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Find the range of values of x which satisfies the following inequalities simultaneously 4x-9 <= x+6 4+x >= 8-3x

Mathematics

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hapa kuna masuluhisho ya maswali:

1. Bila kutumia jedwali za hisabati au kikokotoo, tathmini:

2log⁡5−12log⁡16+2log⁡402 \log 5 - \frac{1}{2} \log 16 + 2 \log 40
  1. Tumia sheria ya logariti nlog⁡a=log⁡ann \log a = \log a^n.

    2log⁡5=log⁡52=log⁡252 \log 5 = \log 5^2 = \log 25 12log⁡16=log⁡161/2=log⁡16=log⁡4\frac{1}{2} \log 16 = \log 16^{1/2} = \log \sqrt{16} = \log 4 2log⁡40=log⁡402=log⁡16002 \log 40 = \log 40^2 = \log 1600
  2. Badilisha maneno haya kwenye usemi wa asili.

    log⁡25−log⁡4+log⁡1600\log 25 - \log 4 + \log 1600
  3. Tumia sheria za logariti log⁡a−log⁡b=log⁡(ab)\log a - \log b = \log \left(\frac{a}{b}\right) na log⁡a+log⁡b=log⁡(ab)\log a + \log b = \log (ab).

    log⁡(254)+log⁡1600\log \left(\frac{25}{4}\right) + \log 1600 log⁡(254×1600)\log \left(\frac{25}{4} \times 1600\right) log⁡(25×400)\log (25 \times 400) log⁡(10000)\log (10000)
  4. Tathmini logariti (tukichukulia msingi ni 10).

    log⁡10(104)=4\log_{10} (10^4) = 4

    Jibu ni 4\boxed{4}.

    2. a) Panua na rahisha usemi wa binomial (2+x)5(2+x)^5 hadi neno la x3x^3.

  5. Tumia fomula ya upanuzi wa binomial (a+b)n=∑k=0n(nk)an−kbk(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k.

    Hapa a=2a=2, b=xb=x, na n=5n=5. Tunahitaji maneno hadi x3x^3, kwa hivyo k=0,1,2,3k=0, 1, 2, 3.

    • Kwa k=0k=0: (50)25x0=1×32×1=32\binom{5}{0} 2^5 x^0 = 1 \times 32 \times 1 = 32
    • Kwa k=1k=1: (51)24x1=5×16×x=80x\binom{5}{1} 2^4 x^1 = 5 \times 16 \times x = 80x
    • Kwa k=2k=2: (52)23x2=10×8×x2=80x2\binom{5}{2} 2^3 x^2 = 10 \times 8 \times x^2 = 80x^2
    • Kwa k=3k=3: (53)22x3=10×4×x3=40x3\binom{5}{3} 2^2 x^3 = 10 \times 4 \times x^3 = 40x^3
  6. Unganisha maneno yaliyopanuliwa.

    (2+x)5≈32+80x+80x2+40x3(2+x)^5 \approx \boxed{32 + 80x + 80x^2 + 40x^3}

    b) Tumia usemi wako kukadiria thamani ya (1.97)5(1.97)^5.

  7. Linganisha (1.97)5(1.97)^5 na (2+x)5(2+x)^5 ili kupata thamani ya xx.

    2+x=1.972+x = 1.97 x=1.97−2=−0.03x = 1.97 - 2 = -0.03
  8. Badilisha x=−0.03x = -0.03 kwenye usemi uliopanuliwa kutoka sehemu (a).

    32+80(−0.03)+80(−0.03)2+40(−0.03)332 + 80(-0.03) + 80(-0.03)^2 + 40(-0.03)^3
  9. Fanya hesabu.

    32−2.4+80(0.0009)+40(−0.000027)32 - 2.4 + 80(0.0009) + 40(-0.000027) 32−2.4+0.072−0.0010832 - 2.4 + 0.072 - 0.00108 29.6+0.072−0.0010829.6 + 0.072 - 0.00108 29.672−0.0010829.672 - 0.00108 29.6709229.67092

    Thamani iliyokadiriwa ni 29.67092\boxed{29.67092}.

    3. Tafuta riba kwenye Ksh. 200,000 kwa miaka 2 kwa 14% kwa mwaka inayojumuishwa nusu-mwaka.

  10. Tambua vigezo.

    Mkuu (P) = Ksh. 200,000
    Muda (t) = miaka 2
    Kiwango cha riba (r) = 14% = 0.14
    Idadi ya mara riba inajumuishwa kwa mwaka (n) = 2 (nusu-mwaka)

  11. Tumia fomula ya kiasi kilichojumuishwa A=P(1+rn)ntA = P\left(1 + \frac{r}{n}\right)^{nt}.

    A=200,000(1+0.142)2×2A = 200,000 \left(1 + \frac{0.14}{2}\right)^{2 \times 2} A=200,000(1+0.07)4A = 200,000 (1 + 0.07)^4 A=200,000(1.07)4A = 200,000 (1.07)^4
  12. Kokotoa thamani ya (1.07)4(1.07)^4.

    (1.07)4≈1.31079601(1.07)^4 \approx 1.31079601
  13. Kokotoa kiasi cha jumla (A).

    A=200,000×1.31079601A = 200,000 \times 1.31079601 A=262,159.202A = 262,159.202
  14. Kokotoa riba (I) kwa kutoa mkuu kutoka kwa kiasi cha jumla.

    I=A−PI = A - P I=262,159.202−200,000I = 262,159.202 - 200,000 I=62,159.202I = 62,159.202

    Riba ni Ksh. 62,159.20\boxed{\text{Ksh. } 62,159.20}.

    4. Katika sehemu A, Grace aliona kilele cha mlingoti wa bendera kwa pembe ya 30∘30^\circ. Baada ya kutembea mita 25 kuelekea chini ya mlingoti wa bendera, alisimama katika sehemu B ambapo aliona tena kwa pembe ya 60∘60^\circ. Tafuta urefu wa mlingoti wa bendera.

  15. Chora mchoro na uweke alama vigezo.

    Acha hh iwe urefu wa mlingoti wa bendera.
    Acha xx iwe umbali kutoka sehemu B hadi chini ya mlingoti wa bendera.
    Umbali kutoka A hadi chini ya mlingoti ni (25+x)(25+x) m.

  16. Tumia uwiano wa tanjenti kwa pembe ya 30∘30^\circ (kutoka A).

    tan⁡30∘=h25+x\tan 30^\circ = \frac{h}{25+x} h=(25+x)tan⁡30∘(1)h = (25+x) \tan 30^\circ \quad (1)
  17. Tumia uwiano wa tanjenti kwa pembe ya 60∘60^\circ (kutoka B).

    tan⁡60∘=hx\tan 60^\circ = \frac{h}{x} h=xtan⁡60∘(2)h = x \tan 60^\circ \quad (2)
  18. Sawazisha milinganyo (1) na (2) na utatue kwa xx.

    Tunajua tan⁡30∘=13\tan 30^\circ = \frac{1}{\sqrt{3}} na tan⁡60∘=3\tan 60^\circ = \sqrt{3}.

    (25+x)13=x3(25+x) \frac{1}{\sqrt{3}} = x \sqrt{3}

    Panua pande zote mbili kwa 3\sqrt{3}:

    25+x=3x25+x = 3x 25=2x25 = 2x x=252=12.5 mx = \frac{25}{2} = 12.5 \text{ m}
  19. Badilisha thamani ya xx kwenye mlinganyo (2) ili kupata hh.

    h=12.5×tan⁡60∘h = 12.5 \times \tan 60^\circ h=12.5×3h = 12.5 \times \sqrt{3} h≈12.5×1.73205h \approx 12.5 \times 1.73205 h≈21.650625h \approx 21.650625

    Urefu wa mlingoti wa bendera ni 21.65 m\boxed{21.65 \text{ m}}.

    5. Rahisisha 57−2\frac{\sqrt{5}}{\sqrt{7}-\sqrt{2}}.

  20. Zidisha nambari na denomineta kwa kiambishi cha denomineta.

    Kiambishi cha 7−2\sqrt{7}-\sqrt{2} ni 7+2\sqrt{7}+\sqrt{2}.

    57−2×7+27+2\frac{\sqrt{5}}{\sqrt{7}-\sqrt{2}} \times \frac{\sqrt{7}+\sqrt{2}}{\sqrt{7}+\sqrt{2}}
  21. Zidisha nambari.

    5(7+2)=5×7+5×2=35+10\sqrt{5}(\sqrt{7}+\sqrt{2}) = \sqrt{5 \times 7} + \sqrt{5 \times 2} = \sqrt{35} + \sqrt{10}
  22. Zidisha denomineta (kumbuka (a−b)(a+b)=a2−b2(a-b)(a+b) = a^2 - b^2).

    (7−2)(7+2)=(7)2−(2)2=7−2=5(\sqrt{7}-\sqrt{2})(\sqrt{7}+\sqrt{2}) = (\sqrt{7})^2 - (\sqrt{2})^2 = 7 - 2 = 5
  23. Andika usemi uliorahisishwa.

    35+105\frac{\sqrt{35} + \sqrt{10}}{5}

    Jibu ni 35+105\boxed{\frac{\sqrt{35} + \sqrt{10}}{5}}.

    6. a) Tafuta safu ya thamani za xx zinazokidhi usawa zifuatazo kwa wakati mmoja:

    4x−9≤x+64x - 9 \le x + 6 4+x≥8−3x4 + x \ge 8 - 3x
  24. Tatua usawa wa kwanza.

    4x−9≤x+64x - 9 \le x + 6 4x−x≤6+94x - x \le 6 + 9 3x≤153x \le 15 x≤5x \le 5
  25. Tatua usawa wa pili.

    4+x≥8−3x4 + x \ge 8 - 3x x+3x≥8−4x + 3x \ge 8 - 4 4x≥44x \ge 4 x≥1x \ge 1
  26. Unganisha masuluhisho ya usawa zote mbili.

    1≤x≤51 \le x \le 5

    Safu ya thamani za xx ni 1≤x≤5\boxed{1 \le x \le 5}.

    b) Wakilisha safu ya thamani za xx katika (a) hapo juu kwenye mstari wa nambari.

    (Kama AI, siwezi kuchora mstari wa nambari, lakini maelezo ni kama ifuatavyo):
    Chora mstari wa nambari. Weka alama nambari 1 na 5.
    Kwa kuwa usawa ni "chini au sawa na" (x≤5x \le 5) na "kubwa au sawa na" (x≥1x \ge 1), tumia duara zilizojazwa (closed circles) kwenye nambari 1 na 5.
    Kisha, paka rangi sehemu ya mstari kati ya 1 na 5.

    7. Fanya tt kuwa mada ya fomula: s=ut+12at2s = ut + \frac{1}{2}at^2.

  27. Panga upya fomula ili iwe katika mfumo wa mlinganyo wa quadratic kwa tt.

    12at2+ut−s=0\frac{1}{2}at^2 + ut - s = 0
  28. Ondoa sehemu kwa kuzidisha pande zote mbili kwa 2.

    at2+2ut−2s=0at^2 + 2ut - 2s = 0
  29. Tumia fomula ya quadratic t=−B±B2−4AC2At = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}, ambapo A=aA=a, B=2uB=2u, na C=−2sC=-2s.

    t=−(2u)±(2u)2−4(a)(−2s)2at = \frac{-(2u) \pm \sqrt{(2u)^2 - 4(a)(-2s)}}{2a} t=−2u±4u2+8as2at = \frac{-2u \pm \sqrt{4u^2 + 8as}}{2a}
  30. Rahisisha usemi chini ya mzizi wa mraba na ugawanye kwa 2.

    t=−2u±4(u2+2as)2at = \frac{-2u \pm \sqrt{4(u^2 + 2as)}}{2a} t=−2u±2u2+2as2at = \frac{-2u \pm 2\sqrt{u^2 + 2as}}{2a} t=−u±u2+2asat = \frac{-u \pm \sqrt{u^2 + 2as}}{a}

    tt kama mada ya fomula ni t=−u±u2+2asa\boxed{t = \frac{-u \pm \sqrt{u^2 + 2as}}{a}}.

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