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
__
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)
__
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:
48 mm ^ 2
Step-by-step explanation:
We know we have a cube of side x, therefore the volume of that cube will be equal to:
V = x ^ 3, because it is a cube and each side is the same.
Now the derivative of the volume with respect to the length (that is, x) would be:
dV / dx = 3 * (x ^ 2)
They tell us that x takes a value of 4, replacing we have:
dV / dx = 3 * (4 ^ 2) = 48
That is, 48 mm ^ 2 would be the value of dV / dx when x is equal to 4 mm.
The answer to this problem is simple all you have to do is pemdas and you get the answer 0.6
Answer:
see explanation
Step-by-step explanation:
Note that (x² + 5)(x² - 5) are the factors of a difference of squares
a² - b² = (a + b)(a - b)
with a = x² and b = 5, hence
(x² + 5)(x² - 5) = (x²)² - 5² = 
- 25, thus
3( - 25) = 3 - 75
