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3 years ago

9

Find the 50th term of the sequence 5, -2, -9, -16....

1 answer:

antiseptic1488 [7]3 years ago

4 0

Yes, it would be. Because if u keep misusing 7 or -7 50 times starting from 5 it will equal -338.

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Expand and simplify<br>2(2y + 1) + 2(5y - 3)

GarryVolchara [31]

Answer:

14y-4

Step-by-step explanation:

Distribute 2 to both the left and right side.

Combine like terms.

14y-4

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

The table shows the number of miles that a car travels and the gallons of gasoline it uses to go those distances.

PtichkaEL [24]

Answer:

4 -.-

Step-by-step explanation:

8 0

2 years ago

Ruth wants to buy a skateboard for $56. She has $90 in her account. She spent $11.65 to buy stationary. She also wants to buy so

nevsk [136]

The answer is the second one I dont have that symbol on my keyboard but it’s the second answer on the multi choice
90-11.65-56=22.35
divide that by 3.72 u get 6 so she can only get six comics or any whole positive number lower than that. Therefore it’s #2

5 0

3 years ago

Read 2 more answers

What is the solution set to the following system?

elena-s [515]

The correct answer is C
3+ 0 is 3
3^2 + 0 is 9

3 0

3 years ago