Here's the question & my work:
for #16.a) the tb answer is 6 km & for #16.b) it's 9 km. firstly, i would like to know if my approach for a is correct? secondly, i'm stuck on how to solve for b!
+26367 Finding distance using absolute values?
b)
\(\text{Let $x$ is the kilometre marker at the end } \\ \text{Because the markers become big in the direction of east $x$ is greater than $3$ km }:\)
\(\begin{array}{|r|r|r|r|} \hline \text{part} & \text{from marker} & \text{to marker} & \text{distance} \\ \hline 1 & 2\ \text{km} & 7\ \text{km} & 7-2 = 5\ \text{km} \\ \hline 2 & 7\ \text{km} & 3\ \text{km} & 7-3 = 4\ \text{km} \\ \hline 3 & 3\ \text{km} & x\ \text{km} & x-3 \ \text{km} \\ \hline & & & \text{sum} = 15\ \text{km} \\ \hline \end{array} \)
\(\begin{array}{|rcll|} \hline 5 + 4 + x-3 &=& 15 \\ 9+x-3 &=& 15 \\ 6+x &=& 15 \\ x &=& 15 - 6 \\ \mathbf{x}& \mathbf{=}& \mathbf{9\ \text{km}} \\ \hline \end{array}\)
At the end of the scanvenger hunt Toby is at the 9-km marker.
a)
\(\begin{array}{|r|r|r|r|} \hline \text{part} & \text{from marker} & \text{to marker} & \text{distance} \\ \hline 1 & 2\ \text{km} & 7\ \text{km} & 7-2 = 5\ \text{km} \\ \hline 2 & 7\ \text{km} & 3\ \text{km} & 7-3 = 4\ \text{km} \\ \hline 3 & 3\ \text{km} & 9\ \text{km} & 9-3 = 6\ \text{km} \\ \hline \end{array}\)
In the last interval Toby travel 6 km.
