how many gallons of a 80% antifreeze solution must be mixed with 70 gallons of 30% antifreeze to get a mixture that is 70% antifreeze
Answers
x+80=gallons go final 70% mixture
..
80%*x+30%*80=70%*(x+80)
.8x+24=.7x+56
.1x=32
x=320
gallons of 80% solution to be mixed=320
Final answer:
To obtain a mixture that is 70% antifreeze, you need to mix 70 gallons of an 80% antifreeze solution with 70 gallons of a 30% antifreeze solution.
Explanation:
To determine how many gallons of an 80% antifreeze solution must be mixed with 70 gallons of 30% antifreeze to get a mixture that is 70% antifreeze, we can use the concept of the concentration of solutions.
Let's assume x gallons of the 80% antifreeze solution need to be added.
The amount of antifreeze in the 80% solution is 0.8x gallons, and the amount of antifreeze in the 30% solution is 0.3 * 70 = 21 gallons. The total amount of antifreeze in the final mixture is 0.7 * (70 + x) gallons.
Setting up the equation: 0.8x + 21 = 0.7 * (70 + x).
Solving for x, we get x = 70 gallons. Therefore, you need to mix 70 gallons of the 80% antifreeze solution with 70 gallons of the 30% antifreeze solution to get a mixture that is 70% antifreeze.
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2. 1's or 0's
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Answer:
Step-by-step explanation:
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3. Identity matrix,invertible matrix, triangular matrix or zero matrix.
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Answer:
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Answers
Answer:
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Step-by-step explanation: hope this helps!!! :)
Answer:
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Step-by-step explanation:
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