69 Practice Exercises 1. The distance from the Earth to Mercury is approximately 57,000,000 miles. Express this distance in scientific notation. 2. The distance between the Earth and the moon is approximately 384000 km. Express this distance in meters in exponential notation. 3. During the first six years of its operation, the Hubble Space Telescope circled Earth 37,000 times, for a total of 1,280,000,000 km. Use scientific notation to find the number of km in one orbit. 4. Haloo send mu kd dn picture dn model yu edkawni... JAZY
Answer (1)
Answer:Let's solve these scientific notation problems:**1. The distance from the Earth to Mercury is approximately 57,000,000 miles. Express this distance in scientific notation.**Scientific notation expresses a number as a product of a number between 1 and 10 and a power of 10.* Move the decimal point 7 places to the left (since there are 7 zeros).* This gives us 5.7.* The exponent of 10 is 7 (the number of places the decimal point was moved).Therefore, the distance in scientific notation is $\boxed{5.7 \times 10^7}$ miles.**2. The distance between the Earth and the moon is approximately 384,000 km. Express this distance in meters in exponential notation.**There are 1000 meters in 1 kilometer.* First, convert kilometers to meters: 384,000 km * 1000 m/km = 384,000,000 m* Move the decimal point 8 places to the left.* This gives us 3.84.* The exponent of 10 is 8.Therefore, the distance in exponential notation is $\boxed{3.84 \times 10^8}$ meters.**3. During the first six years of its operation, the Hubble Space Telescope circled Earth 37,000 times, for a total of 1,280,000,000 km. Use scientific notation to find the number of km in one orbit.*** Express the total distance in scientific notation: $1.28 \times 10^9$ km* Express the number of orbits in scientific notation: $3.7 \times 10^4$ orbits* Divide the total distance by the number of orbits: $(1.28 \times 10^9) / (3.7 \times 10^4)$* Divide the coefficients: 1.28 / 3.7 ≈ 0.346* Subtract the exponents: 9 - 4 = 5* Combine the results: $0.346 \times 10^5$* Adjust to proper scientific notation: $3.46 \times 10^4$Therefore, the number of km in one orbit is approximately $\boxed{3.46 \times 10^4}$ km.**4. Haloo send mu kd dn picture dn model yu edkawni... JAZY**
