if the sphere has a diameter of 5, then its radius is half that, or 2.5.
![\bf \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=2.5 \end{cases}\implies V=\cfrac{4\pi (2.5)^3}{3}\implies V=\cfrac{62.5\pi }{3} \\\\\\ V\approx 65.44984694978736\implies V=\stackrel{\textit{rounded up}}{65.45}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20sphere%7D%5C%5C%5C%5C%20V%3D%5Ccfrac%7B4%5Cpi%20r%5E3%7D%7B3%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D2.5%20%5Cend%7Bcases%7D%5Cimplies%20V%3D%5Ccfrac%7B4%5Cpi%20%282.5%29%5E3%7D%7B3%7D%5Cimplies%20V%3D%5Ccfrac%7B62.5%5Cpi%20%7D%7B3%7D%20%5C%5C%5C%5C%5C%5C%20V%5Capprox%2065.44984694978736%5Cimplies%20V%3D%5Cstackrel%7B%5Ctextit%7Brounded%20up%7D%7D%7B65.45%7D)
13t-13t=0
so your left with
6=12
which is a false statement which makes this equation have no solutions
$17.49 because you multiply 22 by .25 getting 5.5. You subtract that from the original 22 getting 16.5 which you then multiply by the sales tax .06. Add the tax amount to the 16.5 to get 17.49.
Check the picture below.
the piecewise function, has two subfunctions or behaviours, one if x < 1, meaning less, not equals, but less than 1, so it doesn't include one, so it has a "hole" on that endpoint.
and the second behaviour of the piecewise is if x ⩾ 1, larger or equals than 1, so it includes 1, so it has a solid ball on that endpoint.