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3 years ago

The amount of water in a bottle is reduced by 85% to 120 ml. How much water was originally in the bottle?

1 answer:

4 0

At first it was 100%. When it was reduced by 85%, it became 25%. The 25% represents 120 ml. So 100% would be (120/25) × 100 = 480 ml.

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Find the equation (in standard form) whose center is (5, -3) and goes through (2, 5).

yanalaym [24]

We can write this as

(x - 5)^2 + (y + 3)^2 = r^2

where r = radius

Plugging in the point (2,5) we have

(2-5)^2 + (5+3)^2 = r^2
r^2 = 9 + 64 = 73
so the required equation is

(x - 5)^2 + (y + 3)^2 = 73

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A student did a 340 page essay for 3/4 hours. how much would it take to do 1 page

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drek231 [11]

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3 years ago

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Aleonysh [2.5K]

D. Will be the correct answer

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3 years ago

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Find the solution of the differential equation dy/dt = ky, k a constant, that satisfies the given conditions. y(0) = 50, y(5) =

irga5000 [103]

Answer: The required solution is $y=50e^{0.1386t}.$

Step-by-step explanation:

We are given to solve the following differential equation :

$\dfrac{dy}{dt}=ky~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

where k is a constant and the equation satisfies the conditions y(0) = 50, y(5) = 100.

From equation (i), we have

$\dfrac{dy}{y}=kdt.$

Integrating both sides, we get

$\int\dfrac{dy}{y}=\int kdt\\\\\Rightarrow \log y=kt+c~~~~~~[\textup{c is a constant of integration}]\\\\\Rightarrow y=e^{kt+c}\\\\\Rightarrow y=ae^{kt}~~~~[\textup{where }a=e^c\textup{ is another constant}]$

Also, the conditions are

$y(0)=50\\\\\Rightarrow ae^0=50\\\\\Rightarrow a=50$

and

$y(5)=100\\\\\Rightarrow 50e^{5k}=100\\\\\Rightarrow e^{5k}=2\\\\\Rightarrow 5k=\log_e2\\\\\Rightarrow 5k=0.6931\\\\\Rightarrow k=0.1386.$

Thus, the required solution is $y=50e^{0.1386t}.$

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3 years ago

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