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vodka [1.7K]
3 years ago
8

Determine the amplitude, period, and phase shift of the function f(x)=cos (8(x-1))+2​

Mathematics
1 answer:
SCORPION-xisa [38]3 years ago
4 0

We have been given a function f(x)=\text{cos}(8(x-1))+2. We are asked to find the amplitude, period and phase shift of the function

We can see that our given function is in form f(x)=A\cdot \cos(B(x-C))+D, where,

|A| = Amplitude.

Period = \frac{2\pi}{|B|}

C = Horizontal shift,

D = Vertical shift.

We can see that value of a is 1, therefore, the amplitude of given function would be 1.

We can see that B is equal to 8, so we will get:

\text{Period}=\frac{2\pi}{|B|}= \frac{2\pi}{|8|}=\frac{\pi}{4}

Therefore, the period of given function is \frac{\pi}{4}.

Since the value of C is 1, therefore, horizontal shift is 1.

Since the value of D is 2, therefore, vertical shift would be 2.

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A bus driver started her day with an empty bus. At her first stop she picked up
lidiya [134]

Answer:

12 people.

Step-by-step explanation:

Let's do this one at a time.

First stop:

0 + 11 = 11

Second stop:

11 + 5 = 16.

16 - 7 = 9.

Third stop:

9 + 5 = 14.

14 - 2 = 12.

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8 0
2 years ago
11) If A = {19, 20, 21, 24} and B = {17, 19, 20, 22), list the elements of A U B.
Elodia [21]

Answer:

A U B = {17, 19, 20, 21, 22, 24}

Step-by-step explanation:

The symbol U is the union symbol, which denotes the union of two sets.

This means that given two sets A and B, the expression

A U B

indicates all the elements that belong to either set A or set B.

In this problem, we have the following two sets:

A = {19, 20, 21, 24}

B = {17, 19, 20, 22}

We can therefore list all the elements belonging to either set A or set B, they are:

17, 19, 20, 21, 22, 24

So, the expression A U B indicates the set:

A U B = {17, 19, 20, 21, 22, 24}

3 0
3 years ago
What equals to 25 in multiplication?
Masteriza [31]
5 times 5..........................
5 0
3 years ago
Find the maximum power of 5 which can divide 49!
JulijaS [17]

Answer:

10

Step-by-step explanation:

8 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E%7B2%7D-5x%2B6%7D%7B2x%5E%7B2%7D-7x%2B6%20%7D" id="TexFormula1" title="\frac{x^{
il63 [147K]

Answer:

\frac{x-3}{2x-3}. hole or removable discontinuity at x=2

Step-by-step explanation:

Well generally if you want the simplest form, you factor each the denominator and numerator and then see if you can cancel any of the factors out (because they're in the denominator and numerator)

So let's start by factoring the first equation:

x^2-5x+6

Now let's find what ac is (it's just c since a=1...)

AC= 6

List factors of -6

\pm1, \pm2, \pm3, \pm6.

Now we have to look for two numbers that add up to -5. It's a bit obvious here since there isn't many factors, but it's -2 and -3, and they're both negative since 6 is positive, and -5 is negative...

So using these two factors we get

(x-2)(x-3)

Ok now let's factor the second equation:

2x^2-7x+6

Multiply a and c

AC = 12

List factors of 12:

\pm1, \pm2, \pm3, \pm4, \pm6, \pm12.

Factors that add up to -7 and multiply to 12:

-3\ and\ -4

Rewrite equation:

2x^2-4x-3x+6

Group terms:

(2x^2-4x)+(-3x+6)

Factor out GCF:

2x(x-2)-3(x-2)

Rewrite:

(2x-3)(x-2)

Now let's write out the equation using these factors:

\frac{(x-2)(x-3)}{(2x-3)(x-2)}.

Here we can factor out the x-2 and the simplified form is:

\frac{x-3}{2x-3}

So we can "technically" define f(2) using the most simplified form, but it's removable discontinuity, so it has a hole as x=2. since it makes (x-2) equal to 0 (2-2) = 0.

8 0
2 years ago
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