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Ch. R - Algebra Review
Lial - Trigonometry 12th Edition
Lial12th EditionTrigonometryISBN: 9780136552161, 9780135440780, 9780136881117, 9780135440827, 9780135924891Non è quello che usi tu?Cambia libro di testo
Capitolo 1, Problema 29

Find each sum or difference. See Example 1. 9⁄10 - ( -4⁄3)

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1
Recognize that the problem involves subtracting a negative fraction: \(\frac{9}{10} - \left(-\frac{4}{3}\right)\). Subtracting a negative is equivalent to adding the positive, so rewrite the expression as \(\frac{9}{10} + \frac{4}{3}\).
To add the fractions \(\frac{9}{10}\) and \(\frac{4}{3}\), find a common denominator. The denominators are 10 and 3, so the least common denominator (LCD) is the least common multiple of 10 and 3.
Calculate the least common multiple (LCM) of 10 and 3. Since 10 = 2 × 5 and 3 is prime, the LCM is \(2 \times 5 \times 3 = 30\). So, the common denominator is 30.
Convert each fraction to an equivalent fraction with denominator 30: multiply numerator and denominator of \(\frac{9}{10}\) by 3 to get \(\frac{27}{30}\), and multiply numerator and denominator of \(\frac{4}{3}\) by 10 to get \(\frac{40}{30}\).
Now add the fractions: \(\frac{27}{30} + \frac{40}{30} = \frac{27 + 40}{30} = \frac{67}{30}\). This is the sum expressed as an improper fraction.

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