Question

carry out the following operations as if it were a calculation of real experimental results. provide units where applicable, and use the appropriate number of significant digits. part 1 of 4 5.157 km²÷7.21 km = part 2 of 4 (5.51×10⁻⁴ mg)−(8.04×10⁻⁵ mg)= part 3 of 4 (6.88×10⁸ dm)+(9.06×10⁷ dm)=

Explanation:

Step1: Perform division for Part 1

$5.157\div7.21 = 0.7152566$ and $\text{km}^2\div\text{km}=\text{km}$, so the result is $0.715\text{ km}$ (3 significant - digits).

Step2: Rewrite numbers for Part 2

Rewrite $8.04\times 10^{-5}\text{ mg}$ as $0.804\times 10^{-4}\text{ mg}$. Then $(5.51\times 10^{-4}\text{ mg})-(0.804\times 10^{-4}\text{ mg})=(5.51 - 0.804)\times10^{-4}\text{ mg}=4.706\times 10^{-4}\text{ mg}\approx4.71\times 10^{-4}\text{ mg}$ (3 significant - digits).

Step3: Rewrite numbers for Part 3

Rewrite $9.06\times 10^{7}\text{ dm}$ as $0.906\times 10^{8}\text{ dm}$. Then $(6.88\times 10^{8}\text{ dm})+(0.906\times 10^{8}\text{ dm})=(6.88 + 0.906)\times10^{8}\text{ dm}=7.786\times 10^{8}\text{ dm}\approx7.79\times 10^{8}\text{ dm}$ (3 significant - digits).

Answer:

Part 1: $0.715\text{ km}$
Part 2: $4.71\times 10^{-4}\text{ mg}$
Part 3: $7.79\times 10^{8}\text{ dm}$