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Serggg [28]
2 years ago
14

31

Mathematics
1 answer:
Viktor [21]2 years ago
6 0

Answer:

no sure about the answer

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Round your answer to two decimal places log 6e
Andru [333]

Solution: We know that e is numerical constant and is equal to 2.71828.

Therefore log( 6 x 2.71828) = 1.21

4 0
3 years ago
Help please and thanks
12345 [234]

Answer:

C

Step-by-step explanation:

6/9 simplified is 2/3 they're equal to each other so they're proportional

Hope this helps :)

6 0
3 years ago
The product of 5 and a number x is 14 .
Elena-2011 [213]
5x = 14  where x is the unknown number

divide both sides by 5:-

x = 14/5  or  2 4/5
8 0
3 years ago
Use the following matrices, A, B, C and D to perform each operation.
wolverine [178]

Step-by-step explanation:

\bold{40.}\\\\A+B=\left[\begin{array}{ccc}3&1\\5&7\end{array}\right] +\left[\begin{array}{ccc}4&1\\6&0\end{array}\right] =\left[\begin{array}{ccc}3+4&1+1\\5+6&7+0\end{array}\right]=\left[\begin{array}{ccc}7&1\\11&7\end{array}\right]

\bold{41.}\\\\B-A=\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]-\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]=\left[\begin{array}{ccc}4-3&1-1\\6-5&0-7\end{array}\right]=\left[\begin{array}{ccc}1&0\\1&-7\end{array}\right]

\bold{42.}\\\\3C=3\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]=\left[\begin{array}{ccc}(3)(-2)&(3)(3)&(3)(1)\\(3)(-1)&(3)(0)&(3)(4)\end{array}\right]=\left[\begin{array}{ccc}-6&9&3\\-3&0&12\end{array}\right]

\bold{43.}\\\\CD=\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]\left[\begin{array}{ccc}-2&3&4\\0&-2&1\\3&4&-1\end{array}\right]\\\\=\left[\begin{array}{ccc}(-2)(-2)+(3)(0)+(1)(3)&(-2)(3)+(3)(-2)+(1)(4)&(-2)(4)+(3)(1)+(1)(-1)\\(-1)(-2)+(0)(0)+(4)(3)&(-1)(3)+(0)(-2)+(4)(4)&(-1)(4)+(0)(1)+(4)(-1)\end{array}\right]\\\\=\left[\begin{array}{ccc}4+0+1&-6-6+4&-8+3-1\\2+0+12&-3+0+16&-4+0-4\end{array}\right]\\\\=\left[\begin{array}{ccc}5&-8&-6\\14&13&-8\end{array}\right]

\bold{44.}\\\\2D+3C=2\left[\begin{array}{ccc}-2&3&4\\0&-2&1\\3&4&-1\end{array}\right]+3\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]\\\\=\left[\begin{array}{ccc}(2)(-2)&(2)(3)&(2)(4)\\(2)(0)&(2)(-2)&(2)(1)\\(2)(3)&(2)(4)&(2)(-1)\end{array}\right]+\left[\begin{array}{ccc}(3)(-2)&(3)(3)&(3)(1)\\(3)(-1)&(3)(0)&(3)(4)\end{array}\right]\\\\=\left[\begin{array}{ccc}-4&6&8\\0&-4&2\\6&8&-2\end{array}\right]+\left[\begin{array}{ccc}-6&9&3\\-3&0&12\end{array}\right]

\large\bold{You\ can\ not\ add\ matrices\ of\ different\ dimensions!!!}

4 0
3 years ago
I can help with math so ask me a math question and I will answer
Sliva [168]
What does it mean when root mean square equals 0
3 0
3 years ago
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