Ask question

Ask question

All categories

3 years ago

5

30th term: a = 4, d = 12

1 answer:

mezya [45]3 years ago

8 0

Answer:

120

Step-by-step explanation:

4*30=120

You might be interested in

Translate this sentence into an equation.

Aliun [14]

14+m=56
m=56-14(minus 56 from both side)
M=42

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:

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

Need help right now !!!!! <br> Find the missing part

oksian1 [2.3K]

What kind of mathematics are you learning?

3 0

3 years ago

Suppose you pick a tile out of a bag containing 26 tiles marked with the letters A through Z. What is the probability of picking

Liula [17]

Given that <span>a bag contains 26 tiles marked with the letters A through Z.

The probability of picking a letter from the name JACK is 4 / 26

The probability of picking a letter from the name BEN is 3 / 26.

Therefore, the probability of picking a letter from the name JACK or from the name BEN iis given by 4 / 26 + 3 / 26 = 7 / 26 </span>

6 0

3 years ago

Evaluate the expression for b = 0 and c= -11.<br> bc + b

avanturin [10]

Answer:

If b = 0 and c = -11, all you have to do is plug in the values into bc + b:

Step-by-step explanation:

(0)(-11) + 0 = 0

3 0

2 years ago

Read 2 more answers