First determine how many buttons each machine can produce in a day.

Machine A: \(\displaystyle \dpi{100} \frac{4}{2}=2\ buttons\ per\ day\)

Machine B: \(\displaystyle \dpi{100} \frac{20}{4}=5\ buttons\ per\ day\)

Machine A and B can produce 7 buttons per day when they are working together.

\(\displaystyle \dpi{100} 7\times 40=280\ buttons\)

Brian eats \(\displaystyle \dpi{100} \frac{1}{3}\times \frac{1}{4}\times 12\ slices\ of\ pizza\).

So Brian eats

\(\displaystyle \dpi{100} \frac{1}{12}\cdot 12=1\ slice\ of\ pizza.\)

\(\displaystyle \dpi{100} 4x^{2}=256\)

\(\displaystyle \dpi{100} \frac{4x^{2}}{4}=\frac{256}{4}\)

\(\displaystyle \dpi{100} x^{2}=64\)

\(\displaystyle \dpi{100} \sqrt{x^{2}}=\sqrt{64}\)

\(\displaystyle \dpi{100} x=\pm 8\)

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

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

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

\(\displaystyle = 0.6^{2}x^{2} +2\cdot 0.6 \cdot 0.4 \cdot xy + 0.4^{2} y ^{2}\)

\(\displaystyle = 0.36x^{2} + 0.48 xy + 0.16 y ^{2}\)

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

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

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

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

\(\displaystyle = 0.36x^{2} - 0.48 xy + 0.16 y ^{2}\)

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

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

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

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

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

\(\displaystyle (5x - 3y) (3x + 8y)\)

\(\displaystyle = 5x \cdot 3x+5x \cdot 8y - 3y \cdot 3x - 3y \cdot 8y\)

\(\displaystyle = 5\cdot3 \cdot x \cdot x+5\cdot 8 \cdot x \cdot y - 3\cdot 3\cdot y\cdot x - 3\cdot 8 \cdot y \cdot y\)

\(\displaystyle = 15 x ^{2}+40xy - 9xy -24 y^{2}\)

\(\displaystyle = 15 x ^{2}+31xy -24 y^{2}\)

Use the FOIL method:

\(\displaystyle (5x - 3y) (3x - 8y)\)

\(\displaystyle = 5x \cdot 3x-5x \cdot 8y - 3y \cdot 3x + 3y \cdot 8y\)

\(\displaystyle = 5\cdot3 \cdot x \cdot x-5\cdot 8 \cdot x \cdot y - 3\cdot 3\cdot y\cdot x + 3\cdot 8 \cdot y \cdot y\)

\(\displaystyle = 15 x ^{2}-40xy - 9xy +24 y^{2}\)

\(\displaystyle = 15 x ^{2}-49xy +24 y^{2}\)

Substitute \(\displaystyle 3-2x\) for \(\displaystyle x\):

\(\displaystyle f(x) = 5 - 2x\)

\(\displaystyle f(3-2x) = 5 - 2 \left ( 3-2x \right )\)

\(\displaystyle f(3-2x) = 5 - 2 \cdot 3- \left ( - 2 \right )\cdot 2x \right )\)

\(\displaystyle f(3-2x) = 5 - 6 +4x\)

\(\displaystyle f(3-2x) =4x -1\)

First, recognize that raising the fraction to a negative power is the same as raising the inverted fraction to a positive power.

\(\displaystyle (\frac{4x^2}{81x^6})^{-\frac{1}{2}}=(\frac{81x^6}{4x^2})^{\frac{1}{2}}\)

Apply the exponent within the parentheses and simplify. Remember that fractional exponents can be written as roots.

\(\displaystyle \frac{(81x^6)^{\frac{1}{2}}}{(4x^2)^{\frac{1}{2}}}=\frac{\sqrt{81x^6}}{\sqrt{4x^2}}\)

Simplify by taking the roots and canceling common factors.

\(\displaystyle \frac{\sqrt{81x^6}}{\sqrt{4x^2}}=\frac{9x^3}{2x}=\frac{9x^2}{2}\)