Question

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)

Explanation:

Step1: Recall the distance - rate - time formula

Distance \(d = r\times t\), where \(r = 0.006\) m/year and \(t=2\times 10^{6}\) years.

Step2: Substitute the values into the formula

\(d=0.006\times(2\times 10^{6})\). First, rewrite \(0.006\) in scientific - notation as \(6\times 10^{-3}\). Then \(d=(6\times 10^{-3})\times(2\times 10^{6})\).

Step3: Use the rule of exponents \(a^{m}\times a^{n}=a^{m + n}\)

\((6\times 10^{-3})\times(2\times 10^{6})=(6\times2)\times(10^{-3 + 6})\).

Step4: Calculate the product of the coefficients and the exponents

\(6\times2 = 12\) and \(-3+6 = 3\), so \(d = 12\times10^{3}\). But in scientific notation, the coefficient should be between \(1\) and \(10\). Rewrite \(12\times10^{3}\) as \(1.2\times10^{4}\).

Answer:

\(1.2\times10^{4}\) m