<h2>○=> <u>Correct answer</u> :</h2>
of the drink is fruit juice
<h3>○=> <u>Steps to derive correct answer</u> :</h3>
Quantity of water in a drink = 650 ml
Quantity of fruit juice in a drink = 150 ml
Total quantity of both items in the drink :


Thus, the total quantity of both items in the drink = 800 ml
Let the percentage of of fruit juice in the drink be x%.
Which means :






Then, the percentage of water in the drink :


Thus, the percentage of water in the drink = 81.25%
<h3>○=> <u>Therefore</u> :</h3>
▪︎Percentage of fruit juice in the drink = 18.75%
▪︎Percentage of water in the drink = 81.75%
Answer:
a
Step-by-step explanation:
25,678 rounded to the nearest 10,000 would be 30,000
Convert to equivalent fractions to see
Both bottom s can be 6
So times the half by 3
And times the third by 2
1/2=3/6
1/3=2/6
So she has a total of 5/6 of the movie accounted for...
So the left over is 1/6
1/6 at nighttime
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)
