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A pharmacist has 40% and 80% of iodine solutions on hand. How many liters of each iodine solution will be required to produce 4 liters of a 50% iodine mixture?
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Let x = liters of 40% iodine
then
4-x = liters of 80% iodine
Using algebra:
.40x + .80(4-x) = .50(4)
.40x + 3.20-.80x = 2
3.20-.40x = 2
x = 4 liters (40% iodine)
80% iodine:
4-x = 4-4 = 0 liters needed (80% iodine)
0.6n = n + 31.8
0.6n - n = 31.8
-0.4n = 31.8
n = 31.8 / -0.4
n = -79.5
Answer:
ok.......what do you want me to answer
have a good day:)
Step-by-step explanation:
F = 30
g = 15
Yeah, hope this helps, etc.
