In modulo 9 arithmetic, a number is congruent to the remainder of its division by 9. 

Since 

\(\displaystyle 5 + 6 + 5 + 4 + 3 = 23\)

and 

\(\displaystyle 23 \div 9 = 2\textrm{ R }5\),

\(\displaystyle 5 + 6 + 5 + 4 + 3 \equiv 5 \mod 9\),

making "5" the correct response.

\(\displaystyle V\) varies directly as \(\displaystyle T\) and inversely as \(\displaystyle P\). This means that for some constant of variation \(\displaystyle k\),

\(\displaystyle \frac{V P }{T } = k\)

Alternatively, 

\(\displaystyle \frac{V_{1}P_{1}}{T_{1}} = \frac{V_{2}P_{2}}{T_{2}}\)

We can substitute the initial conditions for thevariables on the left side and the final conditions for those on the right side, then solve for \(\displaystyle V_{2}\):

\(\displaystyle \frac{1,000 \cdot 100}{273} = \frac{V_{2} \cdot 110}{293}\)

\(\displaystyle \frac{V_{2} \cdot 110}{293} \cdot \frac{293}{ 110} = \frac{1,000 \cdot 100}{273} \cdot \frac{293}{ 110}\)

\(\displaystyle V_{2} = \frac{1,000 \cdot 100 \cdot 293}{273\cdot 110} \approx 976\)

\(\displaystyle E\) varies directly as both \(\displaystyle m\) and the square of \(\displaystyle v\). This means that for some constant of variation \(\displaystyle k\),

\(\displaystyle \frac{E}{mv^{2}} = k\).

Alternatively stated,

\(\displaystyle \frac{E_{1}}{m_{1}v_{1}^{2}} = \frac{E_{2}}{m_{2}v_{2}^{2}}\).

We can substitute the initial conditions for the variables on the left side and the final conditions for those on the right side, then solve for \(\displaystyle E_{2}\):

\(\displaystyle \frac{8,000}{10 \cdot 40^{2}} = \frac{E_{2}}{15 \cdot 30^{2}}\)

\(\displaystyle \frac{8,000}{10 \cdot 40^{2}} \cdot 15 \cdot 30^{2} = \frac{E_{2}}{15 \cdot 30^{2}} \cdot 15 \cdot 30^{2}\)

\(\displaystyle E_{2} = \frac{8,000 \cdot 15 \cdot 30^{2}}{10 \cdot 40^{2}} = \frac{108,000,000}{16,000}= 6,750\)

Rewrite the two equations by setting \(\displaystyle x = \sqrt [4] {a}\) and \(\displaystyle y = \sqrt [4] {b}\) and substituting:

\(\displaystyle 4 \sqrt[4] {a}+ 7 \sqrt[4] {b} =18\)

\(\displaystyle 4x+7y = 18\)

\(\displaystyle 3\sqrt[4] {a} -5 \sqrt[4] {b} = 34\) 

\(\displaystyle 3x-5y = 34\)

The result is a two-by-two linear system in terms of \(\displaystyle x\) and \(\displaystyle y\):

\(\displaystyle 4x+7y = 18\)

\(\displaystyle 3x-5y = 34\)

This can be solved, among other ways, using Gaussian elimination on an augmented matrix:

\(\displaystyle \begin{bmatrix} 4& 7\\ 3& -5 \end{matrix} \begin{vmatrix} 18 \\ 34 \\ \end{bmatrix}\)

Perform row operations until the left two columns show identity matrix \(\displaystyle \begin{bmatrix} 1 & 0\\ 0 &1 \end{bmatrix}\). One possible sequence:

\(\displaystyle R1-R2 \rightarrow R1\)

\(\displaystyle \begin{bmatrix} 4 - 3 & 7- (-5 )\\ 3& -5 \end{matrix} \begin{vmatrix} 18 - 34 \\ 34 \\ \end{bmatrix}\)

\(\displaystyle \begin{bmatrix} 1 & 12 \\ 3& -5 \end{matrix} \begin{vmatrix} -16 \\ 34 \\ \end{bmatrix}\)

\(\displaystyle R2 - 3R1 \rightarrow R2\)

\(\displaystyle \begin{bmatrix} 1 & 12 \\ 3 - 3(1)& -5 - 3(12) \end{matrix} \begin{vmatrix} -16 \\ 34 - 3(-16) \\ \end{bmatrix}\)

\(\displaystyle \begin{bmatrix} 1 & 12 \\ 0& -41\end{matrix} \begin{vmatrix} -16 \\ 82\\ \end{bmatrix}\)

\(\displaystyle R2 \div (-41 )\rightarrow R2\)

\(\displaystyle \begin{bmatrix} 1 & 12 \\ 0\div (-41 )& -41\div (-41 )\end{matrix} \begin{vmatrix} -16 \\ 82\div (-41 )\\ \end{bmatrix}\)

\(\displaystyle \begin{bmatrix} 1 & 12 \\ 0& 1 \end{matrix} \begin{vmatrix} -16 \\-2\\ \end{bmatrix}\)

\(\displaystyle R1 - 12 R2 \rightarrow R1\)

\(\displaystyle \begin{bmatrix} 1 - 12(0) & 12 - 12(1) \\ 0& 1 \end{matrix} \begin{vmatrix} -16 - 12(-2) \\ 2\\ \end{bmatrix}\)

\(\displaystyle \begin{bmatrix} 1 & 0 \\ 0& 1 \end{matrix} \begin{vmatrix} 8 \\ 2\\ \end{bmatrix}\)

\(\displaystyle x= 8\) and \(\displaystyle y = 2\). In the former equation, substitute \(\displaystyle \sqrt [4] {a}\) back for \(\displaystyle x\), and raise both sides to the power of 4:

\(\displaystyle \sqrt [4] {a} = 8\)

\(\displaystyle (\sqrt [4] {a}) ^{4}= 8^{4}\)

\(\displaystyle a = 4,096\)