Hi there!
To find x, you would need to use the inverse of subtraction (addition) and inverse of division (multiplication), since you'd need to isolate x to one side of the equation to set x equal to a number.
Hope this helps!
Answer:
ok ok ok ok ok ok o kom ko ko ko ko ko komk ko ko ko k kokok ok ok ok ok ok ok ok ok o-k ok ok
Step-by-step explanation:
The coefficient matrix is build with its rows representing each equation, and its columns representing each variable.
So, you may write the matrix as
![\left[\begin{array}{cc}\text{x-coefficient, 1st equation}&\text{y-coefficient, 1st equation}\\\text{x-coefficient, 2nd equation}&\text{y-coefficient, 2nd equation} \end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Ctext%7Bx-coefficient%2C%201st%20equation%7D%26%5Ctext%7By-coefficient%2C%201st%20equation%7D%5C%5C%5Ctext%7Bx-coefficient%2C%202nd%20equation%7D%26%5Ctext%7By-coefficient%2C%202nd%20equation%7D%20%5Cend%7Barray%7D%5Cright%5D%20%20)
which means
![\left[\begin{array}{cc}4&-3\\8&-3\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26-3%5C%5C8%26-3%5Cend%7Barray%7D%5Cright%5D%20%20)
The determinant is computed subtracting diagonals:
![\left | \left[ \begin{array}{cc}a&b\\c&d\end{array}\right]\right | = ad-bc](https://tex.z-dn.net/?f=%20%5Cleft%20%7C%20%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%5Cright%20%7C%20%3D%20ad-bc%20)
So, we have
![\left | \left[\begin{array}{cc}4&-3\\8&-3\end{array}\right] \right | = 4(-3) - 8(-3) = -4(-3) = 12](https://tex.z-dn.net/?f=%20%5Cleft%20%7C%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26-3%5C%5C8%26-3%5Cend%7Barray%7D%5Cright%5D%20%5Cright%20%7C%20%3D%204%28-3%29%20-%208%28-3%29%20%3D%20-4%28-3%29%20%3D%2012%20%20)
Answer:
256
Step-by-step explanation:
A calculator works well for this.
_____
None of the minus signs are subject to the exponents (because they are not in parentheses, as (-1)^5, for example. Since there are an even number of them in the product, their product is +1 and they can be ignored.
1 to any power is still 1, so the factors (1^n) can be ignored.
After you ignore all of the things that can be ignored, your problem simplifies to ...
(2^2)(2^-3)^-2
The rules of exponents applicable to this are ...
(a^b)^c = a^(b·c)
(a^b)(a^c) = a^(b+c)
Then your product simplifies to ...
(2^2)(2^((-3)(-2)) = (2^2)(2^6)
= 2^(2+6)
= 2^8 = 256
Answer:
the sequence is by -5
therefore the next numbers are
-13,-18,-23,-28