we know that b ⩾ 1, so b is never 0.
a)
notice, the first figure starts off with 4 tiles and has 10 bordering tiles.
the second figure starts off with 14 tiles and has 14 bordering tiles
the third figure has starts off with 28 tiles and has 20 bordering tiles.
[tex]\bf \stackrel{\textit{first row, b = 1}}{4(1-1)+10}\implies \stackrel{\textit{border tiles}}{4(0)+10\implies 10}
\\\\\\
\stackrel{\textit{second row, b = 2}}{4(2-1)+10}\implies \stackrel{\textit{border tiles}}{4(1)+10\implies 14}
\\\\\\
\stackrel{\textit{third row, b = 3}}{4(3-1)+10}\implies \stackrel{\textit{border tiles}}{4(2)+10\implies 18}[/tex]
b)
She's correct, check a).
[tex]\bf \stackrel{\stackrel{starting~tiles}{\downarrow }}{4}(\stackrel{\stackrel{row~number}{\downarrow }}{b}-1)+10~~=~~\textit{number of tiles per border}[/tex]
c)
n(b - 1) + 10
