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balu736 [363]
3 years ago
12

Jodi poured herself a cold soda that had an initial temperature 36 degrees F and immediately went outside to sunbathe where the

temperature was a steady 99 degree F. After 5minutes the temperature of the soda was 46 degree F .Jodi had to run back into the house to answer the phone . What is the expected temperature of the soda after an additional 13 minutes ?
Mathematics
1 answer:
mr Goodwill [35]3 years ago
4 0

this can be solve using newtons heating of cooling

(Ts – T) =(Ts – To)*e^(-kt)

Where Ts is the ambient temperature

To is initial temperature

T is the temperature at time t

t is the time

k is constant

fisrt solve the constant k for the given first scenario

(99 – 36) = (99 – 46)*e(-5k)

K = -0.0346

Using k, solve T at t = 13 min

(99 – 46) = (99 – T)*e(-13*(-0.0346)

T = 58.82 degree F

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4 0
3 years ago
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(4x+2)=102+44 solve for x
sergiy2304 [10]
<h3>Solution:</h3>
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vladimir2022 [97]
(2x + 3y = 12) x (-2)
(4x - 3y = 6) x 1

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You can cancel out the x values by adding the two equations together. 
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Solve for x now...
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Check... (x = 3, y = 2)
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