Answer: 3x + 6 -4x = 1
Simplify to -x + 6 = 1 because 3x - 4x = -1x
Answer:y=-3
Step-by-step explanation:yes firs multiply everything by -3 in parenthesies
Then you would subtract 15 from both sides
Next you would divide 9 by -3 and you would get -3
So y =-3
Answer:
A) x = (-8log(6)-2log(17))/(-2log(17)+log(6))
Step-by-step explanation:
Taking the logarithm of the equation, you have ...
(x+8)log(6) = (2x-2)log(17)
Subtracting the right side from the equation gives ...
x(log(6) -2log(17)) +8log(6)+2log(17) = 0
Subtracting the constant and dividing by the coefficient of x gives ...
x = -(8log(6) +2log(17))/(log(6) -2log(17)) . . . . . matches selection A
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You don't need to work out the whole solution to determine the correct answer choice. Once you take the initial log, you find that the x-coefficient includes a multiplier of 2log(17). This term only appears in the denominator of choice A. (The value of x will be found after dividing by the x-coefficient, so you know this must show up in the denominator of the answer.)
Answer:
Matrix multiplication is not conmutative
Step-by-step explanation:
The matrix multiplication can be performed if the number of columns of the first matrix is equal to the number of rows of the second matrix
Let A with dimension mxn and B with dimension nxp represent two matrix
The multiplication of A by B is a matrix C with dimension mxp, but the multiplication of B by A is can't be calculated because the number of columns of B is not the number of rows of A. Therefore, you can notice that is not conmutative in general.
But even if the multiplication of AB and BA is defined (For example if A and B are squared matrix of 2x2) the multiplication is not necessary conmutative.
The matrix multiplication result is a matrix which entries are given by dot product of the corresponding row of the first matrix and the corresponding column of the second matrix:
![A=\left[\begin{array}{ccc}a11&a12\\a21&a22\end{array}\right]\\B= \left[\begin{array}{ccc}b11&b12\\b21&b22\end{array}\right]\\AB = \left[\begin{array}{ccc}a11b11+a12b21&a11b12+a12b22\\a21b11+a22b21&a21b12+a22b22\end{array}\right]\\\\BA=\left[\begin{array}{ccc}b11a11+b12a21&b11a12+b12a22\\b21a11+b22ba21&b21a12+b22a22\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da11%26a12%5C%5Ca21%26a22%5Cend%7Barray%7D%5Cright%5D%5C%5CB%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Db11%26b12%5C%5Cb21%26b22%5Cend%7Barray%7D%5Cright%5D%5C%5CAB%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da11b11%2Ba12b21%26a11b12%2Ba12b22%5C%5Ca21b11%2Ba22b21%26a21b12%2Ba22b22%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5CBA%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Db11a11%2Bb12a21%26b11a12%2Bb12a22%5C%5Cb21a11%2Bb22ba21%26b21a12%2Bb22a22%5Cend%7Barray%7D%5Cright%5D)
Notice that in general, the result is not the same. It could be the same for very specific values of the elements of each matrix.