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expeople1 [14]
2 years ago
5

It takes 3 eggs to make two dozen cookies. About what fraction of an egg are in two cookies

Mathematics
1 answer:
wariber [46]2 years ago
4 0

keeping in mind that two dozen cookies is just 24 cookies, then

\begin{array}{ccll} eggs&cookies\\ \cline{1-2} 3 & 24\\ x& 2 \end{array} \implies \cfrac{3}{x}=\cfrac{24}{2}\implies \cfrac{3}{x}=12\implies 3=12x \\\\\\ \cfrac{3}{12}=x\implies \cfrac{1}{4}=x

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Step-by-step explanation:

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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

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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

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Separating the variables, we have

dA/(L - Ak) = dt

Integrating, we have

∫-kdA/-k(L - Ak) = ∫dt

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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}

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A = \frac{L}{k}  - \frac{L}{k} e^{kt}

So, D = R - A =

D = \frac{L}{k} - \frac{L}{k}  - \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{kt}

when t = 0(at initial time), the initial value of D =

D = \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{k0}\\D = \frac{L}{k} e^{0}\\D = \frac{L}{k}

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3 years ago
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