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Kobotan [32]
4 years ago
7

The graph of f(x) is show below.

Mathematics
2 answers:
vlada-n [284]4 years ago
5 0

ANSWER

(2,2)

EXPLANATION

Since f(x) and its inverse function are symmetric about the line y=x, if (a,b) lies on the graph of f, then (b,a) must lie on f inverse.

The point (2,2) lies on f and the same time on f inverse.

The solution is a point that satisfies both equations.

Hence the correct choice is:

(2,2)

jok3333 [9.3K]4 years ago
3 0

Answer:

Did the test. Answer is (2,2).

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3 years ago
In New York City at the spring equinox there are 12 hours 8 minutes of daylight. The longest and the shortes days of the year va
gayaneshka [121]

Question:

In New York City at the spring equinox there are 12 hours 8 minutes of daylight. The  longest and the shortest days of the year vary by 2 hours 53 minutes from the equinox.  In this year, the equinox falls on March 21. In this task, you'll use a trigonometric function  to model the hours of daylight hours on certain days of the year in New York City.

- Find amplitude and the period of the function

- Create a trigonometric function that describes the hours of sunlight for each day of the year

- Then use the function you built to find how fewer daylight hours February 10 will have then March 21

Answer:

(a)

A = 2.883  --- Amplitude

T = 365 ---- Period

(b) Trigonometry function

f(x) = 12.133 + 2.883sin(\frac{2\pi}{365}[x - 80])

(c) Hours= 1.794

Step-by-step explanation:

Given

Average\ Sunlight = 12hr\ 8 min

Variance = 2hr\ 53min

Solving (a): Amplitude (P) and Period (T)

The amplitude is the amount of time the longest and the shortest day vary.

So

A = 2\ hr\ 53\ min

Convert to hours

A = 2\ + \frac{53}{60}

A = 2+0.883

A = 2.883

The period (T) is the duration i.e 1 year

T = 1\ year

Assume no leap year

T = 365

Solving (b): Trigonometry function

The function follows a sinusoidal pattern and the general form is:

f(x) = \mu+ Asin(\frac{2\pi}{T}(x -n))

Where

\mu = Average\ Value

\mu = 12\ hr 8\ min

Convert to hours

\mu = 12 + \frac{8}{60}

\mu = 12 + 0.133

\mu = 12.133

A = 2.883  --- Amplitude

T = 365 ---- Period

n = Equinox

n = March\ 21

March\ 21st = 80th\ day

So:

n= 80

The function becomes:

f(x) = \mu+ Asin(\frac{2\pi}{T}(x -n))

f(x) = 12.133 + 2.883sin(\frac{2\pi}{365}[x - 80])

Solving (c): Fewer daylight hours will Feb. 10 have.

Feb\ 10 = 41st\ day

So:

f(x) = 12.133 + 2.883sin(\frac{2\pi}{365}[x - 80])

f(41) = 12.133 + 2.883sin(\frac{2\pi}{365}[41 - 80])

f(41) = 12.133 + 2.883sin(\frac{2\pi}{365}[-39])

2\pi = 360^\circ

So:

f(41) = 12.133 + 2.883sin(\frac{360}{365}[-39])

f(41) = 12.133 + 2.883sin(-38.466)

f(41) = 12.133 - 2.883*0.6221

f(41) = 10.339

The fewer daylight hours is the calculated as:

Hours= Average - f(41)

Hours= \mu - f(41)

Hours= 12.133 - 10.339

Hours= 1.794

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3 years ago
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Answer:

17 + (9 - 6)(12 ÷ 3) =

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17 + 12 =

29

7 0
3 years ago
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