20%2B%20%5Ccos%285x%20%2B%20p%29%20%3D%20%5Csin%283x%29%20%2B%20%5Ccos%285x%29)
On the left side, we have
![\sin(3x + p) = \sin(3x + 2\pi + p - 2\pi) \\\\ ~~~~~~~~ = \sin(3x+2\pi) \cos(p-2\pi) + \cos(3x+2\pi) \sin(p-2\pi) \\\\ ~~~~~~~~ = \sin(3x) \cos(p-2\pi) + \cos(3x) \sin(p - 2\pi)](https://tex.z-dn.net/?f=%5Csin%283x%20%2B%20p%29%20%3D%20%5Csin%283x%20%2B%202%5Cpi%20%2B%20p%20-%202%5Cpi%29%20%5C%5C%5C%5C%20~~~~~~~~%20%3D%20%5Csin%283x%2B2%5Cpi%29%20%5Ccos%28p-2%5Cpi%29%20%2B%20%5Ccos%283x%2B2%5Cpi%29%20%5Csin%28p-2%5Cpi%29%20%5C%5C%5C%5C%20~~~~~~~~%20%3D%20%5Csin%283x%29%20%5Ccos%28p-2%5Cpi%29%20%2B%20%5Ccos%283x%29%20%5Csin%28p%20-%202%5Cpi%29)
and
![\cos(5x + p) = \cos(5x + 2\pi + p - 2\pi) \\\\ ~~~~~~~~ = \cos(5x+2\pi) \cos(p-2\pi) - \sin(5x+2\pi) \sin(p-2\pi) \\\\ ~~~~~~~~ = \cos(5x) \cos(p-2\pi) - \sin(5x) \sin(p-2\pi)](https://tex.z-dn.net/?f=%5Ccos%285x%20%2B%20p%29%20%3D%20%5Ccos%285x%20%2B%202%5Cpi%20%2B%20p%20-%202%5Cpi%29%20%5C%5C%5C%5C%20~~~~~~~~%20%3D%20%5Ccos%285x%2B2%5Cpi%29%20%5Ccos%28p-2%5Cpi%29%20-%20%5Csin%285x%2B2%5Cpi%29%20%5Csin%28p-2%5Cpi%29%20%5C%5C%5C%5C%20~~~~~~~~%20%3D%20%5Ccos%285x%29%20%5Ccos%28p-2%5Cpi%29%20-%20%5Csin%285x%29%20%5Csin%28p-2%5Cpi%29)
So, in terms of its period, we have
![f(x) = \sin(3x) \cos(p - 2\pi) + \cos(3x) \sin(p - 2\pi) \\\\ ~~~~~~~~ ~~~~+ \cos(5x) \cos(p - 2\pi) - \sin(5x) \sin(p - 2\pi)](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Csin%283x%29%20%5Ccos%28p%20-%202%5Cpi%29%20%2B%20%5Ccos%283x%29%20%5Csin%28p%20-%202%5Cpi%29%20%20%5C%5C%5C%5C%20~~~~~~~~%20~~~~%2B%20%5Ccos%285x%29%20%5Ccos%28p%20-%202%5Cpi%29%20-%20%5Csin%285x%29%20%5Csin%28p%20-%202%5Cpi%29)
and we need to find the smallest positive
such that
![\begin{cases} \cos(p - 2\pi) = 1 \\ \sin(p - 2\pi) = 0 \end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%20%5Ccos%28p%20-%202%5Cpi%29%20%3D%201%20%5C%5C%20%5Csin%28p%20-%202%5Cpi%29%20%3D%200%20%5Cend%7Bcases%7D)
which points to
, since
![\cos(2\pi-2\pi) = \cos(0) = 1](https://tex.z-dn.net/?f=%5Ccos%282%5Cpi-2%5Cpi%29%20%3D%20%5Ccos%280%29%20%3D%201)
![\sin(2\pi - 2\pi) = \sin(0) = 0](https://tex.z-dn.net/?f=%5Csin%282%5Cpi%20-%202%5Cpi%29%20%3D%20%5Csin%280%29%20%3D%200)
Answer:
In mathematics, a complex number is a number that can be expressed in the form a + bi, where a and b are real numbers, and i is a symbol, called the imaginary unit, that satisfies the equation i² = −1. Because no real number satisfies this equation, i was called an imaginary number by René Descartes.
Step-by-step explanation:
Complex Integer
(or Gaussian integer), a number of the form a + bi, where a and b are integers. An example is 4 – 7i. Geometrically, complex integers are represented by the points of the complex plane that have integral coordinates.
Complex integers were introduced by K. Gauss in 1831 in his investigation of the theory of biquadratic residues. The advances made in such areas of number theory as the theory of higher-degree residues and Fermat’s theorem through the use of complex integers helped clarify the role of complex numbers in mathematics. The further development of the theory of complex integers led to the creation of the theory of algebraic integers.
The arithmetic of complex integers is similar to that of integers. The sum, difference, and product of complex integers are complex integers; in other words, the complex integers form a ring.
When two variables have an inverse relationship, the slope is negatively correlated.
Exact Form: 16/15
Decimal Form: 1.06666666666
Mixed Number: 1 1/15