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Simora [160]
11 months ago
7

What does the constant 0.9955 reveal about the rate of change of the quantity?

Mathematics
1 answer:
cricket20 [7]11 months ago
7 0

The model given is:

f(t)=290(0.9955)^{60t}

The term in the parentheses represents the change of rate of the model. If the number is the parentheses is less than 1, we call it <em>exponential decay</em>.

If the number in parentheses is bigger than 1, we call it exponential growth.

In this case, the correct option is "decaying"

The number in parentheses let's call it R, is:

R=1-r

Where r is the rate of change. To find it:

\begin{gathered} 0.9955=1-r \\ . \\ r=1-0.9955=0.0045 \end{gathered}

To convert to percentage, we convert by multiplying by 100::

0.0045\cdot100=0.45\%

The second answer is 0.45%

And since t is the time passed in hours, and has a "60" multiplying it, the last answer is: Every 60 hours

<em>The final answer is:</em>

The function is exponentially decaying at a rate of 0.45% evary 60 hours.

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A is a graph that is unbroken
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3 years ago
(1 point) A fish tank initially contains 15 liters of pure water. Brine of constant, but unknown, concentration of salt is flowi
Drupady [299]

Answer:

a. \dfrac{dx}{dt} = 6\frac{2}{3} \cdot c

b. x(t) = 6\frac{2}{3} \cdot c \cdot t

c. c  = \dfrac{3}{8}  \ g/L

Step-by-step explanation:

a. The volume of water initially in the fish tank = 15 liters

The volume of brine added per minute = 5 liters per minute

The rate at which the mixture is drained = 5 liters per minute

The amount of salt in the fish tank after t minutes = x

Where the volume of water with x grams of salt = 15 liters

dx =  (5·c - 5·c/3)×dt = 20/3·c = 6\frac{2}{3} \cdot c \cdot dt

\dfrac{dx}{dt} = 6\frac{2}{3} \cdot c

b. The amount of salt, x after t minutes is given by the relation

\dfrac{dx}{dt} = 6\frac{2}{3} \cdot c

dx = 6\frac{2}{3} \cdot c \cdot dt

x(t) = \int\limits \, dx  = \int\limits \left ( 6\frac{2}{3} \cdot c \right) \cdot dt

x(t) = 6\frac{2}{3} \cdot c \cdot t

c. Given that in 10 minutes, the amount of salt in the tank = 25 grams, and the volume is 15 liters, we have;

x(10) = 25 \ grams(15 \ in \ liters) = 6\frac{2}{3} \times c \times 10

6\frac{2}{3} \times c  =\dfrac{25 \ grams }{10}

c  =\dfrac{25 \ g/L }{10 \times 6\frac{2}{3} }  = \dfrac{25 \ g/L}{10 \times \dfrac{20}{3} } =\dfrac{3}{200} \times 25 \ g/L= \dfrac{75}{200}  \ g/L = \dfrac{3}{8}  \ g/L

c  = \dfrac{3}{8}  \ g/L

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2 years ago
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