All categories

3 years ago

Classify each pair of lines as parallel, perpendicular, coinciding, or intersecting at a non-right angle:

1 answer:

7 0

M= y2-y1/x2-x1

Line 1: 3-(-5) / 0-4 = -2
Line 2: 5-11 / (-1)-(-4) = -2

Since the slopes are the same they are parallel

You might be interested in

What is the number for x

FinnZ [79.3K]

Wut??????????????????

5 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:

$y(t) = y_o e^{\beta t}$

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

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

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 $e^c = C$

$y(t) = C e^{\beta t}$

Now from the question we are told that

$y(0) = y_o$

Hence

$y(0) = y_o = Ce^{\beta * 0}$

=> $y_o = C$

$y(t) = y_o e^{\beta t}$

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

$x(0) = x_o$

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

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

divide both side by $(K * x_o)$

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

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

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

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

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

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

5 0

3 years ago

If K is the midpoint of JL, JK = 8x+11 and KL = 14x-1, find JL.

gladu [14]

Step-by-step explanation:

Since, K is the midpoint of JL.

$\therefore \: JK = KL \\ 8x + 11 = 14x - 1 \\ 11 + 1 = 14x - 8x \\ 12 = 6x \\ \frac{12}{6} = x \\ 2 = x \\ \\ \because \: JL = JK + KL \\ \therefore \: JL = 8x + 11 + 14x - 1 \\ \therefore \: JL = 8x + 14x + 11 - 1 \\ \therefore \: JL = 22x + 10\\ \therefore \: JL = 22 \times 2 + 10..(plug \: x = 2) \\ \therefore \:JL = 44 + 10 \\ \huge \red{ \boxed{\therefore \:JL = 54}}$