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}}

7 0

4 years ago

PLZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZSelect all of the choices that are equal to (-5) - (-12).

valentina_108 [34]

Answer:

2 and 3

Step-by-step explanation:

subtract (-5) - (-12) and get 7 plus if you see a - that is finding the difference. therefore your answers are 2 and 3. Hope this helps! ^w^

4 0

3 years ago