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tamaranim1 [39]

2 years ago

5

Taylor’s gymnastics team is selling 5 raffle tickets for $8.00 What is the constant of proportionality of dollars per ticket? (o

ptions below)
$40.00 per ticket

$8.00 per ticket

$1.60 per ticket

$0.63 pet ticket

1 answer:

choli [55]2 years ago

5 0

It is $1.60 per ticket because you would have to divide 5 by 8 which would give you 1.6

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Find the slope of the line that contains the following pairs of points (2,4) (0,8)

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Answer:

The slope is -2

Step-by-step explanation:

The slope of a line is given by

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= (8-4)/(0-2)

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3 years ago

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Add all the fractions together:
to do this find the lcd:
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3 years ago

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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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3 years ago

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goldenfox [79]

Answer:

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5/24 = 0.208 percentage of defective bolts

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A spring is attatched at one end to support B and at the other end to collar A, as represented in the figure. Collar A slides al

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Step-by-step explanation:

cos θ = 3 / AB

When θ = 28°:

cos 28° = 3 / AB

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θ = cos⁻¹(3 / 5.2)

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3 years ago