Ask question

Ask question

All categories

3 years ago

5

What is k in terms of n?

2 answers:

soldier1979 [14.2K]3 years ago

4 0

Answer:

i think its newton.

Step-by-step explanation:

topjm [15]3 years ago

4 0

My answer is Newton

You might be interested in

20 POINTS help me with this question please

Leto [7]

<span>1.) Factor the denominator into 2 expressions leaves you with:

(x-2)/(x+5)(x-2)

x-2 cancels out, you are left with 1/(x+5), excluding x=2
</span>

7 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

Plz help?!!? it's due by 11

Jobisdone [24]

Step-by-step explanation:

1. x^2 + 10x + 25

2. x^2 - 25

.........

7 0

2 years ago

Read 2 more answers

Use matrix addition to solve this equation: B + 15 −7 4 0 1 2 = 1 2 12 4 0 2 b11 = b12 = b13 = b21 = 4 b22 = −1 b23 = 0

Arada [10]

Answer:

$b_{11}=-14,b_{12}=9,b_{13}=8,b_{21}=4,b_{22}=-1,b_{23}=0$

Step-by-step explanation:

The given matrix addition is

$B+\begin{bmatrix}15&-7&4\\ 0&1&2\end{bmatrix}=\begin{bmatrix}1&2&12\\ 4&0&2\end{bmatrix}$

We need to find the elements of matrix B.

Let $B=\begin{bmatrix}b_{11}&b_{12}&b_{13}\\ b_{21}&b_{22}&b_{23}\end{bmatrix}$

Substitute the value of matrix.

$\begin{bmatrix}b_{11}&b_{12}&b_{13}\\ b_{21}&b_{22}&b_{23}\end{bmatrix}+\begin{bmatrix}15&-7&4\\ 0&1&2\end{bmatrix}=\begin{bmatrix}1&2&12\\ 4&0&2\end{bmatrix}$

After addition of two matrix we get

$\begin{bmatrix}b_{11}+15&b_{12}-7&b_{13}+4\\ b_{21}+0&b_{22}+1&b_{23}+2\end{bmatrix}=\begin{bmatrix}1&2&12\\ 4&0&2\end{bmatrix}$

On equating both sides.

$b_{11}+15=1\Rightarrow b_{11}=-14$

$b_{12}-7=2\Rightarrow b_{12}=9$

$b_{13}+4=12\Rightarrow b_{13}=8$

$b_{21}+0=4\Rightarrow b_{21}=4$

$b_{22}+1=0\Rightarrow b_{22}=-1$

$b_{23}+2=2\Rightarrow b_{23}=0$

Therefore, the elements of matrix B are $b_{11}=-14,b_{12}=9,b_{13}=8,b_{21}=4,b_{22}=-1,b_{23}=0$ .

3 0

3 years ago

The circumference of a circle is . What is the area of the circle in terms of ?

LenKa [72]

Answer:

$\frac{225}{4}$ π

Step-by-step explanation:

Formula for the circumference of a circle

2r(π)

Area of a circle

πr²

2r(π)=x

x/2(π)=r

π(x/2(π))²=Area

π(x²/4(π)²)=Area

(x²/4(π))=Area

(15π)²÷4π=Area

$\frac{225}{4}$ π

7 0

3 years ago