This can be solved using the pattern for the sum times difference of two terms:

\(\displaystyle (0.6x-0.4y)(0.6x+0.4y)\)

\(\displaystyle =\left ( 0.6x \right )^{2} -\left ( 0.4y \right ) ^{2}\)

\(\displaystyle =0.6^{2}x^{2}- 0.4^{2}y^{2}\)

\(\displaystyle =0.36x^{2}- 0.16y^{2}\)

This can be solved using the pattern for the sum times difference of two terms:

\(\displaystyle \left (\frac{4}{3}x + \frac{5}{3} \right )\left (\frac{4}{3}x - \frac{5}{3} \right )\)

\(\displaystyle =\left (\frac{4}{3}x \right ) ^{2}- \left (\frac{5}{3} \right )^{2}\)

\(\displaystyle =\frac{4 ^{2}}{3 ^{2}}\cdot x ^{2}- \frac{5^{2}}{3^{2}}\)

\(\displaystyle =\frac{16}{9} x ^{2}- \frac{25}{9}\)

This can be solved using the pattern for the square of a binomial:

\(\displaystyle \left (\frac{4}{3}x - \frac{5}{3} \right )^{2}\)

\(\displaystyle =\left (\frac{4}{3}x \right ) ^{2} -2 \cdot \frac{4}{3}x \cdot \frac{5}{3} + \left ( \frac{5}{3} \right ) ^{2}\)

\(\displaystyle =\frac{4^{2}}{3^{2}}\cdot x^{2} -2 \cdot \frac{4}{3}\cdot \frac{5}{3} \cdot x +\frac{5^{2}}{3^{2}}\)

\(\displaystyle =\frac{16}{9}x^{2} - \frac{40}{9}x +\frac{25}{9}\)

\(\displaystyle N^{2} - 2N + 1 = (N - 1)^{2}\). Since \(\displaystyle N > 1\), \(\displaystyle N - 1 > 0\), and  \(\displaystyle N^{2} - 2N + 1 = (N - 1)^{2} > 0\). This makes (a) greater.

If \(\displaystyle N = 2\), then \(\displaystyle N^{2} - 4N + 4 = 2 ^{2} - 4 \cdot 2 + 4= 4 - 8 + 4 = 0\).

If \(\displaystyle N = 3\), then \(\displaystyle N^{2} - 4N + 4 = 3 ^{2} - 4 \cdot 3 + 4= 9 - 12 + 4 = 1\).

Therefore, at least two possibilities can be demonstrated. 

\(\displaystyle -8 (x^{2}-x-6 )\)

\(\displaystyle =( -8 ) x^{2}-( -8 )x-( -8 )6\)

\(\displaystyle =-8 x^{2}-( -8x )-( -48 )\)

\(\displaystyle =-8 x^{2}+ 8x + 48\)

Since \(\displaystyle 48 > -48\),

\(\displaystyle -8 x^{2}+ 8x + 48 > -8 x^{2}+ 8x - 48\).

Each expression can be simplified using the properties of exponents, among others:

\(\displaystyle \frac{32y^{8}}{2y^{4}} = \frac{32 }{2 } \cdot \frac{ y^{8}}{ y^{4}} = 16 y^{8-4} =16y^{4}\)

\(\displaystyle 8 y^{2} \cdot 2y^{2} = 8 \cdot 2 \cdot y^{2} \cdot y^{2} = 16 y ^{2+2}= 16 y ^{4}\)

\(\displaystyle 4y^{4} + 4y^{4} +4y^{4} + 4y^{4} = (4+4+4+4)y^{4} = 16y^{4}\)

\(\displaystyle \left ( 4y^{4}\right )^{4} = \left ( 4 \right )^{4} \cdot \left ( y^{4}\right )^{4} = 256y^{4 \cdot 4} = 256y^{16}\)

The correct choice is \(\displaystyle \left ( 4y^{4}\right )^{4}\), as it is the only choice not equivalent to \(\displaystyle 16 y^{4}\).

\(\displaystyle -4 > -5\),

So by the multiplication property of inequality, when each is multiplied by the negative number \(\displaystyle y\),

\(\displaystyle -4y < -5y\)

and (B) is greater.

117% of \(\displaystyle N\) is equal to \(\displaystyle 1.17 N\).

One-ninth of this is \(\displaystyle \frac{1}{9} \cdot 1.17 N = \frac{1.17 }{9} \cdot N = 0.13N\)

143% of \(\displaystyle N\) is equal to \(\displaystyle 1.43N\)

One-eleventh of this is \(\displaystyle \frac{1}{11} \cdot 1.43 N = \frac{1.43 }{11} \cdot N = 0.13N\)

Regardless of the value of \(\displaystyle N\), the quanitites are equal.

Use the difference of squares pattern as follows:

\(\displaystyle \left ( 11x - \sqrt {11x}\right )\left ( 11x +\sqrt {11x}\right )\)

\(\displaystyle =\left ( 11x \right) ^{2} - \left ( \sqrt {11x}\right ) ^{2}\)

\(\displaystyle =11^{2}x^{2} - 11x\)

\(\displaystyle =121x^{2} - 11x\)