The average tectonic plate moves at the rate of 0.006 m per year. How many meters would it move in 2×10^6 years? Write the answer in scientific notation, expressed to the exact decimal place.(1 point)
To find the distance the tectonic plate would move in 2×10^6 years, we can multiply the rate of movement (0.006 m/year) by the number of years.
0.006 m/year * 2×10^6 years = 0.006 * 2 * 10^6 = 1.2 * 10^4 m.
Therefore, the tectonic plate would move 1.2 * 10^4 meters in 2×10^6 years.
To find out how many meters the tectonic plate would move in 2×10^6 years, we can multiply the rate of movement by the number of years.
Rate of movement = 0.006 m/year
Number of years = 2×10^6 years
To multiply these two values, we can simply multiply the numerical parts and add the exponents:
0.006 m/year * 2×10^6 years = 0.006 * 2 * 10^6 m
Multiplying the numerical parts gives us:
0.006 * 2 = 0.012
And combining the exponents gives us:
10^6 * 10^0 = 10^6
Therefore, the tectonic plate would move 0.012 * 10^6 meters in 2×10^6 years.
Expressing it in scientific notation to the exact decimal place means that we combine the numerical part (0.012) with the exponent (6):
0.012 * 10^6 = 1.2 * 10^(-2) * 10^6
Combining the exponents gives us:
10^(-2) * 10^6 = 10^(6-2) = 10^4
Therefore, the tectonic plate would move 1.2 * 10^4 meters in 2×10^6 years.
To find the distance the tectonic plate would move in 2x10^6 years, we can multiply the rate of 0.006 m/year by the time, 2x10^6 years.
0.006 m/year * 2x10^6 years = 0.006 * 2 * 10^6 m
To multiply the numbers, we simply multiply the coefficients together and add the exponents of 10:
0.006 * 2 = 0.012
10^6 * 10^-3 = 10^3
Therefore, the tectonic plate would move 0.012 * 10^3 meters in 2x10^6 years.
In scientific notation, we express the result as a number between 1 and 10 multiplied by a power of 10, so we can rewrite 0.012 * 10^3 as 1.2 * 10^2 (since 0.012 is equivalent to 1.2 * 10^-1).
Therefore, the tectonic plate would move 1.2 * 10^2 meters in 2x10^6 years.