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kvasek [131]
3 years ago
5

Given the function ƒ(x) = 3x + 5, find ƒ(4) and x such that ƒ(x) = 38.

Mathematics
1 answer:
MissTica3 years ago
5 0

Answer: A.17;11

f(4)= 3(4) + 5

=12 + 5

=17

And, f(x)= 38

=>3x+5=38

=>3x= 38-5

=>3x=33

=>x = 33÷3

=>x = 11

So, A.17;11

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igomit [66]

Answer:

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I hope I helped you^_^

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3 years ago
Part 1 You will need to measure five different people. Record your measurements on a piece of paper. Using a tape measure or rul
Jlenok [28]

Answer:

Part 1 and 2:  

A. Forearm (in)        Foot (in)                      

Sarah   10                9

Sam   11.               10

Emma  8                  7

Mom   10                 9

Dad  11                 10  

B. M=9-7 over 8-10 the answer is 2 over 2.

 

C.  For every forearm the foot is one inch smaller.

Part 3:

A. 0.860(17)+3.302

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Step-by-step explanation:

This is just an example dont you the names

8 0
3 years ago
The use of mathematical methods to study the spread of contagious diseases goes back at least to some work by Daniel Bernoulli i
harina [27]

Answer:

a

   y(t) = y_o e^{\beta t}

b

      x(t) =  x_o e^{\frac{-\alpha y_o }{\beta }[e^{-\beta t} - 1] }

c

      \lim_{t \to \infty} x(t) = x_oe^{\frac{-\alpha y_o}{\beta } }

Step-by-step explanation:

From the question we are told that

    \frac{dy}{y} =  -\beta dt

Now integrating both sides

     ln y  =  \beta t + c

Now taking the exponent of both sides

       y(t) =  e^{\beta t + c}

=>     y(t) =  e^{\beta t} e^c

Let  e^c =  C

So

      y(t) = C e^{\beta t}

Now  from the question we are told that

      y(0) =  y_o

Hence

        y(0) = y_o  = Ce^{\beta * 0}

=>     y_o = C

So

        y(t) = y_o e^{\beta t}

From the question we are told that

      \frac{dx}{dt}  = -\alpha xy

substituting for y

      \frac{dx}{dt}  = - \alpha x(y_o e^{-\beta t })

=>   \frac{dx}{x}  = -\alpha y_oe^{-\beta t} dt

Now integrating both sides

         lnx = \alpha \frac{y_o}{\beta } e^{-\beta t} + c

Now taking the exponent of both sides

        x(t) = e^{\alpha \frac{y_o}{\beta } e^{-\beta t} + c}

=>     x(t) = e^{\alpha \frac{y_o}{\beta } e^{-\beta t} } e^c

Let  e^c  =  A

=>  x(t) =K e^{\alpha \frac{y_o}{\beta } e^{-\beta t} }

Now  from the question we are told that

      x(0) =  x_o

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      x(0)=x_o =K e^{\alpha \frac{y_o}{\beta } e^{-\beta * 0} }

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divide both side  by    (K * x_o)

=>    K = x_o e^{\frac {\alpha y_o  }{\beta } }

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=>    x(t) =  x_o e^{\frac{\alpha y_o }{\beta }[e^{-\beta t} - 1] }

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so

    \lim_{t \to \infty} x(t) = x_oe^{\frac{-\alpha y_o}{\beta } }

5 0
3 years ago
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Answer :

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Step-by-step explanation:

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A two and a half pound of grass seeds covers one eight of an acre.
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Answer: 60 lbs?

Step-by-step explanation:

3 0
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