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3 years ago

Consider the following function. f(x)=x^2+2x+1.

Mathematics

1 answer:

Gekata [30.6K]3 years ago

4 0

Answer:

A) x² + 2x + 6

B) x² + 2x - 7

C) ¼(x²+2x+1))

D) 6x²+12x+6

E) -x²-2x-1

Step-by-step explanation:

A) f(x) + 5 =x²+2x+1 + 5

= x² + 2x + 6

B) f(x)-8,=x^2+2x+1-8

= x² + 2x - 7

C) ¼f(x) = ¼(x²+2x+1)

D) 6f(x) = 6(x²+2x+1) = 6x²+12x+6

E) -f(x) = -(x²+2x+1) = -x²-2x-1

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Leno4ka [110]

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Step-by-step explanation:

We are given system of equations

$5x+4y+5z=-1$

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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= $\left[\begin{array}{ccc}-1\\1\\-3\end{array}\right]$

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We are finding rank of matrix

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

Apply $R_2\rightarrow R_2-R_1$

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

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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