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harina [27]
3 years ago
9

A sauna has an initial temperature of 80° If the opening to the sauna is left open, and the temperature therefore decreases by 3

% every minute, what is the temperature of the sauna after 210 seconds?
Mathematics
1 answer:
ddd [48]3 years ago
5 0

Answer:

71.6°

Step-by-step explanation:

=210s÷60s(1 minutes)

=3.5mins

=3.5mins×3%(=1min)

=10.5%

=100%-10.5%

=89.5%

=89.5%/100×80°

=71.6°

Hope this will help ya!

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Y''+y'+y=0, y(0)=1, y'(0)=0
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Answer:

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Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form ay''+by'+cy=0.

Given: y''+y'+y=0

Let y=e^{rt} be it's solution.

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Applying conditions y(0)=1 on e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right ), c_1=1

So, equation becomes y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

On differentiating with respect to t, we get

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Applying condition: y'(0)=0, we get 0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}

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