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Maksim231197 [3]

3 years ago

12

What is the value of 9 (12) + 5

2 answers:

kati45 [8]3 years ago

6 0

The answer to this is 113

ASHA 777 [7]3 years ago

3 0

9 (12) +5

9 x 12= 108

108+ 5= 113

Hope this helps!

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Andrei weighs 32 kg, which is 4/7 of the weight of his older sister and 2/5 of the weight of his father. The mother’s weight is

Leya [2.2K]

Andrei 32kg
Older sister 56kg
Father 80kg
Mother 70kg
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Total 255kg
Answer: Yes they can all go into the elevator together.

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4 years ago

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Someone pleaseee heelp me with this!!!! WILL GIVE BRAINLISEST IF CORRECT!!!!!!!!!!

svlad2 [7]

Wow Briana... sly.... i was just gonna look it up on the website elsner got it from and fine the answeer keys there XD

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5 0

4 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

4 years ago

One end of a ramp is raised to the back of a truck 1 meter

Bogdan [553]

Answer:

26.6°

Step-by-step explanation:

8 0

3 years ago

HELP ASAP MARKING BRAINLEST<br><br> use the diagram to write the standard equation of the circle.

barxatty [35]

Answer:

your answer is:

(x-2)^2 +(y-2)^2=16

Hope this helps.

Brainliest plz :)

8 0

3 years ago