The average tectonic plate moves at the rate of 0.006 m per year. How many meters would it move in 2 x 10^6 years? Write the answer in scientific notation, expressed to the exact decimal place

Answer 1

To find how far a tectonic plate moves in \(2 \times 10^6\) years, you can multiply the rate of movement by the time:

\[ \text{Distance} = \text{Rate} \times \text{Time} \]

Given:

• Rate = \(0.006 , \text{m/year}\)
• Time = \(2 \times 10^6 , \text{years}\)

Now substituting in:

\[ \text{Distance} = 0.006 , \text{m/year} \times 2 \times 10^6 , \text{years} \]

First, calculate \(0.006 \times 2\):

\[ 0.006 \times 2 = 0.012 \]

Next, multiply that by \(10^6\):

\[ 0.012 \times 10^6 = 1.2 \times 10^4 \]

Thus, the tectonic plate would move \(1.2 \times 10^4\) meters in \(2 \times 10^6\) years.

So, the answer in scientific notation, expressed to the exact decimal place is:

\[ \boxed{1.2 \times 10^4} \]

Answer 2

are you sure

Answer 3

Let's double-check the calculations step by step.

1. Rate of movement: \(0.006 , \text{m/year}\)
2. Time: \(2 \times 10^6 , \text{years}\)

To calculate the distance moved:

\[ \text{Distance} = \text{Rate} \times \text{Time} = 0.006 , \text{m/year} \times 2 \times 10^6 , \text{years} \]

3. First, multiply the numerical values:

\[ 0.006 \times 2 = 0.012 \]

4. Now, incorporate the scientific notation:

\[ 0.012 \times 10^6 \]

To express \(0.012\) in scientific notation:

\[ 0.012 = 1.2 \times 10^{-2} \]

5. Thus, we have:

\[ 1.2 \times 10^{-2} \times 10^6 \]

6. Combining the powers of ten:

\[ 1.2 \times 10^{(6 - 2)} = 1.2 \times 10^4 , \text{meters} \]

So the correct distance the tectonic plate would move in \(2 \times 10^6\) years is indeed:

\[ \boxed{1.2 \times 10^4 , \text{meters}} \]

The calculations verify that the answer is correct. Thank you for your patience!