The radioactive substance cesium-137 has a half-life of 30 years. The amount \( A(t) \) (in grams) of a sample of cesium-137 remaining after \( t \) years is given by the following exponential function. \[ A(t)=621\left(\frac{1}{2}\right)^{\frac{t}{30}} \] Find the initial amount in the sample and the amount remaining after 100 years. Round your answers to the nearest gram as necessary. Initial amount: \( \square \) grams Amount after 100 years: \( \square \) grams

II. Dadas las funcione \( \mathrm{f}(\mathrm{x})=5 \mathrm{x}^{2}+4 \mathrm{x}-1 \) \( \mathrm{~g}(\mathrm{x})=\mathrm{x}^{2}-6 \) Encontra a) \( \operatorname{Lim}_{x \rightarrow 1}\left[\frac{f(x)}{g(x)}\right] \) b) \( \operatorname{Lim}_{x \rightarrow 0}[f(x) \cdot g(x)] \)

\( \lim _ { x \rightarrow 0 } ( \frac { a - \cos ( \underline { 3 } x ) } { x - \sin ( b x ) } ) = ? \)

(2) \( f(x)=x^{3}-10 x^{2}-5 x+20 \) a) \( \int_{4}^{7} f(b)-f(a) \) b) \( 1^{\text {ra }} \) derivada c) Velocidad \( [5,8] \quad V_{m}=\frac{f\left(x_{f}\right)-f\left(x_{i}\right)}{V_{f}-x_{i}} \)

Use an appropriate test to determine whether the following series converges. \( \sum_{\mathrm{k}=1}^{\infty} \frac{5}{\sqrt{\mathrm{k}^{3}}} \) Select the correct choice below and fill in the answer box to complete your choice. A. The series diverges. It is a p-series with \( \mathrm{p}=\square \). The series converges. It is a p-series with \( \mathrm{p}=\square \). C. The series diverges by the Integral Test. The value of \( \int_{1}^{\infty} \frac{5}{\sqrt{\mathrm{x}^{3}}} \mathrm{dx} \) is \( \square \). D. The series converges by the Divergence Test. The value of \( \lim _{\mathrm{k} \rightarrow \infty} \frac{5}{\sqrt{\mathrm{k}^{3}}} \) is \( \square \). E. The series diverges by the Divergence Test. The value of \( \lim _{\mathrm{k} \rightarrow \infty} \frac{5}{\sqrt{\mathrm{k}^{3}}} \) is \( \square \).

Find the average rate of change of the function \( f(x)=x^{2}+9 x \) from \( x_{1}=1 \) to \( x_{2}=2 \). The average rate of change is \( \square \). (Simplify your answer.)