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Fittoniya [83]
3 years ago
11

If m(x) = x+5/x-1 and n(x) = x — 3, have the same domain as (m • n) (x)

Mathematics
1 answer:
Yuri [45]3 years ago
8 0

For this case we have the following functions:

m (x) = \frac {x + 5} {x-1}\\n (x) = x-3

By definition we have to:

(f * g) (x) = f (x) * g (x)

So:

(m * n) (x) = m (x) * n (x)\\(m * n) (x) = \frac {x + 5} {x-1} (x-3)\\(m * n) (x) = \frac {(x + 5) (x-3)} {x-1}

By definition, the domain of a function is given by the values for which the function is defined.

The domain of m(x) is given by all reals except 1.

The domain of n(x) is given by all reals.

While the domain of (m * n) (x) is given by:

All reals, except the 1. With x = 1, the denominator is 0 and the function is no longer defined.

Answer:

Domain of(m * n) (x) is given by all reals except 1.

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Answer:

Infinite number of solutions.

Step-by-step explanation:

We are given system of equations

5x+4y+5z=-1

x+y+2z=1

2x+y-z=-3

Firs we find determinant of system of equations

Let a matrix A=\left[\begin{array}{ccc}5&4&5\\1&1&2\\2&1&-1\end{array}\right] and B=\left[\begin{array}{ccc}-1\\1\\-3\end{array}\right]

\mid A\mid=\begin{vmatrix}5&4&5\\1&1&2\\2&1&-1\end{vmatrix}

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Determinant of given system of equation is zero therefore, the general solution of system of equation is many solution or no solution.

We are finding rank of matrix

Apply R_1\rightarrow R_1-4R_2 and R_3\rightarrow R_3-2R_2

\left[\begin{array}{ccc}1&0&1\\1&1&2\\0&-1&-3\end{array}\right]:\left[\begin{array}{ccc}-5\\1\\-5\end{array}\right]

ApplyR_2\rightarrow R_2-R_1

\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&-1&-3\end{array}\right]:\left[\begin{array}{ccc}-5\\6\\-5\end{array}\right]

Apply R_3\rightarrow R_3+R_2

\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&0&-2\end{array}\right]:\left[\begin{array}{ccc}-5\\6\\1\end{array}\right]

Apply R_3\rightarrow- \frac{1}{2} and R_2\rightarrow R_2-R_3

\left[\begin{array}{ccc}1&0&1\\0&1&0\\0&0&1\end{array}\right]:\left[\begin{array}{ccc}-5\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]

Apply R_1\rightarrow R_1-R_3

\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]:\left[\begin{array}{ccc}-\frac{9}{2}\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]

Rank of matrix A and B are equal.Therefore, matrix A has infinite number of solutions.

Therefore, rank of matrix is equal to rank of B.

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