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3 years ago

7

A photographer charges $37 to travel to a client's location, plus $29 per hour for his

1 answer:

Burka [1]3 years ago

8 0

Answer:

2 hours

Step-by-step explanation:

95-37=58

58÷29=2

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Joe is building a fort for his little brother out of boxes. He is wondering how much room there will be inside the fort.

ser-zykov [4K]

Answer:

Depends on how wide and tall it is

Step-by-step explanation:

5 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

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

5 0

3 years ago

Hii please help i’ll give brainliest

pshichka [43]

Answer:

maybe B

Step-by-step explanation:

cuz interquaInterquartile is just a range of q1-q3

7 0

3 years ago

Target is having a sale. All make-up is 30% off. Alissa spends $45 on make-up. How much will Alissa save?

baherus [9]

Answer:

$13.50

Step-by-step explanation:

Alissa will save $13.50 from the sale.

7 0

3 years ago

NEED HELP ASAP!!

devlian [24]

Answer:we multiply by the reciprocal of a fraction when completing a division problem because it is a rule in mathematics that when you divide two fraction, you change the division sign to multiplication and flip the fraction at the right of the multiplication sign.

Step-by-step explanation:

For example if you have 4÷ 1/2. It basically means how many 1/2 can one get in 4. And the answers is 8.

Therefore multiplying by the reciprocal of a fraction when completing a division problem is a short cut method that has been tested and proven to be correct.

4 0

3 years ago