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Sphinxa [80]
3 years ago
12

A large container for water weighs 21 pounds. If a gallon of water weighs 8.34 pounds, what is an equation that relates the weig

ht of the
water and the container, W. in pounds and the amount of water, A, in gallons? (1 point)
W
Mathematics
1 answer:
amid [387]3 years ago
7 0

Answer:

The equation that relates the weight of the water and the container in pounds and the amount of water in gallons is A = \frac{W}{8.34}.

Step-by-step explanation:

From Physics we understand that conversions are represented by the following linear equation:

u' = r\cdot u (1)

Where:

u - Original value, measured in original unit. (pounds)

r - Conversion value, measured in final unit per original unit. (gallons per pound)

u' - Final value, measured in final unit. (gallons)

If we know that u' = A, u = W and r = \frac{1}{8.34} \,\frac{gal}{lbm}, then the equation is:

A = \frac{W}{8.34}

The equation that relates the weight of the water and the container in pounds and the amount of water in gallons is A = \frac{W}{8.34}.

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A triangular parking lot has two sides that are the same length and the third side is 5 m longer. If the perimeter is 71 m, find
Scilla [17]

Answer:

The equation can be expressed as;

x+y=71 where,

Lengths of two equal sides=x=22 m

Lengths of longer side=y=27 m

Step-by-step explanation:

The total perimeter can be expressed as;

Total perimeter=Total length of the equal sides+total length of the longer side

where;

Total perimeter=71 m

Total length of each of the equal sides=x m

Total length of longer side=y=(x+5) m

replacing;

x+y=71...equation 1

71=(2×x)+(x+5)

2 x+x=71-5

3 x=66

x=66/3

x=22 m

y=x+5=22+5=27 m

Lengths of two equal sides=22 m

Lengths of longer side=27 m

Equation;

x+y=71

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3 years ago
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Answer:

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Step-by-step explanation:

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The use of mathematical methods to study the spread of contagious diseases goes back at least to some work by Daniel Bernoulli i
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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

So  

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

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

divide both side  by    (K * x_o)

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

So

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

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

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

Generally as  t tends to infinity ,  e^{- \beta t} tends to zero  

so

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

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