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Harman [31]
3 years ago
7

Given that a function, g, has a domain of -1 ≤ x ≤ 4 and a range of 0 ≤ g(x) ≤ 18 and that g(-1) = 2 and g(2) = 8, select the st

atement that could be true for g.
g(5) = 12


g(1) = -2


g(2) = 4


g(3) = 18
Mathematics
2 answers:
Juli2301 [7.4K]3 years ago
8 0

Answer:

Option 4 is correct.

Step-by-step explanation:

Consider a function g, it has a domain of -1 ≤ x ≤ 4 and a range of 0 ≤ g(x) ≤ 18. It is given that g(-1) = 2 and g(2) = 8.

The statement g(5) = 12 is not true because the value of x is 5 which is not in its domain.

The statement g(1) = -2 is not true because the value of function g(x) is -2 which is not in its range.

The statement g(2) = 4 is not true because g is a function and each function has unique output for each input value.

If g(2)=8 and g(2)=4, then the value of g(x) is 8 and 4 at x=2. It means g(x) is not a function, which is contradiction of given statement.

The statement g(3) = 18 is true because the value of x is 3 which is in the domain and the value of function g(x) is 18 which is in its range.

Therefore, the correct option is 4.

satela [25.4K]3 years ago
6 0

Answer:  g(3) = 18

Step-by-step explanation:

The inputs and outputs of the function need to be within the domain and range of the function.

The statement g(5) = 12 cannot be true, because the input, 5, is not in the domain of the function.

The statement g(1) = -2 cannot be true, because the output, -2, is not in the range of the function.

For a relation to be a function, each input, x, in the domain can have exactly one output, g(x), in the range. It is given that g(x) is a function.

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How many ways are there to rearrange the letters ALIVE in such a way that the word EVIL is not contained in the resulting word
Sergio [31]

Answer: 118 combinations.

Step-by-step explanation:

In ALIVE we have 5 letters, rearranging it we have the possible options of:

5 options for the first letter.

4 options for the second letter, as one was already chosen for the first one

3 options for the third letter, as 2 were already chosen.

2 options for the fourth letter.

1 option for the last letter.

The total number of combinations will be equal to the product of the numbers of options for each letter, then we have:

5*4*3*2*1 = 120 combinations.

And the combinations where the word EVIL is formed are:

AEVIL

EVILA

2 combinations, then the total number of combinations such that the word evil is not formed are:

120 - 2 = 118 combinations.

5 0
3 years ago
If you help me with this easy question, I will give you 1 thanks and 5 stars :) (number 1)
inessss [21]

1 to 45 is 45

90 divided by 2 is 45

135 divided by 4 is 45

180 divided by 4 is 45

all of them have the same relationship and a proportional to each other

8 0
2 years ago
Solve: 2cos(x)-square root 3=0 for 0 less than x less than 2 pi
Leona [35]

Answer:

The general solution of   cos x = cos(\frac{\pi }{6})   is  

                                                x = 2nπ±\frac{\pi }{6}

The general solution values  

                                 x = - \frac{\pi }{6}  and x = \frac{\pi }{6}

Step-by-step explanation:

Explanation:-

Given equation is  

                              2cosx-\sqrt{3} =0  for 0

                              2cosx =\sqrt{3}

Dividing '2' on both sides, we get

                             cos x =\frac{\sqrt{3} }{2}

                             cos x = cos(\frac{\pi }{6})

<em>General solution of cos θ = cos ∝ is θ = 2nπ±∝</em>

<em>Now The general solution of   </em>cos x = cos(\frac{\pi }{6})<em>   is  </em>

<em>                                                 x = 2nπ±</em>\frac{\pi }{6}<em></em>

put n=0

x = - \frac{\pi }{6}  and x = \frac{\pi }{6}

Put n=1  

x = 2\pi +\frac{\pi }{6} = \frac{13\pi }{6}

x = 2\pi -\frac{\pi }{6} = \frac{11\pi }{6}

put n=2

x = 4\pi +\frac{\pi }{6} = \frac{25\pi }{6}

x = 4\pi -\frac{\pi }{6} = \frac{23\pi }{6}

And so on

But given 0 < x< 2π

The general solution values  

                                 x = - \frac{\pi }{6}  and x = \frac{\pi }{6}

                               

6 0
3 years ago
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Sophie [7]

Answer:

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Step-by-step explanation:

3 0
2 years ago
Find the first, fourth, and tenth terms of the arithmetic sequence described by the given rule. A(n)=-6+(n-1)(1/5) A)-6,-5 1/5,
scoundrel [369]

Step-by-step explanation:

An arithmetic sequence is given by relation as follows :

A(n)=-6+(n-1)\dfrac{1}{5}

For the first term, put n = 1. So,

A(1)=-6+(1-1)\dfrac{1}{5}\\\\A(1)=-6

For fourth term, put n = 4. So,

A(4)=-6+(4-1)\dfrac{1}{5}\\\\A(4)=\dfrac{-27}{5}\\\\A(4)=-5\dfrac{2}{5}

For tenth term, put n = 10. So,

A(10)=-6+(10-1)\dfrac{1}{5}\\\\A(10)=-6+\dfrac{9}{5}\\\\A(10)=\dfrac{-21}{5}\\\\A(10)=-4\dfrac{1}{5}

Hence, the correct option is (C).

7 0
3 years ago
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