Answer:
- a(x) = 20 + 0.60x
- domain [0, 50]; range [20, 50]
- maybe
Step-by-step explanation:
a) If x liters are removed from a container with a volume of 50 L, the amount remaining in the container is (50 -x) liters. Of that amount, 40% is acid, so the acid in the container before any more is added will be ...
0.40 × (50 -x)
The x liters are replaced with 100% acid, so the amount of acid that was added to the container is ...
1.00 × (x)
Then after the remove/replace operation, the total amount of acid in the container is ...
a(x) = 0.40(50 -x) +1.00(x)
a(x) = 20 +0.60x . . . . . liters of acid in the final mixture
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b) The quantity removed cannot be less than zero, nor can it be more than 50 liters. The useful domain of the function is 0 ≤ x ≤ 50. (liters)
The associated range is 20 ≤ a ≤ 50. (liters)
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c) As we found in part b, the amount of acid in the final mixture may range from 20 liters to 50 liters. So, the percentage of acid in the final mix will range from 20/50 = 40% to 50/50 = 100%. The mixture could be 50% acid, but is not necessarily.
Answer:
Option A- 1.8x – 10 = –4; x = 1.8 x minus 10 equals negative 4; x equals StartFraction 10 Over 2 EndFraction.
Answer:
$48
Step-by-step explanation:
First Hour: $0+$2
Second Hour: $1+$1
Every Hour is $2
$2 times by 24 equal $48
Answer:
1/2 = r
Step-by-step explanation:
7r + 21 = 49r
Subtract 7r from each side
7r-7r + 21 = 49r-7r
21 = 42r
Divide each side by 2
21/42 = 42r/42
1/2 = r
Answer:
18 but look at others awnsers aswell
Step-by-step explanation: