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For the following exercises, convert numbers to scientific notation.
\(0.0134\)
\(0.0000761\)
\(3,400\)
\(8,980,000\)
For the following exercises, convert numbers to standard form.
\(9.01 \times {10^5}\)
\(3.78 \times {10^7}\)
\(4.32 \times {10^{ - 3}}\)
\(5.781 \times {10^{ - 5}}\)
For the following exercises, the numbers are not in scientific notation. Convert them to scientific notation.
\(37.65 \times {10^4}\)
\(0.0034 \times {10^6}\)
\(0.0834 \times {10^{ - 7}}\)
\(14.56 \times {10^{ - 3}}\)
For the following exercises, make the conversions required.
The distance from the sun to the star Polaris is about 3,056,000,000,000,000 km. Express that distance in scientific notation.
The distance from us to the next-closest galaxy is about 662,000,000,000,000,000 km. Express that distance in scientific notation.
The mass of a grain of sand is about \(6.66 \times {10^{ - 4}}\) g. Convert that mass to standard form.
The diameter of a cell is about \(2 \times {10^{ - 6}}\) m. Convert that diameter to standard form.
The equatorial circumference of Earth is approximately \(4.007 \times {10^4}\) km. Convert that circumference to standard form.
The straight-line distance from Buffalo, NY, to Buenos Aires, Argentina, is approximately \(8.86 \times {10^6}\) m. Convert that distance to standard form.
The mass of a proton is approximately \(1.67 \times {10^{ - 27}}\) kg. Convert that mass to standard form.
The diameter of a housefly egg is approximately \(1.2 \times {10^{ - 3}}\) m. Convert that diameter to standard form.
The tallest building in the world, the Burj Khalifa in Dubai, stands at 829.8 m tall. Convert that height to scientific notation.
Using the rings of the shell, the age of an Icelandic clam is 507 years. Express that age in scientific notation.
Calculate the following:
\(1.3 \times {10^2} + 3.8 \times {10^2}\)
\(7.8 \times {10^{12}} + 1.1 \times {10^{12}}\)
\(3.36 \times {10^4} + 2.71 \times {10^4}\)
\(4.58 \times {10^9} + 1.93 \times {10^9}\)
\(8.1 \times {10^{ - 17}} + 1.6 \times {10^{ - 17}}\)
\(4.506 \times {10^{ - 3}} + 3.908 \times {10^{ - 3}}\)
\(8.602 \times {10^{ - 25}} + 1.096 \times {10^{ - 25}}\)
\(2.0557 \times {10^{ - 6}} + 1.001 \times {10^{ - 6}}\)
\(5.2 \times {10^4} - 4.1 \times {10^4}\)
\(9.48 \times {10^{15}} - 6.78 \times {10^{15}}\)
\(7.81 \times {10^{ - 7}} - 4.62 \times {10^{ - 7}}\)
\(4.53 \times {10^{ - 3}} - 2.79 \times {10^{ - 3}}\)
\(4.6 \times {10^{ - 5}} + 9.1 \times {10^{ - 5}}\)
\(6.7 \times {10^8} + 5.7 \times {10^8}\)
\(4.13 \times {10^{ - 4}} + 7.93 \times {10^{ - 4}}\)
\(5.671 \times {10^{ - 8}} + 9.073 \times {10^{ - 8}}\)
\(4.513 \times {10^6} + 7.856 \times {10^4}\)
\(7.135 \times {10^8} + 5.143 \times {10^6}\)
\(3.17 \times {10^{ - 3}} + 5.92 \times {10^{ - 5}}\)
\(4.503 \times {10^{ - 6}} + 3.119 \times {10^{ - 4}}\)
\((4.5 \times {10^4}) \times (1.2 \times {10^6})\)
\((3.45 \times {10^7}) \times (2.81 \times {10^{ - 3}})\)
\((3.1 \times {10^8}) \times (2.7 \times {10^{ - 5}})\)
\((6.32 \times {10^{ - 4}}) \times (1.31 \times {10^{ - 5}})\)
\((3.91 \times {10^6}) \times (8.13 \times {10^2})\)
\((7.12 \times {10^{ - 11}}) \times (6.61 \times {10^{ - 5}})\)
\((3.45 \times {10^4}) \div (1.5 \times {10^6})\)
\((1.4 \times {10^3}) \div (5.6 \times {10^{ - 5}})\)
For the following exercises, apply your understanding of scientific notation to real-world applications.
When stretched out, a strand of human DNA is, on average, \(2.066 \times {10^2}\) cm. One centimeter, or 1 cm, is \(1 \times {10^{ - 5}}\) km. Determine the average length of a strand of human DNA in kilometers.
One approximation of the average number of cells in the human body is \(3 \times {10^{13}}\) cells (30 trillion!!!). If the DNA of each cell were stretched out and laid end to end, what would be the total length of the DNA in km? Use your answer from Exercise 51 for the length, in kilometers, of DNA.
The equatorial circumference of Earth is approximately \(4.007 \times {10^4}\) km. Use the answer from Exercise 52 to determine the number of times that the stretched out human DNA would encircle Earth.
The average stride length of a 1.651 m tall woman is \(6.6 \times {10^{ - 1}}\) meters. If such a person could walk from Buffalo, NY, to Buenos Aires, Argentina, in a straight line, how many steps would that person need to take? See Exercise 18 for the distance from Buffalo, NY, to Buenos Aires, Argentina.
3.9.0: Exercises is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
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