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Finger [1]
1 year ago
12

Given that f(x) = x + 3 a) Find f(2)

Mathematics
1 answer:
Nataly_w [17]1 year ago
7 0

{ \qquad\qquad\huge\underline{{\sf Answer}}}

Here we go ~

In the above question, it is given that :

\qquad \sf  \boxed{ \sf f(x) =  \frac{x + 3}{2} }

A.) Find f(2) :

\qquad \sf  \dashrightarrow \: f(2) =  \dfrac{2 + 3}{2}

\qquad \sf  \dashrightarrow \: f(2) =  \dfrac{5}{2}

or

\qquad \sf  \dashrightarrow \: f(2) = 0.5

B.) Find { \sf {f}^{-1}(x) } :

\qquad \sf  \dashrightarrow \: let \: y = f (x)

so, we can write it as :

\qquad \sf  \dashrightarrow \: y =  \dfrac{x + 3}{2}

\qquad \sf  \dashrightarrow \: 2y = x + 3

\qquad \sf  \dashrightarrow \: x = 2y - 3

Now, put x = { \sf {f}^{-1}(x) }, and y = x and we will get our required inverse function ~

\qquad \sf  \dashrightarrow \: f {}^{ - 1}(x) = 2x- 3

C.) Find { \sf {f}^{-1}(12) } :

\qquad \sf  \dashrightarrow \: f {}^{ - 1}(12) = 2(12)- 3

\qquad \sf  \dashrightarrow \: f {}^{ - 1}(12) = 36- 3

\qquad \sf  \dashrightarrow \: f {}^{ - 1}(12) = 33

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Answer:

\dfrac{dy}{dt}=0.27-0.009y(t),$  y(0)=60kg

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<u />

<u>Rate In</u>

A solution of concentration 0.03 kg of salt per liter enters a tank at the rate 9 L/min.

R_{in} =(concentration of salt in inflow)(input rate of solution)

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Concentration, C(t)=\dfrac{Amount}{Volume} =\dfrac{y(t)}{1000}

R_{out} =(concentration of salt in outflow)(output rate of solution)

=\dfrac{y(t)}{1000}* 9\dfrac{liter}{min}=0.009y(t)\dfrac{kg}{min}

Therefore, the differential equation for the amount of Salt in the Tank  at any time t:

\dfrac{dy}{dt}=R_{in}-R_{out}\\\\\dfrac{dy}{dt}=0.27-0.009y(t),$  y(0)=60kg

8 0
4 years ago
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Step-by-step explanation:

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