2. Find the number that lies halfway between these fractions. a) \( \frac{1}{3} \) and \( \frac{5}{9} \) b) \( \frac{7}{15} \) and \( \frac{1}{3} \) 3. Calculate. a) \( \frac{1}{2}-\frac{3}{4} \times \frac{7}{6}+2 \) b) \( \left(-\frac{1}{2}\right)^{2} \times\left(-\frac{3}{4}\right)^{2} \) c) \( \sqrt[3]{-\frac{8}{216}}+\frac{1}{2} \) d) \( \sqrt{\frac{16}{25}} \) 4. Find three fractions between \( \frac{4}{10} \) and \( \frac{4}{5} \). 5. What is the difference between \( \frac{3}{5} \) and \( \frac{2}{3} \) ? 6. What must be subtracted from \( \frac{8}{9} \) to give \( \frac{7}{8} \) ?

**2a. Find the number halfway between \(\frac{1}{3}\) and \(\frac{5}{9}\).** To find the midpoint, calculate the average: \[ \text{Midpoint} = \frac{1}{2}\left(\frac{1}{3}+\frac{5}{9}\right) \] Write \(\frac{1}{3}\) with denominator 9: \[ \frac{1}{3} = \frac{3}{9} \] Now add: \[ \frac{3}{9}+\frac{5}{9}=\frac{8}{9} \] Then multiply by \(\frac{1}{2}\): \[ \frac{1}{2}\times \frac{8}{9}=\frac{8}{18}=\frac{4}{9} \] --- **2b. Find the number halfway between \(\frac{7}{15}\) and \(\frac{1}{3}\).** First, express \(\frac{1}{3}\) with denominator 15: \[ \frac{1}{3}=\frac{5}{15} \] Now compute their sum: \[ \frac{7}{15}+\frac{5}{15}=\frac{12}{15}=\frac{4}{5} \] Then the midpoint is: \[ \text{Midpoint}=\frac{1}{2}\times \frac{4}{5}=\frac{4}{10}=\frac{2}{5} \] --- **3a. Calculate \( \frac{1}{2}-\frac{3}{4} \times \frac{7}{6}+2 \).** First, perform the multiplication: \[ \frac{3}{4} \times \frac{7}{6}=\frac{21}{24}=\frac{7}{8} \] Now substitute back: \[ \frac{1}{2}-\frac{7}{8}+2 \] Express \(\frac{1}{2}\) as \(\frac{4}{8}\): \[ \frac{4}{8}-\frac{7}{8}=-\frac{3}{8} \] Finally, add 2 (written as \(\frac{16}{8}\)): \[ -\frac{3}{8}+\frac{16}{8}=\frac{13}{8} \] --- **3b. Calculate \( \left(-\frac{1}{2}\right)^{2} \times\left(-\frac{3}{4}\right)^{2} \).** Square each term (note that the square of a negative number is positive): \[ \left(-\frac{1}{2}\right)^{2}=\frac{1}{4},\quad \left(-\frac{3}{4}\right)^{2}=\frac{9}{16} \] Multiply the results: \[ \frac{1}{4}\times \frac{9}{16}=\frac{9}{64} \] --- **3c. Calculate \( \sqrt[3]{-\frac{8}{216}}+\frac{1}{2} \).** Simplify the expression inside the cube root: \[ -\frac{8}{216}=-\frac{1}{27}\quad \text{(dividing numerator and denominator by 8)} \] Taking the cube root: \[ \sqrt[3]{-\frac{1}{27}}=-\frac{1}{3} \] Now add \(\frac{1}{2}\) (using a common denominator of 6): \[ -\frac{1}{3}+\frac{1}{2}=-\frac{2}{6}+\frac{3}{6}=\frac{1}{6} \] --- **3d. Calculate \( \sqrt{\frac{16}{25}} \).** Take the square root of the numerator and denominator: \[ \sqrt{\frac{16}{25}}=\frac{\sqrt{16}}{\sqrt{25}}=\frac{4}{5} \] --- **4. Find three fractions between \(\frac{4}{10}\) and \(\frac{4}{5}\).** First, simplify \(\frac{4}{10}\): \[ \frac{4}{10}=\frac{2}{5} \] Express both fractions with a common denominator, for example 10: \[ \frac{2}{5}=\frac{4}{10} \quad \text{and} \quad \frac{4}{5}=\frac{8}{10} \] The fractions between \(\frac{4}{10}\) and \(\frac{8}{10}\) can be: \[ \frac{5}{10},\quad \frac{6}{10},\quad \frac{7}{10} \] In simplest form: \[ \frac{5}{10}=\frac{1}{2},\quad \frac{6}{10}=\frac{3}{5},\quad \frac{7}{10}\text{ remains the same} \] --- **5. What is the difference between \(\frac{3}{5}\) and \(\frac{2}{3}\)?** Subtract the smaller fraction from the larger one. Express both with a common denominator (15): \[ \frac{3}{5}=\frac{9}{15},\quad \frac{2}{3}=\frac{10}{15} \] Subtract: \[ \frac{10}{15}-\frac{9}{15}=\frac{1}{15} \] --- **6. What must be subtracted from \(\frac{8}{9}\) to give \(\frac{7}{8}\)?** Let the number to subtract be \(x\). Then: \[ \frac{8}{9}-x=\frac{7}{8} \] Solve for \(x\): \[ x=\frac{8}{9}-\frac{7}{8} \] Find a common denominator (72): \[ \frac{8}{9}=\frac{64}{72},\quad \frac{7}{8}=\frac{63}{72} \] Subtract: \[ x=\frac{64}{72}-\frac{63}{72}=\frac{1}{72} \]