All categories

3 years ago

What is multiply: -7*4

Mathematics

1 answer:

Luden [163]3 years ago

7 0

Answer:

-28

you welcome ........ :)

You might be interested in

Two data sets have the same mean,

nirvana33 [79]

For Data Set B, we see that the data is more varied. The absolute deviations are 4, 3, 2, 5. The average of these absolute deviations is 3.5. MAD_B = (4+3+2+5)/4 =3.5 M ADB

Hence, The average of these absolute deviations is 3.5.

5 0

2 years ago

3x-5<13 or -2x+4<img src="https://tex.z-dn.net/?f=%5Cleq" id="TexFormula1" title="\leq" alt="\leq" align="absmiddle" class="l

Anton [14]

Answer:

I would go with the second one.

Step-by-step explanation:

3 0

3 years ago

The perimeter of an equilateral triangle is 21cm, find its length

Margarita [4]

Answer:

7cm is the length for each side of an equalateral triangle

7 0

3 years ago

Read 2 more answers

6k squared -kl 4l squared

OleMash [197]

Square equals > [email protected]

8 0

4 years ago

Please help if you can pic attatched

MA_775_DIABLO [31]

Answer:

Step-by-step explanation:

$CI=\left[\begin{array}{ccc}1&6&0\\0&1&2\\1&-1&3\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$

Subtract row 3 from row 1:

$\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&7&-3\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\1&0&-1\end{array}\right]$

Subtract row 3 from 7 times row 2:

$\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&0&17\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\-1&7&1\end{array}\right]$

Divide row 3 by 17:

$\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&0&1\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]$

Subtract 2 of row 3 from row 2:

$\left[\begin{array}{ccc}1&6&0\\0&1&0\\0&0&1\end{array}\right] \left[\begin{array}{ccc}1&0&0\\\frac{2}{17} &\frac{3}{17} &\frac{-2}{17} \\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]$

Subtract 6 of row 2 from row 1:

$\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] \left[\begin{array}{ccc}\frac{5}{17}&\frac{-18}{17}&\frac{12}{17}\\\frac{2}{17} &\frac{3}{17} &\frac{-2}{17} \\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]$

$C^{-1}=\frac{1}{17} \left[\begin{array}{ccc}5&-18&12\\2&3&-2\\-1&7&1\end{array}\right]$

$C^{-1}b=\frac{1}{17} \left[\begin{array}{ccc}5&-18&12\\2&3&-2\\-1&7&1\end{array}\right]\left[\begin{array}{c}10&1&3\end{array}\right]=\left[\begin{array}{c}4&1&0\end{array}\right]$

3 0

3 years ago