The for any number n>1 is |(3/4+2/3i)^n| greater than 1 after simplifying the complex number.
<h3>What is a complex number?</h3>
It is defined as the number which can be written as x+iy where x is the real number or real part of the complex number and y is the imaginary part of the complex number and i is the iota which is nothing but a square root of -1.
We have a complex number:
Here n>1
Plug n = 2
= 145/144
= 1.0069
Which is greater than 1.
Thus, the for any number n>1 is |(3/4+2/3i)^n| greater than 1 after simplifying the complex number.
Learn more about the complex number here:
brainly.com/question/10251853
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Answer:
bye
Step-by-step explanation:
Answer:
The rule is ( x - 4, y + 3)
Step-by-step explanation:
Step-by-step explanation:
im not positive but it should be 12% chance that one is defective and then 11% chance when you pull one out
Let a be the first term in the sequence, and d the common difference between consecutive terms. If aₙ denotes the n-th term in the sequence, then
a₁ = a
a₂ = a₁ + d = a + d
a₃ = a₂ + d = a + 2d
a₄ = a₃ + d = a + 3d
and so on, up to the n-th term
aₙ = a + (n - 1) d
The sum of the first 10 terms is 100, and so
where we use the well-known sum formulas,
The sum of the next 10 terms is 300, so
Solve for a and d. Eliminating a gives
(10a + 145d) - (10a + 45d) = 300 - 100
100d = 200
d = 2
and solving for a gives
10a + 145×2 = 300
10a = 10
a = 1
So, the given sequence is simply the sequence of positive odd integers,
{1, 3, 5, 7, 9, …}
given recursively by the relation
and explicitly by
for n ≥ 1.
