Question

point e is located at $2.6 \times 10^{-6}$. plot point e on the number line below. click on the graph to plot a point. click a point to delete it. 0 $10^{-4}$ answer attempt 2 out of 2

Explanation:

Step 1: Convert to decimal

First, convert \(2.6\times10^{-6}\) to decimal. We know that \(10^{-6}=\frac{1}{10^{6}} = 0.000001\), so \(2.6\times10^{-6}=2.6\times0.000001 = 0.0000026\).

Step 2: Analyze the number line

The number line goes from \(0\) to \(10^{-4}=0.0001\). Let's find the distance between the marks. From \(0\) to \(10^{-4}\), if we assume the number of intervals (the small ticks) between \(0\) and \(10^{-4}\) is, say, \(10\) (since \(10^{-4}\div10 = 10^{-5}\), but let's check the first mark. The first mark after \(0\) – wait, maybe the scale is such that each major tick (but here the first mark after \(0\) is at \(10^{-6}\)? Wait, no, let's see: \(10^{-4}=0.0001\), \(10^{-6}=0.000001\). Wait, \(2.6\times10^{-6}\) is \(2.6\) times \(10^{-6}\). So if the first mark after \(0\) is at \(10^{-6}\) (since \(0\) to \(10^{-4}\) has \(100\) intervals of \(10^{-6}\), because \(10^{-4}\div10^{-6}=100\)). Wait, \(10^{-4}=100\times10^{-6}\), so the number line from \(0\) to \(10^{-4}\) is divided into \(100\) equal parts, each of length \(10^{-6}\). So the first mark after \(0\) is at \(1\times10^{-6}\), the second at \(2\times10^{-6}\), the third at \(3\times10^{-6}\), etc. So \(2.6\times10^{-6}\) is \(2.6\) units of \(10^{-6}\) from \(0\). So we need to plot the point at \(2.6\times10^{-6}\), which is \(2.6\) times the first interval (each interval is \(10^{-6}\)). So on the number line, starting at \(0\), move \(2.6\) intervals of \(10^{-6}\) (each interval is the distance between \(0\) and the first mark, then the first to the second, etc.). So we click at the position that is \(2.6\) times the length of one \(10^{-6}\) interval from \(0\).

Answer:

To plot Point E at \(2.6\times10^{-6}\), we first recognize that \(2.6\times10^{-6}\) is \(2.6\) micro - units (where each micro - unit is \(10^{-6}\)) from \(0\). On the number line (where the total length from \(0\) to \(10^{-4}\) is \(100\) micro - units of \(10^{-6}\)), we plot the point \(2.6\) micro - units from \(0\), which is \(2.6\) times the distance from \(0\) to the first mark (each mark is \(10^{-6}\) apart). So we click at the position corresponding to \(2.6\times10^{-6}\) on the number line (approximately \(2.6\) intervals of \(10^{-6}\) from \(0\)).