It would look like this: O=C=O
Hope this helps you out.
Between 0 and 1, add dashes so you have a total of 6 marks including 0 and 1. Then count 3 over and that would be 3/5.
Answer:r-1
Step-by-step explanation:
(-r-5)-(-2r-4) Or (-r-5)-(-2r-4)
-r+-5+2r+4. -r-5+2r+4
r-1 r-1
You bring down the first Parentheses Then have to multiply the (implied 1) -1•-2=2 and -1•-4=4 and do the math regularly
Answer:
≈ -5.1857
≈ -5.4857
≈ 3.7262
Step-by-step explanation:
Rewrite the equation system as:



Now, write the system in its augmented matrix form:
![\left[\begin{array}{cccc}6&8&0&-75\\-3&6&6&5\\2&-9&0&39\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D6%268%260%26-75%5C%5C-3%266%266%265%5C%5C2%26-9%260%2639%5Cend%7Barray%7D%5Cright%5D)
applying row reduction process to its associated augmented matrix:
Swap R1 and R3, and then Swap R1 and R2:
![\left[\begin{array}{cccc}-3&6&6&5\\2&-9&0&39\\6&8&0&-75\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D-3%266%266%265%5C%5C2%26-9%260%2639%5C%5C6%268%260%26-75%5Cend%7Barray%7D%5Cright%5D)
R3+2R1
![\left[\begin{array}{cccc}-3&6&6&5\\2&-9&0&39\\0&20&12&-65\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D-3%266%266%265%5C%5C2%26-9%260%2639%5C%5C0%2620%2612%26-65%5Cend%7Barray%7D%5Cright%5D)
3R2+2R1
![\left[\begin{array}{cccc}-3&6&6&5\\0&-15&12&127\\0&20&12&-65\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D-3%266%266%265%5C%5C0%26-15%2612%26127%5C%5C0%2620%2612%26-65%5Cend%7Barray%7D%5Cright%5D)
15R3+20R2
![\left[\begin{array}{cccc}-3&6&6&5\\0&-15&12&127\\0&0&420&1565\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D-3%266%266%265%5C%5C0%26-15%2612%26127%5C%5C0%260%26420%261565%5Cend%7Barray%7D%5Cright%5D)
Now we have a simplified system:


From (3):
(4)
Replacing (4) in (2)
(5)
Finally replacing (5) and (4) in (1)

$5.12
Make a chart for this sort of problem it makes it easier. Excel is good for it