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3 years ago

It is 3:00pm here, and Australia is 13 hours ahead of us. What time is it in Australia?

Mathematics

2 answers:

AfilCa [17]3 years ago

5 0

Answer:

4 am

Step-by-step explanation:

Murrr4er [49]3 years ago

5 0

Answer:

4:00 AM

Step-by-step explanation:

15:00 + 13:00

28:00

24:00+4:00

00:00+4:00

4:00 AM

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Pls help ill give brainliest if correct<br><br> no scams or links

soldi70 [24.7K]

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3rd one. Amsterdam is the same distance from sea level as Drachten.

Step-by-step explanation:

because Amsterdam is 2 below, and drachten is 2 above, but they are both 2 meters away from the sea level.

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2 years ago

Convert 12.3 liters to milliliters.

agasfer [191]

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2 years ago

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The sum of two rational numbers is -8 1/4. if one of the numbers is -5 2/3, find the other number. im too lazy to do this

Nataly [62]

Answer:

Step-by-step explanation:

let : x and t this numbers : x +y = -81/4

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2 years ago

Write the equation of the line that passes through the points (0,8) and (-1, –4).

andrey2020 [161]

Answer:

Step-by-step explanation

First, let's remember the equation for point-slope form and how to find the 'm' value:

$y-y1 = m (x-x1)\\m = rise/run\\m = \frac{-4-(8)}{-1-(0)} = 12$

Now plug your m value and (x1, y1) into point-slope form:

$y - 8 = 12(x-0)\\reduce\\y=12x+8$

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2 years ago

Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de

Ganezh [65]

Answer:

a. $\mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

b. $\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

Step-by-step explanation:

The initial value problem is given as:

$y' +y = 7+\delta (t-3) \\ \\ y(0)=0$

Applying laplace transformation on the expression $y' +y = 7+\delta (t-3)$

to get $L[{y+y'} ]= L[{7 + \delta (t-3)}]$

$l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

Taking inverse of Laplace transformation

$y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$