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Otrada [13]
3 years ago
14

Identify the asymptote(s) of this function: y=5/(x-4)+1

Mathematics
1 answer:
Dimas [21]3 years ago
6 0
y= \frac{5}{(x-4)+1}
y= \frac{5}{x-3}
So you take the denominator and compare to zero. 
x-4+1=0
x-3=0
x=3
Then the horizontal asymptote is y = 0
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A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years

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\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2}  \approx 669\ years

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