Write 3.29 as a fraction or a mixed number

Alchen [17]

(B) 8 less than the product of 3 and n

Here, we are given an equation 3n - 8

Let us consider each of the options one by one

(A) the product of 3, n and 8 will be written as 24n. Hence this isn't the correct answer.

(B) 8 less than the product of 3 and n will be written as 3n - 8. This matches our equation and hence is the correct answer.

(C) n minus 8 times 3 will be written as 24 - n. Hence, this is also incorrect.

(D) 3 times n less than 8 will be written as 8 - 3n. Thus, this is also eliminated.

Therefore, option B is the correct answer.

Learn more about expressing equations here-

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6 0

2 years ago

The question is in the picture

GrogVix [38]

Ten-thousand AFAFANDIUAHDIUAHIUDHAIDh (had to write 20 more "letters")

8 0

3 years ago

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)

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