This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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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. Step 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 × XD Step 2: Tambua urefu wa kamba CD. Tunajua CD = CX + XD. CD = 21cm Hivyo, CX = CD - XD = 21 - XD. Step 3: Badilisha thamani zinazojulikana kwenye fomula. 10cm × 8cm = (21 - XD)cm × XDcm 80 = (21 - XD)XD Step 4: Panua na panga upya mlinganyo kuwa mlinganyo wa quadratic. 80 = 21XD - XD^2 XD^2 - 21XD + 80 = 0 Step 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. XD^2 - 5XD - 16XD + 80 = 0 XD(XD - 5) - 16(XD - 5) = 0 (XD - 16)(XD - 5) = 0 Hivyo, XD - 16 = 0 au XD - 5 = 0. XD = 16cm au XD = 5cm Step 6: Angalia majibu. Ikiwa XD = 16cm, basi CX = 21 - 16 = 5cm. Ikiwa XD = 5cm, basi CX = 21 - 5 = 16cm. 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 15. Kokotoa eneo lililofungwa na curve y = x^2 + 2x, mistari x = 1, x = 3 na mhimili wa x. Step 1: Kwanza, angalia kama curve y = x^2 + 2x iko juu au chini ya mhimili wa x katika kipindi [1, 3]. Tafuta mizizi ya y = x^2 + 2x = x(x+2). Mizizi ni x=0 na x=-2. Katika kipindi [1, 3], x ni chanya na x+2 ni chanya, hivyo y = x(x+2) ni chanya. Hii inamaanisha curve iko juu ya mhimili wa x. Step 2: Weka integral kamili ili kukokotoa eneo. Eneo = _1^3 (x^2 + 2x)\,dx Step 3: Tafuta antiderivative ya x^2 + 2x. (x^2 + 2x)\,dx = x^2+12+1 + 2x^1+11+1 + C = (x^3)/(3) + (2x^2)/(2) + C = (x^3)/(3) + x^2 + C Step 4: Tathmini integral kamili kwa kutumia Theorem ya Msingi ya Calculus. Eneo = [(x^3)/(3) + x^2]_1^3 Eneo = ((3^3)/(3) + 3^2) - ((1^3)/(3) + 1^2) Eneo = ((27)/(3) + 9) - ((1)/(3) + 1) Eneo = (9 + 9) - ((1)/(3) + (3)/(3)) Eneo = 18 - (4)/(3) Eneo = (54)/(3) - (4)/(3) Eneo = (50)/(3) Jibu la mwisho ni: (50)/(3) vitengo vya mraba 16. Pointi P ambayo kuratibu zake ni (8, 4, -1) inagawanya mstari AB ndani kwa uwiano 4:1. Veta ya nafasi ya pointi B inatolewa na OB = i + 3j - 2k. Bainisha kuratibu za pointi A. Step 1: Andika veta za nafasi za pointi P na B. OP = 8i + 4j - k OB = i + 3j - 2k Step 2: Tumia fomula ya sehemu kwa mgawanyo wa ndani. Ikiwa pointi P inagawanya mstari AB kwa uwiano m:n, basi OP = nOA + mOBm+n. Katika kesi hii, P inagawanya AB kwa uwiano 4:1, hivyo m=4 na n=1. OP = 1OA + 4OB1+4 OP = OA + 4OB5 Step 3: Panga upya fomula ili kutatua kwa OA. 5OP = OA + 4OB OA = 5OP - 4OB Step 4: Badilisha veta zinazojulikana na kokotoa OA. OA = 5(8i + 4j - k) - 4(i + 3j - 2k) OA = (40i + 20j - 5k) - (4i + 12j - 8k) OA = (40 - 4)i + (20 - 12)j + (-5 - (-8))k OA = 36i + 8j + (-5 + 8)k OA = 36i + 8j + 3k Step 5: Badilisha veta ya nafasi OA kuwa kuratibu za pointi A. Kuratibu za pointi A ni coefficients za i, j, k. Jibu la mwisho ni: A = (36, 8, 3) Drop the next question.