Question

Butter is taken out of the fridge and left to warm up. The temperature of the butter, T (in Celsius degrees), after m minutes is given by the equation T=c-k*1.2^{-m}. The initial temperature of the butter is 5º, and after 5 minutes its temperature has increased to 15º, as shown on the graph
a) Find the values of k and c

Answer 1

Yeaa...nope.

@Vocaloid

Answer 2

\(T(m) = c- k \cdot 1.2^{-m}\)

\( T(0) = 5 \implies c = k + 5\)

\(T(5) = 15 \implies c - k ~ 1.2^{-5} = 5 \)

solve from there

Answer 3

\(T(5) = c - k ~1.2^{-15} = 15\)

Answer 4

oh balls :( you get the picture. just plug in the numbers and solve :)

Answer 5

For T(5), I got
\[c=15+\frac{1}{2.48832}k\]

Answer 6

So then I would do
\[k+5=15+\frac{1}{2.48832}k?\]

Answer 7

the key is to write both equations and solve by elimination

5 = c - k(1.2)^(0)
15 = c - k(1.2)^(-5)

you can subtract these two equations to eliminate c

10 = c - k(1.2)^(-5) - [c - k(1.2)^(0)]

after you eliminate c you should be able to solve for k, then back-substitute to get c

Answer 8

ok I got it now

10 = c - k(1.2)^(-5) - [c - k(1.2)^(0)]

10 = -k(1.2)^(-5) + k

k is about 16.719, then you can select one of the initial points, plug in and solve for c

Answer 9

\[ 5=c-16.719*1.2^{-0}\\
c=21.719\]

Answer 10

good

just to be on the safe side I would double-check those k and c values by plugging them back into the points to see if you still get 5 and 15 ;;

Answer 11

\[ 5=21.719-16.719*1.2^{0}\\
5=5\]

Answer 12

Thank you Vocaloid :P

Answer 13

So: here, Completing the Square is a real timesaver.

Not too shabby  😀

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I don't see a subject for this.
I just need help not an answer

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