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3 years ago

-4 1/3. Divided by -2 3/5

2 answers:

8 0

$- 4 \frac{1}{3} \div ( - 2 \frac{3}{5}) \\ \\ - \frac{13 }{3} \div ( - \frac{13}{5} ) \\ \\ - \frac{13}{3} \times ( - \frac{5}{13} ) \\ \\ = \frac{5}{3} =1 \frac{2}{3} = 1.6$

I hope I helped you^_^

Marrrta [24]3 years ago

5 0

Answer:

1 2/3

Step-by-step explanation:

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12) g = xy/c , for x

Paladinen [302]

Answer:

c=xy/g happy to help ya:)

Step-by-step explanation:

6 0

2 years ago

What is 2 3/12 minus 1 8/12 equal to?

ArbitrLikvidat [17]

Answer:

7/12

Step-by-step explanation:

3/12=1/4

2 1/4=9/4

8/12=2/3

1 2/3=5/3

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9/4-5/3=27/12-20/12=7/12

6 0

3 years ago

Read 2 more answers

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)

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