Question

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.

answer attempt 1 out of 2

Explanation:

Step1: Convert scientific notation to decimal

To convert \(1.2\times10^{-4}\) to decimal, we know that \(10^{-4}=\frac{1}{10^{4}} = 0.0001\). So, \(1.2\times10^{-4}=1.2\times0.0001 = 0.00012\).

Step2: Analyze the number line

The number line goes from \(0\) to \(10^{-2}=0.01\). Let's find the scale of the number line. The distance from \(0\) to \(10^{-2}\) (which is \(0.01\)) is divided into, let's see, from \(0\) to \(0.01\), if we assume the number of intervals between \(0\) and \(0.01\) is, say, \(10\) intervals (since \(0.01\div0.001 = 10\), but let's check the value of \(1.2\times10^{-4}=0.00012\)). Wait, maybe a better way: \(10^{-2}=0.01\), \(10^{-4}=0.0001\), so \(1.2\times10^{-4}=1.2\times10^{-4}\), and \(10^{-2}=100\times10^{-4}\). So the number line from \(0\) to \(10^{-2}\) is a line where each unit on the line (in terms of \(10^{-4}\)): from \(0\) (which is \(0\times10^{-4}\)) to \(10^{-2}\) (which is \(100\times10^{-4}\)). So the position of \(1.2\times10^{-4}\) is \(1.2\) units from \(0\) when we consider the scale of \(10^{-4}\). So on the number line, we plot the point at \(1.2\times10^{-4}\), which is \(0.00012\), between \(0\) and \(10^{-2}\) (0.01), closer to \(0\).

Answer:

To plot \(1.2\times10^{-4}\) (which is \(0.00012\)) on the number line from \(0\) to \(10^{-2}\) (0.01), we identify its position relative to \(0\) and \(10^{-2}\). Since \(1.2\times10^{-4}\) is much smaller than \(10^{-2}\) (as \(10^{-2}=100\times10^{-4}\) and \(1.2\times10^{-4}\) is \(1.2\) times \(10^{-4}\)), we plot the point near \(0\), at a distance corresponding to \(1.2\times10^{-4}\) on the number line. (Note: Since the actual plotting is done by clicking on the graph, the key is to recognize the value \(1.2\times10^{-4}=0.00012\) and place it correctly between \(0\) and \(0.01\) on the number line.)