add or subtract. write in simplest form.\n1. $2\\frac{3}{5}+1\\frac{4}{5}$\n2. $3\\frac{5}{6}-1\\frac{1}{6}$\…

add or subtract. write in simplest form.\n1. $2\\frac{3}{5}+1\\frac{4}{5}$\n2. $3\\frac{5}{6}-1\\frac{1}{6}$\n3. $4\\frac{3}{4}+3\\frac{1}{2}$\n4. $6\\frac{3}{8}-2\\frac{1}{4}$\n5. $5\\frac{9}{10}+8\\frac{2}{5}$\n6. $3\\frac{5}{8}-2\\frac{7}{8}$\n7. $7\\frac{5}{12}-3\\frac{3}{4}$\n8. $1\\frac{3}{5}+2\\frac{5}{6}$\n9. $6 - 2\\frac{3}{4}$\n10. $3\\frac{1}{2}+2\\frac{5}{8}-4\\frac{1}{4}$\n11. geometry find the perimeter of the triangle.
Answer
Explanation:
Step1: Convert mixed - numbers to improper fractions
- For (2\frac{3}{5}+1\frac{4}{5}), (2\frac{3}{5}=\frac{2\times5 + 3}{5}=\frac{13}{5}) and (1\frac{4}{5}=\frac{1\times5+4}{5}=\frac{9}{5}).
Step2: Add the fractions
(\frac{13}{5}+\frac{9}{5}=\frac{13 + 9}{5}=\frac{22}{5}=4\frac{2}{5})
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For (3\frac{5}{6}-1\frac{1}{6}), (3\frac{5}{6}=\frac{3\times6 + 5}{6}=\frac{23}{6}) and (1\frac{1}{6}=\frac{1\times6+1}{6}=\frac{7}{6}). Then (\frac{23}{6}-\frac{7}{6}=\frac{23 - 7}{6}=\frac{16}{6}=\frac{8}{3}=2\frac{2}{3})
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For (4\frac{3}{4}+3\frac{1}{2}), (4\frac{3}{4}=\frac{4\times4 + 3}{4}=\frac{19}{4}) and (3\frac{1}{2}=\frac{3\times2+1}{2}=\frac{7}{2}=\frac{14}{4}). Then (\frac{19}{4}+\frac{14}{4}=\frac{19 + 14}{4}=\frac{33}{4}=8\frac{1}{4})
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For (6\frac{3}{8}-2\frac{1}{4}), (6\frac{3}{8}=\frac{6\times8 + 3}{8}=\frac{51}{8}) and (2\frac{1}{4}=\frac{2\times4+1}{4}=\frac{9}{4}=\frac{18}{8}). Then (\frac{51}{8}-\frac{18}{8}=\frac{51 - 18}{8}=\frac{33}{8}=4\frac{1}{8})
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For (5\frac{9}{10}+8\frac{2}{5}), (5\frac{9}{10}=\frac{5\times10 + 9}{10}=\frac{59}{10}) and (8\frac{2}{5}=\frac{8\times5+2}{5}=\frac{42}{5}=\frac{84}{10}). Then (\frac{59}{10}+\frac{84}{10}=\frac{59 + 84}{10}=\frac{143}{10}=14\frac{3}{10})
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For (3\frac{5}{8}-2\frac{7}{8}), (3\frac{5}{8}=\frac{3\times8 + 5}{8}=\frac{29}{8}) and (2\frac{7}{8}=\frac{2\times8+7}{8}=\frac{23}{8}). Then (\frac{29}{8}-\frac{23}{8}=\frac{29 - 23}{8}=\frac{6}{8}=\frac{3}{4})
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For (7\frac{5}{12}-3\frac{3}{4}), (7\frac{5}{12}=\frac{7\times12 + 5}{12}=\frac{89}{12}) and (3\frac{3}{4}=\frac{3\times4+3}{4}=\frac{15}{4}=\frac{45}{12}). Then (\frac{89}{12}-\frac{45}{12}=\frac{89 - 45}{12}=\frac{44}{12}=\frac{11}{3}=3\frac{2}{3})
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For (1\frac{3}{5}+2\frac{5}{6}), (1\frac{3}{5}=\frac{1\times5 + 3}{5}=\frac{8}{5}) and (2\frac{5}{6}=\frac{2\times6+5}{6}=\frac{17}{6}). The common denominator is (30), (\frac{8}{5}=\frac{8\times6}{5\times6}=\frac{48}{30}) and (\frac{17}{6}=\frac{17\times5}{6\times5}=\frac{85}{30}). Then (\frac{48}{30}+\frac{85}{30}=\frac{48 + 85}{30}=\frac{133}{30}=4\frac{13}{30})
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For (6 - 2\frac{3}{4}), (6=\frac{24}{4}) and (2\frac{3}{4}=\frac{2\times4+3}{4}=\frac{11}{4}). Then (\frac{24}{4}-\frac{11}{4}=\frac{24 - 11}{4}=\frac{13}{4}=3\frac{1}{4})
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For (3\frac{1}{2}+2\frac{5}{8}-4\frac{1}{4}), (3\frac{1}{2}=\frac{7}{2}=\frac{28}{8}), (2\frac{5}{8}=\frac{21}{8}), (4\frac{1}{4}=\frac{17}{4}=\frac{34}{8}). (\frac{28}{8}+\frac{21}{8}-\frac{34}{8}=\frac{28 + 21-34}{8}=\frac{15}{8}=1\frac{7}{8})
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For the perimeter of the triangle with side - lengths (1\frac{5}{6}\text{ in}), (1\frac{1}{6}\text{ in}), and (2\frac{2}{3}\text{ in}).
- (1\frac{5}{6}=\frac{11}{6}), (1\frac{1}{6}=\frac{7}{6}), (2\frac{2}{3}=\frac{8}{3}=\frac{16}{6})
- The perimeter (P=\frac{11}{6}+\frac{7}{6}+\frac{16}{6}=\frac{11 + 7+16}{6}=\frac{34}{6}=\frac{17}{3}=5\frac{2}{3}\text{ in})
Answer:
- (4\frac{2}{5})
- (2\frac{2}{3})
- (8\frac{1}{4})
- (4\frac{1}{8})
- (14\frac{3}{10})
- (\frac{3}{4})
- (3\frac{2}{3})
- (4\frac{13}{30})
- (3\frac{1}{4})
- (1\frac{7}{8})
- (5\frac{2}{3}\text{ in})