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olga nikolaevna [1]

3 years ago

14

Quick!! I need help with this question!

1 answer:

d1i1m1o1n [39]3 years ago

3 0

1. 60 / 1/8 * 3/4

2. 3/4 * 15

3. 7/8 / 3/4

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Which number is NOT in the solution set of x-8 < 14?<br> A)17<br> B)19<br> C)21<br> D) 23

laiz [17]

answer

d.2e

Step-by-step explanation:

x<22

a,b,c are all in except d.

4 0

2 years ago

X ≥ 84 how do i do this explain please.

pogonyaev

Answer:

you can put anything to replace x that is greater or equal to 84.

For example:

84 ≥ 84

91 ≥ 84
102 ≥ 84

It just has to be greater or equal to 84.

> = greater than

< = less than

≥ = greater than or equal to

≤ = less than or equal to

Hope this helpss! :)

3 0

3 years ago

Can someone explain how this is done?

Musya8 [376]

Answer:

Just put in the values in the equation for example 4(1)-5<7 and solve it if the eqution is true at the end than that the right answer

Step-by-step explanation:

3 0

3 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

Find x and y<br><br> Don’t argue in the comments I actually need help

igor_vitrenko [27]

X=132

To find y
180-132= 48
180-68-48=64
y=64

7 0

2 years ago

Read 2 more answers