1)
you can think of the cup's sleeve as a cylinder with say a height of 4 and a diameter of 5. Now, if the diameter is 5 inches, the radius is half that, or 2.5,
[tex]\bf \textit{total surface area of a cylinder}\\\\
s=2\pi r(h+r)\quad
\begin{cases}
r=2.5\\
h=4
\end{cases}\implies s=2\pi (2.5)(4+2.5)
\\\\\\
s=5\pi (6.5)\implies s=32.5\pi [/tex]
2)
same here, you can think of the tire as a cylinder, since it's a cylindrical tire, with a height of 8 and a diameter of 29, therefore a radius of 14.5,
[tex]\bf \textit{total surface area of a cylinder}\\\\
s=2\pi r(h+r)\quad
\begin{cases}
r=14.5\\
h=8
\end{cases}\implies s=2\pi (14.5)(8+14.5)
\\\\\\
s=29\pi (22.5)\implies s=652.5\pi [/tex]
now, that's how much spaces it covers one time only, so 5 times around is just 5 times that much.
3)
same goes for the mugs,
[tex]\bf \textit{total surface area of a cylinder}\\\\
s=2\pi r(h+r)\quad
\begin{cases}
r=2\\
h=3.75
\end{cases}\implies s=2\pi (2)(3.75+2)
\\\\\\
s=4\pi (5.75)\implies s=\stackrel{one~mug}{23\pi }\qquad\qquad \stackrel{4~mugs}{4\cdot 23\pi }[/tex]
4)
and the same goes for the can of peas, which is also a cylinder,
a)
[tex]\bf \textit{total surface area of a cylinder}\\\\
s=2\pi r(h+r)\quad
\begin{cases}
r=3.8\\
h=10
\end{cases}\implies s=2\pi (3.8)(10+3.8)
\\\\\\
s=7.6\pi (13.8)\implies s=101.08\pi [/tex]
b)
the label is only the sides, and thus only the lateral area,
[tex]\bf \textit{\underline{lateral} surface area of a cylinder}\\\\
s=2\pi rh\quad
\begin{cases}
r=3.8\\
h=10
\end{cases}\implies s=2\pi (3.8)(10)\implies s=76\pi [/tex]
