1. **Evaluate the expression $5^{-3}$ as a fraction or integer.** Recall the rule for negative exponents: $a^{-n} = \frac{1}{a^n}$. So, $$5^{-3} = \frac{1}{5^3} = \frac{1}{125}.$$ 2. **Evaluate the expression $\left(\frac{1}{6}\right)^{-2}$.** Using the negative exponent rule: $$\left(\frac{1}{6}\right)^{-2} = \left(\frac{6}{1}\right)^2 = 6^2 = 36.$$ 3. **Evaluate the expression $\left(\frac{3}{2}\right)^{-3} \times \frac{2}{3}$.** First, apply the negative exponent: $$\left(\frac{3}{2}\right)^{-3} = \left(\frac{2}{3}\right)^3 = \frac{2^3}{3^3} = \frac{8}{27}.$$ Now multiply by $\frac{2}{3}$: $$\frac{8}{27} \times \frac{2}{3} = \frac{8 \times 2}{27 \times 3} = \frac{16}{81}.$$ 4. **Calculate the growth rate of a population which increases by 105%.** Growth rate is the percentage increase expressed as a decimal: $$\text{Growth rate} = \frac{105}{100} = 1.05.$$ 5. **Calculate the decay rate of a substance which decreases by 15%.** Decay rate is the percentage decrease expressed as a decimal: $$\text{Decay rate} = \frac{15}{100} = 0.15.$$ 6. **Find the amount of sodium bicarbonate left after 100 days if the half-life is 25 days and initial amount is 20 grams.** The formula for remaining amount after time $t$ with half-life $T_{1/2}$ is: $$A = A_0 \times \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$$ Substitute values: $$A = 20 \times \left(\frac{1}{2}\right)^{\frac{100}{25}} = 20 \times \left(\frac{1}{2}\right)^4 = 20 \times \frac{1}{16} = \frac{20}{16} = \frac{5}{4}.$$ So, $\frac{5}{4}$ grams remain after 100 days. **Concluding statement:** The amount of sodium bicarbonate left after 100 days is $\frac{5}{4}$ grams. --- **Slug:** exponent growth **Subject:** algebra **Desmos:** {"latex":"","features":{"intercepts":true,"extrema":true}} **q_count:** 6