Question

point t is located at $1.2 \times 10^{-4}$. plot point t on the number line below.
click on the graph to plot a point. click a point to delete it.
0 $10^{-2}$

Explanation:

Step1: Convert \(1.2\times10^{-4}\) to decimal

\(1.2\times10^{-4}=\frac{1.2}{10^{4}} = 0.00012\)

Step2: Analyze the number line scale

The number line goes from \(0\) to \(10^{-2}=0.01\). The distance between \(0\) and \(0.01\) is divided into, let's see, from \(0\) to \(10^{-2}\), the number of intervals: since \(10^{-2}-0 = 0.01\), and if we assume each small tick is \(10^{-4}\) (because \(0.01\div100 = 10^{-4}\), as there are 100 intervals from \(0\) to \(0.01\) if each is \(10^{-4}\)). Wait, \(10^{-2}=100\times10^{-4}\), so the number line from \(0\) to \(10^{-2}\) has 100 equal parts, each part is \(10^{-4}\). So \(1.2\times10^{-4}\) is \(1.2\) times \(10^{-4}\), so it's 1.2 ticks to the right of \(0\).

Answer:

To plot Point T at \(1.2\times10^{-4}\) (which is \(0.00012\)) on the number line from \(0\) to \(10^{-2}\) (\(0.01\)): since the interval between \(0\) and \(10^{-2}\) can be considered as having 100 equal sub - intervals (each of length \(10^{-4}\)), move \(1.2\) sub - intervals to the right of \(0\) and plot the point.