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A) In a survey at airport, 55 travelers said that last year they had been to Spain, 53 to France and 30 had been to France and Spain. Use Venn diagram to find the number of travelers who took part in the survey.

Mathematics

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Haya hapa masuluhisho ya maswali yako:

2. b) Kutafuta thamani ya log⁡45\log 45:

  1. Tumia sheria za logariti kueleza log⁡45\log 45 kwa kutumia log⁡3\log 3 na log⁡5\log 5.

    Tunajua kuwa 45=9×5=32×545 = 9 \times 5 = 3^2 \times 5.
    Hivyo, log⁡45=log⁡(32×5)\log 45 = \log (3^2 \times 5).

  2. Tumia sheria ya kuzidisha ya logariti (log⁡(ab)=log⁡a+log⁡b\log(ab) = \log a + \log b) na sheria ya nguvu (log⁡(an)=nlog⁡a\log(a^n) = n \log a).

    log⁡45=log⁡(32)+log⁡5\log 45 = \log (3^2) + \log 5 log⁡45=2log⁡3+log⁡5\log 45 = 2 \log 3 + \log 5
  3. Badilisha thamani ulizopewa za log⁡3\log 3 na log⁡5\log 5.

    log⁡45=2(0.4771)+0.6990\log 45 = 2(0.4771) + 0.6990
  4. Fanya hesabu.

    log⁡45=0.9542+0.6990\log 45 = 0.9542 + 0.6990 log⁡45=1.6532\log 45 = 1.6532

    Thamani ya log⁡45\log 45 ni 1.6532\boxed{1.6532}.

    3. a) Swali hili halijasomeka kikamilifu kutokana na picha iliyokatika. Tafadhali tuma picha kamili ya swali hili.

    3. b) Kutafuta seti A′A' na (A∩B)′(A \cap B)':

    Kwanza, orodhesha vipengele vya kila seti:
    Seti ya ulimwengu U={1,2,3,...,25}U = \{1, 2, 3, ..., 25\}
    Seti A={nambari asilia chini ya 20}={1,2,3,...,19}A = \{\text{nambari asilia chini ya 20}\} = \{1, 2, 3, ..., 19\}
    Seti B={nambari shufwa chini ya 10}={2,4,6,8}B = \{\text{nambari shufwa chini ya 10}\} = \{2, 4, 6, 8\}

    i) A′A' (Kikamilisho cha A):

  5. A′A' inajumuisha vipengele vyote vilivyo katika UU lakini havipo katika AA.

    A′=U−AA' = U - A A′={1,2,...,25}−{1,2,...,19}A' = \{1, 2, ..., 25\} - \{1, 2, ..., 19\}
  6. Orodhesha vipengele vilivyobaki.

    A′={20,21,22,23,24,25}A' = \{20, 21, 22, 23, 24, 25\}

    Seti A′A' ni {20,21,22,23,24,25}\boxed{\{20, 21, 22, 23, 24, 25\}}.

    ii) (A∩B)′(A \cap B)' (Kikamilisho cha makutano ya A na B):

  7. Kwanza, tafuta makutano ya AA na BB, yaani A∩BA \cap B.

    Hivi ni vipengele vilivyo katika AA na BB.

    A∩B={1,2,...,19}∩{2,4,6,8}A \cap B = \{1, 2, ..., 19\} \cap \{2, 4, 6, 8\} A∩B={2,4,6,8}A \cap B = \{2, 4, 6, 8\}
  8. Kisha, tafuta kikamilisho cha A∩BA \cap B, yaani (A∩B)′(A \cap B)'.

    Hivi ni vipengele vyote vilivyo katika UU lakini havipo katika A∩BA \cap B.

    (A∩B)′=U−(A∩B)(A \cap B)' = U - (A \cap B) (A∩B)′={1,2,...,25}−{2,4,6,8}(A \cap B)' = \{1, 2, ..., 25\} - \{2, 4, 6, 8\}
  9. Orodhesha vipengele vilivyobaki.

    (A∩B)′={1,3,5,7,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25}(A \cap B)' = \{1, 3, 5, 7, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25\}

    Seti (A∩B)′(A \cap B)' ni {1,3,5,7,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25}\boxed{\{1, 3, 5, 7, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25\}}.

    4. a) Kutafuta mlinganyo wa mstari ulionyooka unaopita kwenye pointi P(−4,5)P(-4, 5) na Q(−2,1)Q(-2, 1):

  10. Tafuta mteremko (mm) wa mstari kwa kutumia formula m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}.

    Kwa pointi P(−4,5)P(-4, 5) na Q(−2,1)Q(-2, 1):

    m=1−5−2−(−4)m = \frac{1 - 5}{-2 - (-4)} m=−4−2+4m = \frac{-4}{-2 + 4} m=−42m = \frac{-4}{2} m=−2m = -2
  11. Tumia mlinganyo wa mstari y−y1=m(x−x1)y - y_1 = m(x - x_1) na pointi P(−4,5)P(-4, 5) na mteremko m=−2m = -2.

    y−5=−2(x−(−4))y - 5 = -2(x - (-4)) y−5=−2(x+4)y - 5 = -2(x + 4)
  12. Rahisisha mlinganyo katika mfumo wa y=mx+cy = mx + c.

    y−5=−2x−8y - 5 = -2x - 8 y=−2x−8+5y = -2x - 8 + 5 y=−2x−3y = -2x - 3

    Mlinganyo wa mstari ni y=−2x−3\boxed{y = -2x - 3}.

    4. b) Mstari unaopita kwenye pointi A(2,−k)A(2, -k) na B(4,3)B(4, 3) una mteremko wa 2:

    i) Thamani ya kk:

  13. Tumia formula ya mteremko m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1} na mteremko uliotolewa m=2m=2.

    Kwa pointi A(2,−k)A(2, -k) na B(4,3)B(4, 3):

    2=3−(−k)4−22 = \frac{3 - (-k)}{4 - 2} 2=3+k22 = \frac{3 + k}{2}
  14. Tatua kwa kk.

    2×2=3+k2 \times 2 = 3 + k 4=3+k4 = 3 + k k=4−3k = 4 - 3 k=1k = 1

    Thamani ya kk ni 1\boxed{1}.

    ii) Mlinganyo wa mstari ABAB:

  15. Sasa tunajua k=1k=1, hivyo pointi AA ni (2,−1)(2, -1).

    Tuna pointi A(2,−1)A(2, -1) na B(4,3)B(4, 3), na mteremko m=2m=2.
    Tumia mlinganyo wa mstari y−y1=m(x−x1)y - y_1 = m(x - x_1) na pointi A(2,−1)A(2, -1) na mteremko m=2m=2.

    y−(−1)=2(x−2)y - (-1) = 2(x - 2) y+1=2x−4y + 1 = 2x - 4
  16. Rahisisha mlinganyo katika mfumo wa y=mx+cy = mx + c.

    y=2x−4−1y = 2x - 4 - 1 y=2x−5y = 2x - 5

    Mlinganyo wa mstari ABAB ni y=2x−5\boxed{y = 2x - 5}.

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