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Nookie1986 [14]
3 years ago
6

Which exponential function is represented by the values in the table?

Mathematics
2 answers:
Masja [62]3 years ago
8 0

Answer:

The function f(x) = 3( {2}^{x} ).

Step-by-step explanation:

Given : Table for x and f(x).

To find : Which exponential function is represented by the values in the table.

Solution :  Let the exponential function that is represented by the values in the table be of the form,

f(x) = a( {b}^{x} )

On plugging the values from table (0,3)

3 = a( {b}^{0} ) we get ,

3 = a.

tex]f(x) = 3( {b}^{x} )[/tex]

To find value of b we substituting other point from table (1,6)

tex]6 = 3( {b}^{1} )[/tex].

2 = b.

Therefore , the function f(x) = 3( {2}^{x} ).

astraxan [27]3 years ago
3 0
ANSWER

The exponential function is
f(x) = 3( {2}^{x} )

EXPLANATION

Let the exponential function that is represented by the values in the table be of the form,

f(x) = a( {b}^{x} )

The points in the table must satisfy this exponential function.


We substitute the point,

(0,3)

to get,


3 = a( {b}^{0} )



This implies that,


3 = a( 1 )


3 = a


Our function now becomes,

f(x) = 3( {b}^{x} )


We gain, plug in another point yo find the value of b too.


Let us substitute
(1,6)


This implies that,

6= 3( {b}^{1} )


We divide through by 3 to obtain,

2 = b


Therefore the function is,


f(x) = 3( {2}^{x} )



The correct answer is A.
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Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

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Step-by-step explanation:

We have the following system of equations

x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2

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\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]

The reduction of this matrix to row-echelon form is outlined below.

R_2\rightarrow R_2-R_1

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R_3\rightarrow R_3-2R_1

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]

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\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]

The last row determines, if there are solutions or not. To be consistent, we must have k such that

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If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

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