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rosijanka [135]
3 years ago
5

Given this function: f ( x )=−2cos(4 πx ) Find the following:

Mathematics
1 answer:
Neporo4naja [7]3 years ago
4 0

Answer:

Period =½

Equation of midline, y=0

Maximum =2

Minimum=-2

Step-by-step explanation:

The given function is

f(x) =  - 2 \cos(4 \pi \: x)

The period is given by:

\frac{2\pi}{b}  =   \frac{2\pi}{4\pi} =  \frac{1}{2}

The equation of the midline is y=0 since there is no vertical shift

The amplitude of this function is 2 so the range is -2≤y≤2.

Hence the maximum value is 2 and minimum value is -2

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What is the solution set of {x | x &lt; -5} ∩ {x | x &gt; 5}?
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3 years ago
The nth term of a sequence is 3n²-1
Anarel [89]

Answer: The number is 26.

Step-by-step explanation:

We know that:

The nth term of a sequence is 3n²-1

The nth term of a different sequence is 30–n²

We want to find a number that belongs to both sequences (it is not necessarily for the same value of n) then we can use n in one term (first one), and m in the other (second one), such that n and m must be integer numbers.

we get:

3n²- 1 =  30–m²

Notice that as n increases, the terms of the first sequence also increase.

And as n increases, the terms of the second sequence decrease.

One way to solve this, is to give different values to m (m = 1, m = 2, etc) and see if we can find an integer value for n.

if m = 1, then:

3n²- 1 =  30–1²

3n²- 1 = 29

3n² = 30

n² = 30/3 = 10

n² = 10

There is no integer n such that n² = 10

now let's try with m = 2, then:

3n²- 1 =  30–2² = 30 - 4

3n²- 1 = 26

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n² = 27/3 = 9

n² = 9

n = √9 = 3

So here we have m = 2, and n = 3, both integers as we wanted, so we just found the term that belongs to both sequences.

the number is:

3*(3)² - 1 = 26

30 - 2² = 26

The number that belongs to both sequences is 26.

6 0
2 years ago
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