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)
...6 and 10 are roots, so x – 6 and x – 10 are factors.
y = a(x – 6)(x – 10).....plug in the point (8, 2) and solve for a:
2 = a(8 – 6)(8 – 10)
2 = –4a
a = –1/2
...y = (–1/2)(x – 6)(x – 10)
...y = (–1/2)(x² – 16x + 60)
...y = (–x²/2) + 8x – 30 <<<------Answer, or:
...y = (–1/2)(x – 8)² + 2 <<<------Answer
<em>Given - a+b+c = 0</em>
<em>To prove that- </em>
<em>a²/bc + b²/ac + c²/ab = 3</em>
<em>Now we know that</em>
<em>when x+y+z = 0,</em>
<em>then x³+y³+z³ = 3xyz</em>
<em>that means</em>
<em> (x³+y³+z³)/xyz = 3 ---- eq 1)</em>
<em>Lets solve for LHS</em>
<em>LHS = a²/bc + b²/ac + c²/ab</em>
<em>we can write it as LHS = a³/abc + b³/abc + c</em><em>³</em><em>/abc</em>
<em>by multiplying missing denominators,</em>
<em>now take common abc from denominator and you'll get,</em>
<em>LHS = (a³+b³+c³)/abc --- eq (2)</em>
<em>Comparing one and two we can say that</em>
<em>(a³+b³+c³)/abc = 3</em>
<em>Hence proved,</em>
<em>a²/bc + b²/ac + c²/ab = 3</em>
