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Doss [256]
3 years ago
15

Consider the function below. x -1 0 1 2 f(x) -2 3 8 13 Which of the following functions could be the inverse of function f? A. x

1 0 -1 -2 s(x) -2 3 8 13 B. x -2 3 8 13 q(x) -1 0 1 2 C. x -2 -3 -8 -13 r(x) 1 0 -1 -2 D. x -1 0 1 2 p(x) 2 -3 -8 -13
Mathematics
2 answers:
uysha [10]3 years ago
8 0

Answer: B

x         q(x)

-2          -1

3            0

8            1

13           2

Step-by-step explanation:

An inverse of a function is a reflection across the y=x line. This results in each (x,y) point becoming (y,x).

x         f(x)

-1          -2

0           3

1            8

2           13

So the inverse becomes:

x         Inverse

-2          -1

3            0

8            1

13           2

Morgarella [4.7K]3 years ago
7 0

Answer:

The correct option is B.

Step-by-step explanation:

The table of values of a function is given below:

x    :     -1    0    1    2

f(x) :    -2    3     8   13

If a function is defined f:R→R

f(x)=\{(x,y):x\in R,y\in R\}

them the inverse of function is defined as

f^{-1}(x)=\{(y,x):x\in R,y\in R\}

The table of values of inverse function is

x      :    -2    3     8   13

f⁻¹(x):     -1    0    1    2

We can say that q(x) is the inverse function of f(x). Therefore the correct option is B.

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From the question we can say that the Hexagon has three shapes inside it,

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Also it is given that,

An equilateral triangle is shown inside a square inside a regular pentagon inside a regular hexagon.

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Show Work Please Thank You
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Answer:

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Step-by-step explanation:

We are given the trigonometric equation of:

\displaystyle{\sin 4x = \dfrac{\sqrt{3}}{2}}

Let u = 4x then:

\displaystyle{\sin u = \dfrac{\sqrt{3}}{2}}\\\\\displaystyle{\arcsin (\sin u) = \arcsin \left(\dfrac{\sqrt{3}}{2}\right)}\\\\\displaystyle{u= \arcsin \left(\dfrac{\sqrt{3}}{2}\right)}

Find a measurement that makes sin(u) = √3/2 true within [0, π) which are u = 60° (π/3) and u = 120° (2π/3).

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Convert u-term back to 4x:

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