What is the ratio of the outputs for x = 6 and x = 2?

mel-nik [20]

Check the picture below.

well, it's noteworthy to say that when dropping a perpendicular line from a right angle in a right triangleto the hypotenuse, we'd end up with 3 similar triangles, a Large a Medium and a Small one, all three similar.

3a)

is the Small similar to the Large one? well, let's notice, they both have a 90° angle and also they share the purple one, similar triangles by AA.

3b)

are the Medium and the Large one similar? well, let's notice, just like before, they both have a 90° and they also share the green one, similarity by AA.

3c)

are the Small and Medium similar?

if Large ~ Medium

and

Large ~ Small

then

Medium ~ Small.

4 0

3 years ago

What is the value of x?

Alexxx [7]

Answer:

To find the value of x, bring the variable to the left side and bring all the remaining values to the right side. Simplify the values to find the result.

Step-by-step explanation:

5 0

2 years ago

If a tables patterns is that the f(x) values are being divided by four, what is the common ratio

9966 [12]

Just divide it and it will get your answer

6 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

$x(0) = x_o$

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

The base of a prism is a regular hexagon with the perimeter of 78mm the height of the prism is 16mm what is the area of one of t

kifflom [539]

The perimeter is the sides of the hexagon. One of the hexagon edges is shared with a rectangular face. Solve for that edge. 78/6=13. The height is the length and edge is the width. Area is length(16) times width(13). 16•13=208. The answer is 208mm^2

3 0

4 years ago