Simplify (2/3)^2∙81 please explain 2

Mathematics

1 answer:

SVETLANKA909090 [29]3 years ago

8 0

Answer:

Step-by-step explanation:

Simplify the following:

(2/3)^2 81

Hint: | Simplify (2/3)^2 using the rule (p/q)^n = p^n/q^n.

(2/3)^2 = 2^2/3^2:

2^2/3^2 81

Hint: | Evaluate 2^2.

2^2 = 4:

4/3^2 81

Hint: | Evaluate 3^2.

3^2 = 9:

4/9×81

Hint: | Express 4/9×81 as a single fraction.

4/9×81 = (4×81)/9:

(4×81)/9

Hint: | In (4×81)/9, divide 81 in the numerator by 9 in the denominator.

81/9 = (9×9)/9 = 9:

4×9

Hint: | Multiply 4 and 9 together.

4×9 = 36:

Answer: 36

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Litter such as leaves falls to the forest floor, where the action of insects and bacteria initiates the decay process. Let A be

Travka [436]

Answer:

D = L/k

Step-by-step explanation:

Since A represents the amount of litter present in grams per square meter as a function of time in years, the net rate of litter present is

dA/dt = in flow - out flow

Since litter falls at a constant rate of L grams per square meter per year, in flow = L

Since litter decays at a constant proportional rate of k per year, the total amount of litter decay per square meter per year is A × k = Ak = out flow

So,

dA/dt = in flow - out flow

dA/dt = L - Ak

Separating the variables, we have

dA/(L - Ak) = dt

Integrating, we have

∫-kdA/-k(L - Ak) = ∫dt

1/k∫-kdA/(L - Ak) = ∫dt

1/k㏑(L - Ak) = t + C

㏑(L - Ak) = kt + kC

㏑(L - Ak) = kt + C' (C' = kC)

taking exponents of both sides, we have

$L - Ak = e^{kt + C'} \\L - Ak = e^{kt}e^{C'}\\L - Ak = C"e^{kt} (C" = e^{C'} )\\Ak = L - C"e^{kt}\\A = \frac{L}{k} - \frac{C"}{k} e^{kt}$

When t = 0, A(0) = 0 (since the forest floor is initially clear)

$A = \frac{L}{k} - \frac{C"}{k} e^{kt}\\0 = \frac{L}{k} - \frac{C"}{k} e^{k0}\\0 = \frac{L}{k} - \frac{C"}{k} e^{0}\\\frac{L}{k} = \frac{C"}{k} \\C" = L$