Question

Please help me answer this question its urgent

Answer 1

Answer:

Step-by-step explanation:

1). $\frac{2}{3}n-\frac{3}{4}n+\frac{1}{6}n+2\frac{2}{9}n$

= $\frac{2}{3}n-\frac{3}{4}n+\frac{1}{6}n+(2+\frac{2}{9})n$

= $(\frac{2}{3}-\frac{3}{4}+\frac{1}{6})n+(2+\frac{2}{9})n$

= $(\frac{8}{12}-\frac{9}{12}+\frac{2}{12})n+(2+\frac{2}{9})n$

= $\frac{(8-9+2)}{12}n+(2+\frac{2}{9})n$

= $\frac{1}{12}n+(2+\frac{2}{9})n$

= $(\frac{1}{12}+2+\frac{2}{9})n$

= $(\frac{3}{36}+\frac{72}{36}+\frac{8}{36})n$

= $\frac{83}{36}n$

= $2\frac{11}{36}n$

2). $\frac{2}{5}g-\frac{1}{6}-g+\frac{3}{10}g-\frac{4}{5}$

= $(\frac{2}{5}-1+\frac{3}{10})g-(\frac{1}{6}+\frac{4}{5})$

= $(\frac{4}{10}-\frac{10}{10}+\frac{3}{10})g-(\frac{5}{30}+ \frac{24}{30})$

= $(\frac{4-10+3}{10})g-(\frac{5+24}{30})$

= $\frac{-3}{10}g-\frac{29}{30}$

= $-\frac{3}{10}g-\frac{29}{30}$

3). $i+6i-\frac{3}{7}i+\frac{1}{3}h+\frac{1}{2}i-h+\frac{1}{4}h$

= $7i-\frac{3}{7}i+\frac{1}{2}i+\frac{1}{3}h-h+\frac{1}{4}h$

= $(7-\frac{3}{7}+\frac{1}{2})i+(\frac{1}{3}-1+\frac{1}{4})h$

= $(\frac{98}{14} -\frac{6}{14}+\frac{7}{14})i+(\frac{4}{12}-\frac{12}{12}+\frac{3}{12})h$

= $(\frac{98-6+7}{14})i+(\frac{4-12+3}{12})h$

= $\frac{99}{14}i-\frac{5}{12}h$