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:
cos R = .38
Step-by-step explanation:
First you need to find the length of the hypotenuse.
Use a squared + b squared = c squared
100 + 576 = c squared
square root of 676 is 26
CosR is adjacent over hypotenuse.
Adjacent is 10
Hypotenuse is 26
10/26 = .38
Answer:
r = - 13 / 15 inches/week
Step-by-step explanation:
The numerator is 13
and the denominator is 15, then we have
r = - 13 / 15 inches/week
r < 0 since the water level of the lake fell
First we have to divide 135,000 by 90, which equals 1,500. Then we divide 1,500 (seconds) by 60 (60 seconds in a minute) and get: 25 Minutes.
