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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Haya, twende tukatatue maswali haya. Swali la 4: Kiwango cha bomba A kujaza tanki: (1)/(10) ya tanki kwa dakika. Kiwango cha bomba B kujaza tanki: (1)/(20) ya tanki kwa dakika. Kiwango cha bomba C kutoa maji kwenye tanki: -(1)/(30) ya tanki kwa dakika. Step 1: Tafuta kiwango cha kujaza tanki wakati mabomba yote matatu yamefunguliwa. Kiwango cha pamoja = Kiwango A + Kiwango B + Kiwango C Kiwango cha pamoja = (1)/(10) + (1)/(20) - (1)/(30) Tafuta kigawo kidogo cha shirika (LCM) cha 10, 20, na 30, ambacho ni 60. Kiwango cha pamoja = (6)/(60) + (3)/(60) - (2)/(60) = (6+3-2)/(60) = (7)/(60) ya tanki kwa dakika Step 2: Tafuta sehemu ya tanki iliyojazwa katika dakika 5 za kwanza. Sehemu iliyojazwa = Kiwango cha pamoja × 5 dakika = (7)/(60) × 5 = (35)/(60) = (7)/(12) Step 3: Tafuta sehemu ya tanki iliyobaki kujazwa. Sehemu iliyobaki = 1 - Sehemu iliyojazwa = 1 - (7)/(12) = (5)/(12) Step 4: Bomba A limefungwa. Tafuta kiwango cha kujaza tanki wakati mabomba B na C yamefunguliwa. Kiwango kipya cha pamoja = Kiwango B + Kiwango C Kiwango kipya cha pamoja = (1)/(20) - (1)/(30) Tafuta LCM ya 20 na 30, ambayo ni 60. Kiwango kipya cha pamoja = (3)/(60) - (2)/(60) = (1)/(60) ya tanki kwa dakika Step 5: Tafuta muda unaohitajika kujaza sehemu iliyobaki ya tanki. Muda = Sehemu iliyobakiKiwango kipya cha pamoja = (5)/(12)(1)/(60) = (5)/(12) × 60 = 5 × 5 = 25 dakika Muda unaohitajika kujaza sehemu iliyobaki ya tanki ni 25 dakika. Swali la 5: Ili kupata usawa unaoelezea eneo lisilo na kivuli, tutapata milinganyo ya mistari mitatu na kisha kuamua usawa sahihi. Step 1: Tafuta mlinganyo wa mstari wa kwanza (kutoka juu kushoto kwenda chini kulia). Mstari huu unapita kwenye pointi (0, 4) na (2, 0). Mteremko m = (0-4)/(2-0) = (-4)/(2) = -2. Y-intercept c = 4. Mlinganyo wa mstari ni y = -2x + 4, au 2x + y = 4. Eneo lisilo na kivuli liko chini ya mstari huu, kwa hivyo usawa ni 2x + y 4. Step 2: Tafuta mlinganyo wa mstari wa pili (kutoka chini kushoto kwenda juu kulia). Mstari huu unapita kwenye pointi (-4, 0) na (0, 2). Mteremko m = (2-0)/(0-(-4)) = (2)/(4) = (1)/(2). Y-intercept c = 2. Mlinganyo wa mstari ni y = (1)/(2)x + 2. Kuzidisha kwa 2 kunatoa 2y = x + 4, au x - 2y = -4. Eneo lisilo na kivuli liko juu ya mstari huu, kwa hivyo usawa ni x - 2y -4. Step 3: Tafuta mlinganyo wa mstari wa tatu (mstari wa x-axis). Mstari huu ni y = 0. Eneo lisilo na kivuli liko juu ya mstari huu, kwa hivyo usawa ni y 0. Usawa unaoelezea eneo lisilo na kivuli ni: 2x + y 4 x - 2y -4 y 0 Swali la 6: Tutatathmini usemi bila kutumia kikokotoo. (1)/(2) of 3(1)/(2) + 1(1)/(2) (2(1)/(2) - (2)/(3))(3)/(4) of 2(1)/(2) ÷ (1)/(2) Step 1: Badilisha namba mchanganyiko kuwa sehemu zisizofaa. 3(1)/(2) = (7)/(2) 1(1)/(2) = (3)/(2) 2(1)/(2) = (5)/(2) Step 2: Tathmini mabano kwenye nambari. 2(1)/(2) - (2)/(3) = (5)/(2) - (2)/(3) = (15)/(6) - (4)/(6) = (11)/(6) Step 3: Tathmini "of" (kuzidisha) na kuzidisha kwenye nambari. (1)/(2) of 3(1)/(2) = (1)/(2) × (7)/(2) = (7)/(4) 1(1)/(2) (2(1)/(2) - (2)/(3)) = (3)/(2) × (11)/(6) = (33)/(12) = (11)/(4) Step 4: Jumlisha nambari. Nambari = (7)/(4) + (11)/(4) = (18)/(4) = (9)/(2) Step 5: Tathmini "of" (kuzidisha) kwenye denominata. (3)/(4) of 2(1)/(2) = (3)/(4) × (5)/(2) = (15)/(8) Step 6: Tathmini kugawa kwenye denominata. Denominata = (15)/(8) ÷ (1)/(2) = (15)/(8) × (2)/(1) = (30)/(8) = (15)/(4) Step 7: Gawanya nambari kwa denominata. Jibu = (9)/(2)(15)/(4) = (9)/(2) × (4)/(15) = (36)/(30) = (6)/(5) Jibu la usemi ni (6)/(5). Tuma swali linalofuata 📸