How do you write (17.5,35.5) in a fraction?

an airplane travels 600 miles against the wind in 5 hours, and makes the return trip with the same wind in 2 hours. find the rat

vesna_86 [32]

Answer:

The speed of wind is 75 miles / hour

Step-by-step explanation:

Let's assume

speed of airplane is x miles /hour

speed of wind is y miles /hour

An airplane travels 600 miles against the wind in 4 hours

against wind means resultant speed will be less

so, speed =x-y

total distance traveled in 4 hour is

and this has to be equal to 600

Divide both sides by 4

makes the return trip with then same wind in 2 hours

with wind means resultant speed will be more

so, speed =x+y

total distance traveled in 2 hour is

and this has to be equal to 600

Divide both sides by 2

So, we get system of equations

now, we can solve it by adding both equations

now, we can solve for x

now, we can find y by plugging x into any one equation

4 0

1 year 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

$x(0) = x_o$

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

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

5 0

3 years ago

HELP QUICK

ASHA 777 [7]

Answer:

Step-by-step explanation:

3x + 6 = 2x

3x - 2x = -6

x = -6

4 0

3 years ago

Angle X measures 110 Degrees if angle x is rotated 60 degrees what is the measure of angle x ?

Flauer [41]

Rotatation could be done to left or right.

110 + 60 = 170 degrees => Rotation to the right side.

110 - 60 = 50 degrees= > Rotation to the left side.

We can see in the choices that 170 degrees is the right choice.

The answer is D.

4 0

3 years ago

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Help me please. I’m getting frustrated. I keep writing out the equation wrong and don’t know where I’m messing up Write all solu

MrMuchimi

Answer:

Infinite Solutions

Step-by-step explanation:

If you distribute the left side of the expression, you will get 3y -32, which is exactly the same as the right side of the equation. Since both lines are the same, that means that they have an infinite amount of solutions. Therefore, the 4th choice is correct.

3 0

3 years ago

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