Explain how complex fractions can be used to solve problems involving ratios?

1 answer:

8 0

Complex fractions are numbers in which the numerator and the denominator are both fractions. in this case, to solve the ratio, we can multiply the numerator fraction by the reciprocal of the denominator fraction. Another way is to solve the fraction separately and then divide eventually.

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A cooler has a temperature of 32 degrees Fahrenheit. A bottled drink is placed in the cooler with an initial temperature of

FinnZ [79.3K]

Answer:

$k \approx 0.44$

Step-by-step explanation:

Given function:

$f(t) = (ce)^{-kt}+32$

As per question statement:

Initial temperature of bottle is 70 $^\circ F$ .

i.e. when time = 0 minutes, f(t) = 70 $^\circ F$

$70 = ce^{-k\times 0}+32\\\Rightarrow 38 = ce^{0}\\\Rightarrow c = 38$

After t = 3, f(t) = 42 $^\circ F$

$42 = 38 \times e^{-k\times 3}+32\\\Rightarrow 42-32 = 38 \times e^{-3k} \\\Rightarrow 10 = 38 \times e^{-3k} \\\Rightarrow e^{3k} = \dfrac{38}{10}\\\Rightarrow e^{3k} = 3.8\\\\\text{Taking } log_e \text{both the sides:}\\\\\Rightarrow log_e {e^3k} = log_e {3.8}\\\Rightarrow 3k \times log_ee=log_e {3.8} (\because log_pq^r=r \times log_pq)\\\Rightarrow 3k \times 1=log_e {3.8}\\\Rightarrow 3k = 1.34\\\Rightarrow k = \dfrac{1.34}{3}\\\Rightarrow k \approx 0.44$