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Zigmanuir [339]

3 years ago

5

PLEASE HELP I WILL GIVE BRAINALIST

1 answer:

Oxana [17]3 years ago

7 0

Answer:

Step-by-step explanation:

Original 873 New 781

The new went down by about 10.5%

873*10.5%= 91.67

873-91.67= 781.33

It decreased 10.5%

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Share £200 between Alice and Brian in the ratio 3 : 5<br>

Pepsi [2]

Answer:

75:125=3:5

Step-by-step explanation:

1. Add the total number of parts

3+5=8

2. Find out how much one part is

200/8=25

3. Multiply each ratio by 25

3*25=75

5*25=125

4. Check they add up to 200

75+125=200

8 0

2 years ago

Read 2 more answers

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

Which property is being used here?<br> 5.2345 x 1 = 5.2345

Mamont248 [21]

Multiplication not to sure hope this helps :)

7 0

2 years ago

Solve for p if 2|p| = 4

ehidna [41]

2p = 4
p = 2

-2p = 4
p = -2

answer p = -2 and p =2

3 0

3 years ago

Read 2 more answers

What is the slope of the line?

damaskus [11]

Answer:

- 7/4

Step-by-step explanation:

1) (y2 - y1)/(x2 - x1)

2) pick two points, mine were (1,-3) and (-3,4)

3) Substitute the values (4+3)/(-3-1)

4) Solve and no matter what points you’ll get -7/4

3 0

2 years ago