♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️ [tex]1 \frac{1}{9} \times 4 \frac{1}{2} + (3 \frac{1}{5} – \frac{1}{3} ) = \\ [/tex] [tex] \frac{9 + 1}{9} \times \frac{8 + 1}{2} + ( \frac{15 + 1}{5} – \frac{1}{3} ) = \\ [/tex] [tex] \frac{10}{9} \times \frac{9}{2} + ( \frac{16 \times 3}{5 \times 3} – \frac{1 \times 5}{3 \times 5} ) = \\ [/tex] [tex] \frac{9 \times 10}{9 \times 2} + ( \frac{48}{15} – \frac{5}{15} ) = \\ [/tex] [tex] \frac{10}{2} +( \frac{48 – 5}{15} ) = \\ [/tex] [tex]5 + ( \frac{43}{15} ) = \\ [/tex] [tex] \frac{75}{15} + \frac{43}{15} = \frac{118}{15} \\ [/tex] ♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️ Log in to Reply
Answer: 118/15 or 7.8666666666666666666666666666666666666666666666666666666666666666… decimal Step-by-step explanation: Simplify the following: (1 + 1/9) (4 + 1/2) + 3 + 1/5 – 1/3 Put 1 + 1/9 over the common denominator 9. 1 + 1/9 = 9/9 + 1/9: 9/9 + 1/9 (4 + 1/2) + 3 + 1/5 – 1/3 9/9 + 1/9 = (9 + 1)/9: (9 + 1)/9 (4 + 1/2) + 3 + 1/5 – 1/3 9 + 1 = 10: 10/9 (4 + 1/2) + 3 + 1/5 – 1/3 Put 4 + 1/2 over the common denominator 2. 4 + 1/2 = (2×4)/2 + 1/2: 10/9 (2×4)/2 + 1/2 + 3 + 1/5 – 1/3 2×4 = 8: 10/9 (8/2 + 1/2) + 3 + 1/5 – 1/3 8/2 + 1/2 = (8 + 1)/2: 10/9 (8 + 1)/2 + 3 + 1/5 – 1/3 8 + 1 = 9: 10/9×9/2 + 3 + 1/5 – 1/3 10/9×9/2 = (10×9)/(9×2): (10×9)/(9×2) + 3 + 1/5 – 1/3 (10×9)/(9×2) = 9/9×10/2 = 10/2: 10/2 + 3 + 1/5 – 1/3 The gcd of 10 and 2 is 2, so 10/2 = (2×5)/(2×1) = 2/2×5 = 5: 5 + 3 + 1/5 – 1/3 Put 5 + 3 + 1/5 – 1/3 over the common denominator 15. 5 + 3 + 1/5 – 1/3 = (15×5)/15 + (15×3)/15 + 3/15 – 5/15: (15×5)/15 + (15×3)/15 + 3/15 – 5/15 15×5 = 75: 75/15 + (15×3)/15 + 3/15 – 5/15 15×3 = 45: 75/15 + 45/15 + 3/15 – 5/15 75/15 + 45/15 + 3/15 – 5/15 = (75 + 45 + 3 – 5)/15: (75 + 45 + 3 – 5)/15 | 1 | | 7 | 5 | 4 | 5 + | | 3 1 | 2 | 3: (123 – 5)/15 | | 1 | 13 | 1 | 2 | 3 – | | | 5 | 1 | 1 | 8: Answer: 118/15 Log in to Reply
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[tex]1 \frac{1}{9} \times 4 \frac{1}{2} + (3 \frac{1}{5} – \frac{1}{3} ) = \\ [/tex]
[tex] \frac{9 + 1}{9} \times \frac{8 + 1}{2} + ( \frac{15 + 1}{5} – \frac{1}{3} ) = \\ [/tex]
[tex] \frac{10}{9} \times \frac{9}{2} + ( \frac{16 \times 3}{5 \times 3} – \frac{1 \times 5}{3 \times 5} ) = \\ [/tex]
[tex] \frac{9 \times 10}{9 \times 2} + ( \frac{48}{15} – \frac{5}{15} ) = \\ [/tex]
[tex] \frac{10}{2} +( \frac{48 – 5}{15} ) = \\ [/tex]
[tex]5 + ( \frac{43}{15} ) = \\ [/tex]
[tex] \frac{75}{15} + \frac{43}{15} = \frac{118}{15} \\ [/tex]
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Answer:
118/15 or 7.8666666666666666666666666666666666666666666666666666666666666666… decimal
Step-by-step explanation:
Simplify the following:
(1 + 1/9) (4 + 1/2) + 3 + 1/5 – 1/3
Put 1 + 1/9 over the common denominator 9. 1 + 1/9 = 9/9 + 1/9:
9/9 + 1/9 (4 + 1/2) + 3 + 1/5 – 1/3
9/9 + 1/9 = (9 + 1)/9:
(9 + 1)/9 (4 + 1/2) + 3 + 1/5 – 1/3
9 + 1 = 10:
10/9 (4 + 1/2) + 3 + 1/5 – 1/3
Put 4 + 1/2 over the common denominator 2. 4 + 1/2 = (2×4)/2 + 1/2:
10/9 (2×4)/2 + 1/2 + 3 + 1/5 – 1/3
2×4 = 8:
10/9 (8/2 + 1/2) + 3 + 1/5 – 1/3
8/2 + 1/2 = (8 + 1)/2:
10/9 (8 + 1)/2 + 3 + 1/5 – 1/3
8 + 1 = 9:
10/9×9/2 + 3 + 1/5 – 1/3
10/9×9/2 = (10×9)/(9×2):
(10×9)/(9×2) + 3 + 1/5 – 1/3
(10×9)/(9×2) = 9/9×10/2 = 10/2:
10/2 + 3 + 1/5 – 1/3
The gcd of 10 and 2 is 2, so 10/2 = (2×5)/(2×1) = 2/2×5 = 5:
5 + 3 + 1/5 – 1/3
Put 5 + 3 + 1/5 – 1/3 over the common denominator 15. 5 + 3 + 1/5 – 1/3 = (15×5)/15 + (15×3)/15 + 3/15 – 5/15:
(15×5)/15 + (15×3)/15 + 3/15 – 5/15
15×5 = 75:
75/15 + (15×3)/15 + 3/15 – 5/15
15×3 = 45:
75/15 + 45/15 + 3/15 – 5/15
75/15 + 45/15 + 3/15 – 5/15 = (75 + 45 + 3 – 5)/15:
(75 + 45 + 3 – 5)/15
| 1 |
| 7 | 5
| 4 | 5
+ | | 3
1 | 2 | 3:
(123 – 5)/15
| | 1 | 13
| 1 | 2 | 3
– | | | 5
| 1 | 1 | 8:
Answer: 118/15