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allochka39001 [22]

2 years ago

8

10. The perimeter of a triangle is 48'. The sides are in the ratio of 1 to 2 to 3. Find the three sides.

1 answer:

pentagon [3]2 years ago

4 0

Answer:

Step-by-step explanation:

The sides are in the ratio 1:2:3, so the lengths of the sides are 1x, 2x, 3x.

1x + 2x + 3x = 48'

6x = 48'

x = 8'

The lengths of the sides are 8', 16', 24'

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What is the product of (-a + 3)(a + 4)?

pantera1 [17]

Answer:

-a^2-a+12

Step-by-step explanation:

thats the product of (-a + 3)(a + 4)

5 0

2 years ago

Alex bought 4 hoodies. Each hoodie cost the same amount and they were on sale for $5 off each. Alex paid a total of $120. How mu

vladimir2022 [97]

Total price equals 4 hoodies times (original price minus $5).

x= original hoodie price

$120= 4(x-5)
Use distributive property to multiply
120=(4*x) - (4*5)
120= 4x - 20
Add 20 to both sides
140=4x
Divide both sides by 4
$35= x

Original hoodie price was $35.

Hope this helps! :)

8 0

2 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

$\lim_{t \to \infty} x(t) = x_oe^{\frac{-\alpha y_o}{\beta } }$

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3 years ago

I need help asap D:

Airida [17]

Answer:

F

Step-by-step explanation:

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2 years ago

Find five rational numbers between −2 and 0

Harlamova29_29 [7]

Answer:

-9/5 , -8/5 , -7/5, -6/5 , -5/5.

Step-by-step explanation:

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3 years ago