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Write an inequality for real numbers x represented on the following number line.

Mathematics

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4 parts
a

2516\frac{25}{16}

b

49360\frac{493}{60}

d
19.063
e

18130\frac{181}{30}

Step-by-step solution

13 steps

Tutatafuta thamani kamili (absolute value) ya kila usemi.

Part a· 3 steps

38−116+54\frac{3}{8} - \frac{1}{16} + \frac{5}{4}

  1. Tafuta denominator ya kawaida (LCM) kwa 8, 16, na 4, ambayo ni 16.

    Badilisha sehemu zote kuwa na denominator 16.

    38=3×28×2=616\frac{3}{8} = \frac{3 \times 2}{8 \times 2} = \frac{6}{16} 54=5×44×4=2016\frac{5}{4} = \frac{5 \times 4}{4 \times 4} = \frac{20}{16}

  2. Fanya hesabu ya kujumlisha na kutoa.

    616−116+2016=6−1+2016=5+2016=2516\frac{6}{16} - \frac{1}{16} + \frac{20}{16} = \frac{6 - 1 + 20}{16} = \frac{5 + 20}{16} = \frac{25}{16}

  3. Tafuta thamani kamili.

    ∣2516∣=2516\left| \frac{25}{16} \right| = \boxed{\frac{25}{16}}

Part b

49360\frac{493}{60}

123+234+3451\frac{2}{3} + 2\frac{3}{4} + 3\frac{4}{5}

  1. Badilisha namba mchanganyiko kuwa sehemu zisizofaa.

    123=1×3+23=531\frac{2}{3} = \frac{1 \times 3 + 2}{3} = \frac{5}{3} 234=2×4+34=1142\frac{3}{4} = \frac{2 \times 4 + 3}{4} = \frac{11}{4} 345=3×5+45=1953\frac{4}{5} = \frac{3 \times 5 + 4}{5} = \frac{19}{5}

  2. Tafuta denominator ya kawaida (LCM) kwa 3, 4, na 5, ambayo ni 60.

    Badilisha sehemu zote kuwa na denominator 60.

    53=5×203×20=10060\frac{5}{3} = \frac{5 \times 20}{3 \times 20} = \frac{100}{60} 114=11×154×15=16560\frac{11}{4} = \frac{11 \times 15}{4 \times 15} = \frac{165}{60} 195=19×125×12=22860\frac{19}{5} = \frac{19 \times 12}{5 \times 12} = \frac{228}{60}

  3. Fanya hesabu ya kujumlisha.

    10060+16560+22860=100+165+22860=49360\frac{100}{60} + \frac{165}{60} + \frac{228}{60} = \frac{100 + 165 + 228}{60} = \frac{493}{60}

  4. Tafuta thamani kamili.

    ∣49360∣=49360\left| \frac{493}{60} \right| = \boxed{\frac{493}{60}}

Part d19.063

62.15−88.765+45.67862.15 - 88.765 + 45.678

  1. Fanya hesabu ya kutoa na kujumlisha.

    62.15−88.765=−26.61562.15 - 88.765 = -26.615 −26.615+45.678=19.063-26.615 + 45.678 = 19.063

  2. Tafuta thamani kamili.

    ∣19.063∣=19.063|19.063| = \boxed{19.063}

Part e

18130\frac{181}{30}

213+415−122\frac{1}{3} + 4\frac{1}{5} - \frac{1}{2}

  1. Badilisha namba mchanganyiko kuwa sehemu zisizofaa.

    213=2×3+13=732\frac{1}{3} = \frac{2 \times 3 + 1}{3} = \frac{7}{3} 415=4×5+15=2154\frac{1}{5} = \frac{4 \times 5 + 1}{5} = \frac{21}{5}

  2. Tafuta denominator ya kawaida (LCM) kwa 3, 5, na 2, ambayo ni 30.

    Badilisha sehemu zote kuwa na denominator 30.

    73=7×103×10=7030\frac{7}{3} = \frac{7 \times 10}{3 \times 10} = \frac{70}{30} 215=21×65×6=12630\frac{21}{5} = \frac{21 \times 6}{5 \times 6} = \frac{126}{30} 12=1×152×15=1530\frac{1}{2} = \frac{1 \times 15}{2 \times 15} = \frac{15}{30}

  3. Fanya hesabu ya kujumlisha na kutoa.

    7030+12630−1530=70+126−1530=196−1530=18130\frac{70}{30} + \frac{126}{30} - \frac{15}{30} = \frac{70 + 126 - 15}{30} = \frac{196 - 15}{30} = \frac{181}{30}

  4. Tafuta thamani kamili.

    ∣18130∣=18130\left| \frac{181}{30} \right| = \boxed{\frac{181}{30}}
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