440 gallons of 80 % antifreeze solution must be mixed with 80 gallons of 15% antifreeze to get a mixture that is 70% antifreeze
<em><u>Solution:</u></em>
Let "x" be the gallons of 80 % antifreeze added
Therefore, "x" gallons of 80 % antifreeze solution must be mixed with 80 gallons of 15% antifreeze
Final mixture is x + 80
Therefore, we can frame a equation as:
"x" gallons of 80 % antifreeze solution must be mixed with 80 gallons of 15% antifreeze to get (x + 80) gallons of 70 % antifreeze
Thus, we get,
x gallons of 80 % + 80 gallons of 15 % = (x + 80) gallons of 70 %
Thus 440 gallons of 80 % antifreeze solution must be mixed
Answer:
3x - 9 - (6 / (x+4))
Step-by-step explanation:
3x - 9
|----------------------------------------
(x +4) | (3x^2 + 3x - 42 )
3x^2 + 12x
------------------------------
-9x - 42
-9x - 36
-----------------------------------
- 6
Answer: 3x - 9 - (6 / (x+4))
You have to subtract the degrees like in this picture:
Answer:
Step-by-step explanation:
Given the system of two equations:
Multiply the first equation and the second equation by 60 to get rid of fractions:
Now multiply the first equation by 4 and the second equation by 5:
Subtract them:
Substitute it into the first equation:
The solution is