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Flura [38]
3 years ago
7

F(x)3x+5/x what is f(a+2)

Mathematics
1 answer:
german3 years ago
7 0

Answer:

3a+6+5/(a+2)

Step-by-step explanation:

To do this you replace x with a+2, meaning that 3x+5/x turns into 3(a+2)+5/a+2. Simplified this is 3a+6+5/(a+2)

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If a = {6 6 4 7 -4 4 -6 9 -3} and B= {1 5 -8 9 -5 -8 9 -5 2}, find 3a +9B Matrix operations
V125BC [204]

Answer:

3A+9B=\begin{bmatrix} 27 & 63 & -60 \\\\ 102 & -57 & -60\\\\ 63 & -18 & 9 \end{bmatrix}

Step-by-step explanation

  • A=\begin{bmatrix} 6 & 6 & 4\\\\7 & -4 & 4\\\\ -6 & 9 & -3\end{bmatrix}\: and \: B=\begin{bmatrix}1 & 5 & -8\\\\ 9 & -5 & -8\\\\ 9 & -5 & 2\end{bmatrix}

  • \rightarrow 3A=3\begin{bmatrix} 6 & 6 & 4\\\\7 & -4 & 4\\\\ -6 & 9 & -3\end{bmatrix},\:  \: 9B =9\begin{bmatrix}1 & 5 & -8\\\\ 9 & -5 & -8\\\\ 9 & -5 & 2\end{bmatrix}

  • \rightarrow 3A=\begin{bmatrix} 3*6 & 3*6 & 3*4\\\\3*7 & 3(-4) & 3*4\\\\ 3(-6) & 3*9 & 3(-3)\end{bmatrix},\:\: 9B =\begin{bmatrix}9*1 &9* 5 &9( -8)\\\\ 9*9 & 9(-5) & 9(-8)\\\\ 9*9 & 9(-5) & 9*2\end{bmatrix}

  • \rightarrow 3A=\begin{bmatrix} 18 & 18 & 12 \\\\ 21 & -12 & 12\\\\ -18 & 27 & -9\end{bmatrix},\:\: 9B =\begin{bmatrix} 9 & 45 & -72 \\\\ 81 & -45 & -72\\\\ 81 & -45 & 18\end{bmatrix}

  • \rightarrow 3A+9B=\begin{bmatrix}18+9 & 18+45 & 12+(-72) \\\\ 21+81 & -12 +(-45) & 12+(-72)\\\\ -18+81 & 27+(-45) & -9+18 \end{bmatrix}

  • \rightarrow 3A+9B=\begin{bmatrix} 18+9 & 18+45 & 12-72 \\\\ 21+81 & -12 -45 & 12-72\\\\ -18+81 & 27-45 & -9+18 \end{bmatrix}

  • \rightarrow \purple{\bold{3A+9B=\begin{bmatrix} 27 & 63 & -60 \\\\ 102 & -57 & -60\\\\ 63 & -18 & 9 \end{bmatrix}}}
6 0
2 years ago
How to do the inverse of a 3x3 matrix gaussian elimination.
nata0808 [166]

As an example, let's invert the matrix

\begin{bmatrix}-3&2&1\\2&1&1\\1&1&1\end{bmatrix}

We construct the augmented matrix,

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

On this augmented matrix, we perform row operations in such a way as to transform the matrix on the left side into the identity matrix, and the matrix on the right will be the inverse that we want to find.

Now we can carry out Gaussian elimination.

• Eliminate the column 1 entry in row 2.

Combine 2 times row 1 with 3 times row 2 :

2 (-3, 2, 1, 1, 0, 0) + 3 (2, 1, 1, 0, 1, 0)

= (-6, 4, 2, 2, 0, 0) + (6, 3, 3, 0, 3, 0)

= (0, 7, 5, 2, 3, 0)

which changes the augmented matrix to

\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array} \right]

• Eliminate the column 1 entry in row 3.

Using the new aug. matrix, combine row 1 and 3 times row 3 :

(-3, 2, 1, 1, 0, 0) + 3 (1, 1, 1, 0, 0, 1)

= (-3, 2, 1, 1, 0, 0) + (3, 3, 3, 0, 0, 3)

= (0, 5, 4, 1, 0, 3)

\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 5 & 4 & 1 & 0 & 3 \end{array} \right]

• Eliminate the column 2 entry in row 3.

Combine -5 times row 2 and 7 times row 3 :

-5 (0, 7, 5, 2, 3, 0) + 7 (0, 5, 4, 1, 0, 3)

= (0, -35, -25, -10, -15, 0) + (0, 35, 28, 7, 0, 21)

= (0, 0, 3, -3, -15, 21)

\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 0 & 3 & -3 & -15 & 21 \end{array} \right]

• Multiply row 3 by 1/3 :

\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]

• Eliminate the column 3 entry in row 2.

Combine row 2 and -5 times row 3 :

(0, 7, 5, 2, 3, 0) - 5 (0, 0, 1, -1, -5, 7)

= (0, 7, 5, 2, 3, 0) + (0, 0, -5, 5, 25, -35)

= (0, 7, 0, 7, 28, -35)

\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 0 & 7 & 28 & -35 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]

• Multiply row 2 by 1/7 :

\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 4 & -5 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]

• Eliminate the column 2 and 3 entries in row 1.

Combine row 1, -2 times row 2, and -1 times row 3 :

(-3, 2, 1, 1, 0, 0) - 2 (0, 1, 0, 1, 4, -5) - (0, 0, 1, -1, -5, 7)

= (-3, 2, 1, 1, 0, 0) + (0, -2, 0, -2, -8, 10) + (0, 0, -1, 1, 5, -7)

= (-3, 0, 0, 0, -3, 3)

\left[ \begin{array}{ccc|ccc} -3 & 0 & 0 & 0 & -3 & 3 \\ 0 & 1 & 0 & 1 & 4 & -5 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]

• Multiply row 1 by -1/3 :

\left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & 1 & -1 \\ 0 & 1 & 0 & 1 & 4 & -5 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]

So, the inverse of our matrix is

\begin{bmatrix}-3&2&1\\2&1&1\\1&1&1\end{bmatrix}^{-1} = \begin{bmatrix}0&1&-1\\1&4&-5\\-1&-5&7\end{bmatrix}

6 0
2 years ago
Express as a trinomial (3x+8) (x+10)
docker41 [41]

Answer:

3x^{2} +38x+80

Step-by-step explanation:

Hello!

A trinomial is a expression consisting of three different terms

To turn this into a trinomial we multiply everything to each other

<u>                     3x                     </u>

3x * x = 3x^{2}

3x * 10 = 30x

<u>                     8                      </u>

8 * x = 8x

8 * 10 = 80

Now we put them all together in an equation

3x^{2} +30x+8x+80

Combine like terms

3x^{2} +38x+80

The answer is 3x^{2} +38x+80

Hope this helps!

7 0
3 years ago
Which of the b-values satisfy the following inequality?5 &lt; b - 3Choose all the answers that apply: A.) b = 8B.) b = 9C.) b =
dusya [7]
[tex]\begin{gathered} \text{The inequality is,} \\ 5
4 0
1 year ago
Need help with this
Kruka [31]

Answer:

ask tiktok i recomend

Step-by-step explanation:

6 0
3 years ago
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