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Eduardwww [97]
2 years ago
15

Can someone help me please

Mathematics
1 answer:
miss Akunina [59]2 years ago
7 0

Answer:

height of b = 280

Step-by-step explanation:

candle b has a height of 280 - 5t, when it comes out of the mold then t=0 so the initial height of b is 280

candle b's height drops by 5 every hour (that's what the -5t part means)

candle a started at 216 and dropped to 192 in 3 hours, so it dropped by 24 in 3 hours so it drops 8 each hour

so candle b starts out taller (280 v. 216) and candle b burns slower (a at 8 per hour and b at 5 per hour), so b will never 'catch up' with a... candle a will be burned out while candle b is still burning

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Find the largest interval which includes x = 0 for which the given initial-value problem has a unique solution. (Enter your answ
Ivan

Answer:

(-\infty,3)

Step-by-step explanation:

We are given that

(x-3)y''+4y=x

y''+\frac{4}{x-3}y=\frac{x}{x-3}

y(0)=0

y'(0)=1

By comparing with

y''+p(x)y'+q(x)y=g(x)

We get

p(x)=\frac{4}{x-3}

g(x)=\frac{x}{x-3}

q(x)=0

p(x),q(x) and g(x) are continuous for all real values of x except 3.

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Hence, the larges interval on which includes x=0 for which given initial value problem has unique solution=(-\infty,3)

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3 years ago
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3 years ago
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<img src="https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7BE%7D%7B%20%5Csqrt%7BR%20%7B%7D%5E%7B2%7D%20%2B%20W%20%7B%7D%5E%7B2%7D%20L%
AleksandrR [38]

Answer:

R = sqrt[(IWL)^2/(E^2 - I^2)] or R = -sqrt[(IWL)^2/(E^2 - I^2)]

Step-by-step explanation:

Squaring both sides of equation:

     I^2 = (ER)^2/(R^2 + (WL)^2)

<=>(ER)^2 = (I^2)*(R^2 + (WL)^2)

<=>(ER)^2 - (IR)^2 = (IWL)^2

<=> R^2(E^2 - I^2) = (IWL)^2

<=> R^2 = (IWL)^2/(E^2 - I^2)

<=> R = sqrt[(IWL)^2/(E^2 - I^2)] or R = -sqrt[(IWL)^2/(E^2 - I^2)]

Hope this helps!

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