Question

1. Type an equation in the equation editor that uses 2 fractions with parentheses around one of them. Example: /tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B3%7D" id="TexFormula1" title="\frac{2}{3}" alt="\frac{2}{3}" align="absmiddle" class="latex-formula"> + (- $\frac{1}{2}$) = $\frac{4}{6} - \frac{3}{6} = \frac{1}{6}$
2. Type an expression that has two terms with exponents, and one with a square root. Example: $2^{3}$ + $9^{2}$ + $\sqrt{16}$

3. Type a compound inequality similar to the one below, but with different numbers. It should be set up the same, with all the symbols in the same places. $(\frac{3}{5} )^{2}$ · $^{3} \sqrt{10} \leq x^{3} - 2x + 5 \leq \sqrt{\frac{1}{3}$

Answer 1

Answer:

i) $\frac{3}{5} + (- \frac{1}{2}) = \frac{6}{10} - \frac{5}{10} = \frac{1}{10}$    $\Rightarrow$ \frac{3}{5} + (- \frac{1}{2}) = \frac{6}{10} - \frac{5}{10} = \frac{1}{10}

ii)$4^{3} + 8^{2} + \sqrt{9}$   $\Rightarrow$  4^{3} + 8^{2} + \sqrt{9}

iii) $(\frac{4}{5})^{2}. \sqrt[3]{8} \leqx^{3} - 3x + 6 \leq \sqrt{\frac{1}{3}} \Rightarrow \hspace{0.2cm}$    (\frac{4}{5})^{2}. \sqrt[3]{8} \leqx^{3} - 3x + 6 \leq \sqrt{\frac{1}{3}}

Step-by-step explanation:

i) $\frac{3}{5} + (- \frac{1}{2}) = \frac{6}{10} - \frac{5}{10} = \frac{1}{10}$    $\Rightarrow$ \frac{3}{5} + (- \frac{1}{2}) = \frac{6}{10} - \frac{5}{10} = \frac{1}{10}

ii)$4^{3} + 8^{2} + \sqrt{9}$   $\Rightarrow$  4^{3} + 8^{2} + \sqrt{9}

iii) $(\frac{4}{5})^{2}. \sqrt[3]{8} \leqx^{3} - 3x + 6 \leq \sqrt{\frac{1}{3}} \Rightarrow \hspace{0.2cm}$    (\frac{4}{5})^{2}. \sqrt[3]{8} \leqx^{3} - 3x + 6 \leq \sqrt{\frac{1}{3}}