# The initial mass of a radioactive goo is 180 grams. After 195 minutes, the sample has decayed to 2.8125 grams. Find the half-life of the substance in minutes? Find a formula for the amount, G(t), remaining at time t.? Find grams remain after 35 min?

Forms of the decay formula that we are aware of are:

Enter the previous value in equation [1]:

Calculate both sides' natural logarithms:

Make use of the logarithmic property, which permits the exponent to be moved to the outside as a coefficient:

The 180-gram sample's remaining quantity after 35 minutes:

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Half-life ((t_{1/2})): (t_{1/2} = \frac{\ln(2)}{\text{slope}})

Exponential decay formula: (G(t) = G_0 \cdot e^{-kt})

Given dataHalf-life ((t_{1/2})): (t_{1/2} = \frac{\ln(2)}{\text{slope}})

Exponential decay formula: (G(t) = G_0 \cdot e^{-kt})

Given Initial mass ((G_0)): 180 g Final mass after 195 min: 2.8125 g

Solve for (k): (2.8125 = 180 \cdot e^{-195k})

Formula for remaining mass: (G(t) = 180 \cdot e^{-kt})

Grams remaining after 35 min: (G(35) = 180 \cdot e^{-35k})

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Half-life = ( t_{\frac{1}{2}} = \frac{t \times \ln(2)}{\ln(\frac{G_0}{G_t})} )

Formula for the remaining amount at time ( t ): ( G(t) = G_0 \times e^{-\frac{t}{t_{\frac{1}{2}}}} )

Grams remaining after 35 minutes: ( G(35) = 180 \times e^{-\frac{35}{t_{\frac{1}{2}}}} )

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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