A tourist converts $5820 USD to Kenya Shillings (Ksh) at a rate of 1 USD = 102.10 Ksh. They spend 450,000 Ksh and then convert the remaining money back to USD at a rate of 1 USD = 103.00 Ksh. How much

The tourist converted $5820 \text{ US dollars}$ at a rate of $1 \text{ USD} = 102.10 \text{ Ksh}$.

$$\text{Amount in Ksh} = 5820 \text{ USD} \times 102.10 \text{ Ksh/USD}$$ $$\text{Amount in Ksh} = 594222 \text{ Ksh}$$

The tourist spent $450,000 \text{ Ksh}$.

$$\text{Remaining Ksh} = 594222 \text{ Ksh} - 450000 \text{ Ksh}$$ $$\text{Remaining Ksh} = 144222 \text{ Ksh}$$

The conversion rate for dollars back is $1 \text{ USD} = 103.00 \text{ Ksh}$.

$$\text{Remaining USD} = \frac{144222 \text{ Ksh}}{103.00 \text{ Ksh/USD}}$$ $$\text{Remaining USD} = 1400.21359... \text{ USD}$$

$$\text{Remaining USD} \approx 1400 \text{ USD}$$

The amount of money that remained, to the nearest dollar, is:

$$\boxed{1400 \text{ USD}}$$

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7. Prime Factorization and Evaluation

$$1728 = 2 \times 864$$ $$= 2 \times 2 \times 432$$ $$= 2 \times 2 \times 2 \times 216$$ $$= 2 \times 2 \times 2 \times 2 \times 108$$ $$= 2 \times 2 \times 2 \times 2 \times 2 \times 54$$ $$= 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 27$$ $$= 2^6 \times 3 \times 9$$ $$= 2^6 \times 3 \times 3 \times 3$$ $$1728 = 2^6 \times 3^3$$

$$2025 = 5 \times 405$$ $$= 5 \times 5 \times 81$$ $$= 5^2 \times 3 \times 27$$ $$= 5^2 \times 3 \times 3 \times 9$$ $$= 5^2 \times 3 \times 3 \times 3 \times 3$$ $$2025 = 3^4 \times 5^2$$

Using the prime factorization:

$$\sqrt[3]{1728} = \sqrt[3]{2^6 \times 3^3}$$ $$= 2^{\frac{6}{3}} \times 3^{\frac{3}{3}}$$ $$= 2^2 \times 3^1$$ $$= 4 \times 3$$ $$\sqrt[3]{1728} = 12$$

Using the prime factorization:

$$\sqrt{2025} = \sqrt{3^4 \times 5^2}$$ $$= 3^{\frac{4}{2}} \times 5^{\frac{2}{2}}$$ $$= 3^2 \times 5^1$$ $$= 9 \times 5$$ $$\sqrt{2025} = 45$$

Divide both the numerator and the denominator by their greatest common divisor, which is 3.

$$\frac{12 \div 3}{45 \div 3} = \frac{4}{15}$$

The evaluated expression is:

$$\boxed{\frac{4}{15}}$$