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 hapa masuluhisho ya maswali yako: 4. Given that 16a^2 + ba + 9 is a perfect square, simplify (16a^2+ba+9)/(16a^2-9). Step 1: Tambua umbo la mraba kamili na kiasi cha chini. Usemi 16a^2 + ba + 9 ni mraba kamili. Hii inamaanisha unaweza kuandikwa kama (Ax ± B)^2. Kulinganisha na A^2x^2 ± 2ABx + B^2: A^2 = 16 A = 4. B^2 = 9 B = 3. Hivyo, 16a^2 + ba + 9 = (4a ± 3)^2. Kiasi cha chini 16a^2-9 ni tofauti ya miraba, ambayo inaweza kuandikwa kama (4a)^2 - 3^2 = (4a-3)(4a+3). Step 2: Bainisha thamani zinazowezekana za ba. Kutoka (4a ± 3)^2, kiasi cha kati ni ± 2(4a)(3) = ± 24a. Hivyo, ba = 24a au ba = -24a. Hii inamaanisha b = 24 au b = -24. Step 3: Badilisha kila uwezekano kwenye usemi na kurahisisha. Kesi ya 1: Ikiwa b = 24 Numerator inakuwa 16a^2+24a+9 = (4a+3)^2. (16a^2+24a+9)/(16a^2-9) = ((4a+3)^2)/((4a-3)(4a+3)) Rahisisha kwa kufuta (4a+3) moja kutoka juu na chini: (4a+3)/(4a-3) Kesi ya 2: Ikiwa b = -24 Numerator inakuwa 16a^2-24a+9 = (4a-3)^2. (16a^2-24a+9)/(16a^2-9) = ((4a-3)^2)/((4a-3)(4a+3)) Rahisisha kwa kufuta (4a-3) moja kutoka juu na chini: (4a-3)/(4a+3) Kwa hivyo, kuna majibu mawili yanayowezekana kulingana na thamani ya b: Majibu: (4a+3)/(4a-3) au (4a-3)/(4a+3) 5. Solve for x, given that 2x = (1)/(2) for 0 x 180^. Step 1: Tafuta thamani kuu za 2x. Tunapewa 2x = (1)/(2). Thamani kuu ya 2x ambapo sine ni (1)/(2) ni 30^. Kwa kuwa sine ni chanya, 2x inaweza kuwa katika roboduara ya kwanza au ya pili. Hivyo, 2x = 30^ au 2x = 180^ - 30^ = 150^. Step 2: Zingatia masafa ya 2x na utafute suluhisho zote ndani ya masafa yaliyotolewa. Masafa ya x ni 0^ x 180^. Hii inamaanisha masafa ya 2x ni 0^ 2x 360^. Kutoka hatua ya 1, tuna suluhisho mbili za kwanza: 2x = 30^ x = (30^)/(2) = 15^. 2x = 150^ x = (150^)/(2) = 75^. Tunahitaji pia kuzingatia mizunguko mingine ndani ya masafa ya 2x. Kwa 2x = 30^ + n · 360^: Ikiwa n=1, 2x = 30^ + 360^ = 390^. Hii inatoa x = 195^, ambayo iko nje ya masafa 0^ x 180^. Kwa 2x = 150^ + n · 360^: Ikiwa n=1, 2x = 150^ + 360^ = 510^. Hii inatoa x = 255^, ambayo iko nje ya masafa 0^ x 180^. Hivyo, suluhisho pekee ndani ya masafa yaliyotolewa ni 15^ na 75^. Jibu: x = 15^, 75^ 6. Find the percentage error in the calculation of 4.5cm × 5.5cm + 1.1cm. Step 1: Kokotoa thamani halisi (nominal) ya hesabu. Thamani halisi ya vipimo ni 4.5cm, 5.5cm, na 1.1cm. Hesabu ni P = 4.5cm × 5.5cm + 1.1cm. P_nominal = (4.5 × 5.5) + 1.1 P_nominal = 24.75 + 1.1 P_nominal = 25.85cm^2 Step 2: Bainisha makosa kamili ya kila kipimo. Kwa kuwa vipimo vimetolewa kwa sehemu moja ya desimali, kosa kamili ni nusu ya kitengo kidogo zaidi, yaani 0.05cm. L_1 = ± 0.05cm kwa 4.5cm. L_2 = ± 0.05cm kwa 5.5cm. L_3 = ± 0.05cm kwa 1.1cm. Step 3: Kokotoa kosa kamili la bidhaa (L_1 × L_2). Hebu Q = L_1 × L_2. Kosa kamili katika bidhaa huhesabiwa kama Q = L_1 L_2 + L_2 L_1. Q = (4.5cm × 0.05cm) + (5.5cm × 0.05cm) Q = 0.225cm^2 + 0.275cm^2 Q = 0.50cm^2 Hivyo, Q = 24.75 ± 0.50cm^2. Step 4: Kokotoa kosa kamili la jumla (Q + L_3). Kwa kujumlisha, makosa kamili huongezwa: P = Q + L_3. P = 0.50cm^2 + 0.05cm^2 P = 0.55cm^2 Step 5: Kokotoa asilimia ya kosa. Asilimia ya kosa = Kosa kamiliThamani halisi × 100\%. Asilimia ya kosa = 0.55cm^225.85cm^2 × 100\% Asilimia ya kosa ≈ 0.0212762 × 100\% Asilimia ya kosa ≈ 2.12762\% Kuzungusha hadi sehemu tatu muhimu (3 significant figures): Jibu: 2.13\% Tuma swali linalofuata 📸