Answer:
dA/dt = k1(M-A) - k2(A)
Step-by-step explanation:
If M denote the total amount of the subject and A is the amount memorized, the amount that is left to be memorized is (M-A)
Then, we can write the sentence "the rate at which a subject is memorized is assumed to be proportional to the amount that is left to be memorized" as:
Rate Memorized = k1(M-A)
Where k1 is the constant of proportionality for the rate at which material is memorized.
At the same way, we can write the sentence: "the rate at which material is forgotten is proportional to the amount memorized" as:
Rate forgotten = k2(A)
Where k2 is the constant of proportionality for the rate at which material is forgotten.
Finally, the differential equation for the amount A(t) is equal to:
dA/dt = Rate Memorized - Rate Forgotten
dA/dt = k1(M-A) - k2(A)
Answer:
(i) (f - g)(x) = x² + 2·x + 1
(ii) (f + g)(x) = x² + 4·x + 3
(iii) (f·g)(x) = x³ + 4·x² + 5·x + 2
Step-by-step explanation:
The given functions are;
f(x) = x² + 3·x + 2
g(x) = x + 1
(i) (f - g)(x) = f(x) - g(x)
∴ (f - g)(x) = x² + 3·x + 2 - (x + 1) = x² + 3·x + 2 - x - 1 = x² + 2·x + 1
(f - g)(x) = x² + 2·x + 1
(ii) (f + g)(x) = f(x) + g(x)
∴ (f + g)(x) = x² + 3·x + 2 + (x + 1) = x² + 3·x + 2 + x + 1 = x² + 4·x + 3
(f + g)(x) = x² + 4·x + 3
(iii) (f·g)(x) = f(x) × g(x)
∴ (f·g)(x) = (x² + 3·x + 2) × (x + 1) = x³ + 3·x² + 2·x + x² + 3·x + 2 = x³ + 4·x² + 5·x + 2
(f·g)(x) = x³ + 4·x² + 5·x + 2
Answer: The third one
Step-by-step explanation: I just did the question on edge. ,and I got it right.
Answer:
160 is the answer please tell me if im wrong
Answer:
6 blooms
Step-by-step explanation:
40 x 15%
40 x 0.15 = 6
