The equation 4(x − 7) − 128 = 9x − 5(x + 6) has no solution
<h3><u>Solution:</u></h3>
Given that 4(x − 7) − 128 = 9x − 5(x + 6)
We have to find the solution for this equation
4(x − 7) − 128 = 9x − 5(x + 6)
Let us use BODMAS rule to perform the sequence of operations
According to Bodmas rule, if an expression contains brackets ((), {}, []) we have to first solve or simplify the bracket followed by of (powers and roots etc.), then division, multiplication, addition and subtraction from left to right.
So let us solve for brackets first
4x - 28 - 128 = 9x - 5x - 30
Now let us perform subtraction
4x - 156 = 4x - 30
Now move the terms from R.H.S to L.H.S
4x - 4x = -30 + 156
Since there is no value of x that will ever make this a true statement, the solution to the equation above is “no solution”
So this equation has no solution
Answer:
$24
Step-by-step explanation:
What you do is you just times 84 x 2/7 to get your answer.
84 x 2/7=$24 is your answer.
Answer: 
Step-by-step explanation:
divided by 
is the same thing as multiplying by 
.
The equation we get is:
With this equation, we can factor each equation:
We can cancel out like terms since they would be dividing each other:
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:
The correct answer to this question is B
