We are given with an equation in <em>variable y</em> and we need to solve for <em>y</em> . So , now let's start !!!
We are given with ;
Take LCM on both sides :
<em>Multiplying</em> both sides by <em>10</em> ;

Can be <em>further written</em> as ;
Transposing <em>6y </em>to<em> LHS</em> and <em>150</em> to<em> RHS </em>


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.
Answer:
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Answer : C
we need to the value of f(–1)
Table is given in the question
From the table ,
f(x) = 4 when x= -5, that is f(-5) = 4
f(x) = 0 when x= -1, that is f(-1) =0
f(x)= -1 when x=6, that is f(6) = -1
f(x)= -3 when x=9, that is f(9) = -3
So, the value of f(–1) = 0