Tumia kanuni ya kamba zinazokutana ndani ya duara, ambayo inasema kwamba zao la sehemu za kamba moja ni sawa na zao la sehemu za kamba nyingine.

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XD=16cm au XD=5cmXD = 16\text{cm} \text{ au } XD = 5\text{cm}

503 vitengo vya mraba\frac{50}{3} \text{ vitengo vya mraba}

A=(36,8,3)A = (36, 8, 3)

Step-by-step solution

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14. Kamba mbili AB na CD zinakutana ndani kwa X, ukipewa AX = 10cm, XB = 8cm na CD = 21cm, bainisha urefu wa XD.

  1. Tumia kanuni ya kamba zinazokutana ndani ya duara, ambayo inasema kwamba zao la sehemu za kamba moja ni sawa na zao la sehemu za kamba nyingine.

    AX×XB=CX×XDAX \times XB = CX \times XD

  2. Tambua urefu wa kamba CD.

    Tunajua CD=CX+XDCD = CX + XD.

    CD=21cmCD = 21\text{cm}

    Hivyo, CX=CD−XD=21−XDCX = CD - XD = 21 - XD.

  3. Badilisha thamani zinazojulikana kwenye fomula.

    10cm×8cm=(21−XD)cm×XDcm10\text{cm} \times 8\text{cm} = (21 - XD)\text{cm} \times XD\text{cm} 80=(21−XD)XD80 = (21 - XD)XD

  4. Panua na panga upya mlinganyo kuwa mlinganyo wa quadratic.

    80=21XD−XD280 = 21XD - XD^2 XD2−21XD+80=0XD^2 - 21XD + 80 = 0

  5. Tatua mlinganyo wa quadratic kwa kutumia factorization.

    Tafuta nambari mbili ambazo zikizidishwa zinatoa 80 na zikijumlishwa zinatoa -21. Nambari hizo ni -5 na -16.

    XD2−5XD−16XD+80=0XD^2 - 5XD - 16XD + 80 = 0 XD(XD−5)−16(XD−5)=0XD(XD - 5) - 16(XD - 5) = 0 (XD−16)(XD−5)=0(XD - 16)(XD - 5) = 0

    Hivyo, XD−16=0XD - 16 = 0 au XD−5=0XD - 5 = 0.

    XD=16cm au XD=5cmXD = 16\text{cm} \text{ au } XD = 5\text{cm}

  6. Angalia majibu.

    Ikiwa XD=16cmXD = 16\text{cm}, basi CX=21−16=5cmCX = 21 - 16 = 5\text{cm}.
    Ikiwa XD=5cmXD = 5\text{cm}, basi CX=21−5=16cmCX = 21 - 5 = 16\text{cm}.
    Majibu yote mawili yanawezekana kulingana na jinsi pointi C na D zilivyopangwa kwenye kamba. Kwa kawaida, tunachukua thamani yoyote inayofanya kazi.

    Jibu la mwisho ni:

    XD=16cm au XD=5cm\boxed{XD = 16\text{cm} \text{ au } XD = 5\text{cm}}

    15. Kokotoa eneo lililofungwa na curve y=x2+2xy = x^2 + 2x, mistari x=1,x=3x = 1, x = 3 na mhimili wa x.

  7. Kwanza, angalia kama curve y=x2+2xy = x^2 + 2x iko juu au chini ya mhimili wa x katika kipindi [1,3][1, 3].

    Tafuta mizizi ya y=x2+2x=x(x+2)y = x^2 + 2x = x(x+2). Mizizi ni x=0x=0 na x=−2x=-2.
    Katika kipindi [1,3][1, 3], xx ni chanya na x+2x+2 ni chanya, hivyo y=x(x+2)y = x(x+2) ni chanya. Hii inamaanisha curve iko juu ya mhimili wa x.

  8. Weka integral kamili ili kukokotoa eneo.

    Eneo=∫13(x2+2x) dx\text{Eneo} = \int_{1}^{3} (x^2 + 2x)\,dx

  9. Tafuta antiderivative ya x2+2xx^2 + 2x.

    ∫(x2+2x) dx=x2+12+1+2x1+11+1+C=x33+2x22+C=x33+x2+C\int (x^2 + 2x)\,dx = \frac{x^{2+1}}{2+1} + \frac{2x^{1+1}}{1+1} + C = \frac{x^3}{3} + \frac{2x^2}{2} + C = \frac{x^3}{3} + x^2 + C

  10. Tathmini integral kamili kwa kutumia Theorem ya Msingi ya Calculus.

    Eneo=[x33+x2]13\text{Eneo} = \left[\frac{x^3}{3} + x^2\right]_{1}^{3} Eneo=(333+32)−(133+12)\text{Eneo} = \left(\frac{3^3}{3} + 3^2\right) - \left(\frac{1^3}{3} + 1^2\right) Eneo=(273+9)−(13+1)\text{Eneo} = \left(\frac{27}{3} + 9\right) - \left(\frac{1}{3} + 1\right) Eneo=(9+9)−(13+33)\text{Eneo} = (9 + 9) - \left(\frac{1}{3} + \frac{3}{3}\right) Eneo=18−43\text{Eneo} = 18 - \frac{4}{3} Eneo=543−43\text{Eneo} = \frac{54}{3} - \frac{4}{3} Eneo=503\text{Eneo} = \frac{50}{3}

    Jibu la mwisho ni:

    503 vitengo vya mraba\boxed{\frac{50}{3} \text{ vitengo vya mraba}}

    16. Pointi P ambayo kuratibu zake ni (8,4,−1)(8, 4, -1) inagawanya mstari AB ndani kwa uwiano 4:1. Veta ya nafasi ya pointi B inatolewa na OB⃗=i⃗+3j⃗−2k⃗\vec{OB} = \vec{i} + 3\vec{j} - 2\vec{k}. Bainisha kuratibu za pointi A.

  11. Andika veta za nafasi za pointi P na B.

    OP⃗=8i⃗+4j⃗−k⃗\vec{OP} = 8\vec{i} + 4\vec{j} - \vec{k} OB⃗=i⃗+3j⃗−2k⃗\vec{OB} = \vec{i} + 3\vec{j} - 2\vec{k}

  12. Tumia fomula ya sehemu kwa mgawanyo wa ndani.

    Ikiwa pointi P inagawanya mstari AB kwa uwiano m:nm:n, basi OP⃗=nOA⃗+mOB⃗m+n\vec{OP} = \frac{n\vec{OA} + m\vec{OB}}{m+n}.
    Katika kesi hii, P inagawanya AB kwa uwiano 4:1, hivyo m=4m=4 na n=1n=1.

    OP⃗=1OA⃗+4OB⃗1+4\vec{OP} = \frac{1\vec{OA} + 4\vec{OB}}{1+4} OP⃗=OA⃗+4OB⃗5\vec{OP} = \frac{\vec{OA} + 4\vec{OB}}{5}

  13. Panga upya fomula ili kutatua kwa OA⃗\vec{OA}.

    5OP⃗=OA⃗+4OB⃗5\vec{OP} = \vec{OA} + 4\vec{OB} OA⃗=5OP⃗−4OB⃗\vec{OA} = 5\vec{OP} - 4\vec{OB}

  14. Badilisha veta zinazojulikana na kokotoa OA⃗\vec{OA}.

    OA⃗=5(8i⃗+4j⃗−k⃗)−4(i⃗+3j⃗−2k⃗)\vec{OA} = 5(8\vec{i} + 4\vec{j} - \vec{k}) - 4(\vec{i} + 3\vec{j} - 2\vec{k}) OA⃗=(40i⃗+20j⃗−5k⃗)−(4i⃗+12j⃗−8k⃗)\vec{OA} = (40\vec{i} + 20\vec{j} - 5\vec{k}) - (4\vec{i} + 12\vec{j} - 8\vec{k}) OA⃗=(40−4)i⃗+(20−12)j⃗+(−5−(−8))k⃗\vec{OA} = (40 - 4)\vec{i} + (20 - 12)\vec{j} + (-5 - (-8))\vec{k} OA⃗=36i⃗+8j⃗+(−5+8)k⃗\vec{OA} = 36\vec{i} + 8\vec{j} + (-5 + 8)\vec{k} OA⃗=36i⃗+8j⃗+3k⃗\vec{OA} = 36\vec{i} + 8\vec{j} + 3\vec{k}

  15. Badilisha veta ya nafasi OA⃗\vec{OA} kuwa kuratibu za pointi A.

    Kuratibu za pointi A ni coefficients za i⃗,j⃗,k⃗\vec{i}, \vec{j}, \vec{k}.

    Jibu la mwisho ni:

    A=(36,8,3)\boxed{A = (36, 8, 3)}
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