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 Chacha, hapa kuna suluhisho la maswali yako: 7. a) Panua kabisa (x - 0.2)^5 Tunatumia Binomial Theorem, ambayo inasema (a+b)^n = _k=0^n nk a^n-k b^k. Hapa, a=x, b=-0.2, na n=5. Step 1: Andika maneno ya upanuzi kwa kutumia Binomial Theorem. (x - 0.2)^5 = 50x^5(-0.2)^0 + 51x^4(-0.2)^1 + 52x^3(-0.2)^2 + 53x^2(-0.2)^3 + 54x^1(-0.2)^4 + 55x^0(-0.2)^5 Step 2: Kokotoa coefficients za binomial na nguvu za -0.2. Coefficients ni: 50=1, 51=5, 52=10, 53=10, 54=5, 55=1. Nguvu za -0.2 ni: (-0.2)^0=1, (-0.2)^1=-0.2, (-0.2)^2=0.04, (-0.2)^3=-0.008, (-0.2)^4=0.0016, (-0.2)^5=-0.00032. Step 3: Badilisha maadili haya kwenye upanuzi na kurahisisha. (x - 0.2)^5 = 1 · x^5 · 1 + 5 · x^4 · (-0.2) + 10 · x^3 · (0.04) + 10 · x^2 · (-0.008) + 5 · x · (0.0016) + 1 · 1 · (-0.00032) (x - 0.2)^5 = x^5 - 1.0x^4 + 0.4x^3 - 0.08x^2 + 0.008x - 0.00032 Upanuzi kamili ni x^5 - x^4 + 0.4x^3 - 0.08x^2 + 0.008x - 0.00032. 7. b) Tumia upanuzi wako kupata thamani kamili ya (9.8)^5 Step 1: Linganisha (9.8)^5 na (x - 0.2)^5 ili kupata thamani ya x. Ikiwa x - 0.2 = 9.8, basi x = 9.8 + 0.2 = 10. Step 2: Badilisha x=10 kwenye upanuzi uliopatikana katika 7a. (9.8)^5 = (10)^5 - (10)^4 + 0.4(10)^3 - 0.08(10)^2 + 0.008(10) - 0.00032 Step 3: Kokotoa kila neno na upate jumla. (9.8)^5 = 100000 - 10000 + 0.4(1000) - 0.08(100) + 0.08 - 0.00032 (9.8)^5 = 100000 - 10000 + 400 - 8 + 0.08 - 0.00032 (9.8)^5 = 90000 + 400 - 8 + 0.08 - 0.00032 (9.8)^5 = 90392 + 0.08 - 0.00032 (9.8)^5 = 90392.07968 Thamani kamili ya (9.8)^5 ni 90392.07968. 8. Fanya b kuwa somo la fomula a = (bd)/(sqrt(b^2 + d)) Step 1: Ondoa mzizi wa mraba kwa kupandisha pande zote mbili kwa mraba baada ya kuhamisha sqrt(b^2+d) upande wa kushoto. asqrt(b^2 + d) = bd Pandisha pande zote mbili kwa mraba: (asqrt(b^2 + d))^2 = (bd)^2 a^2(b^2 + d) = b^2d^2 Step 2: Panua upande wa kushoto. a^2b^2 + a^2d = b^2d^2 Step 3: Kusanya maneno yenye b^2 upande mmoja wa mlinganyo. a^2d = b^2d^2 - a^2b^2 Step 4: Toa b^2 kama kigezo cha kawaida kutoka upande wa kulia. a^2d = b^2(d^2 - a^2) Step 5: Fanya b^2 kuwa somo la fomula. b^2 = (a^2d)/(d^2 - a^2) Step 6: Chukua mzizi wa mraba wa pande zote mbili ili kufanya b kuwa somo. b = ±sqrt((a^2d)/(d^2 - a^2)) Unaweza pia kuandika a nje ya mzizi wa mraba: b = ± asqrt((d)/(d^2 - a^2)) b kama somo la fomula ni b = ± asqrt((d)/(d^2 - a^2)). 9. Neno la kumi na moja la A.P ni mara nne ya neno la pili. Ikiwa jumla ya maneno sita ya kwanza ya A.P ni 175, pata neno la kwanza na tofauti ya kawaida. Acha neno la kwanza liwe a na tofauti ya kawaida iwe d. Fomula ya neno la n-th ni T_n = a + (n-1)d. Fomula ya jumla ya maneno n ya kwanza ni S_n = (n)/(2)(2a + (n-1)d). Step 1: Andika milinganyo miwili kulingana na taarifa iliyotolewa. Kutoka "Neno la kumi na moja la A.P ni mara nne ya neno la pili": T_11 = 4 · T_2 a + (11-1)d = 4(a + (2-1)d) a + 10d = 4(a + d) a + 10d = 4a + 4d Step 2: Rahisisha mlinganyo wa kwanza ili kueleza a kwa kutumia d. 10d - 4d = 4a - a 6d = 3a a = 2d (Mlinganyo 1) Step 3: Tumia taarifa ya pili "jumla ya maneno sita ya kwanza ya A.P ni 175" kuunda mlinganyo wa pili. S_6 = 175 (6)/(2)(2a + (6-1)d) = 175 3(2a + 5d) = 175 6a + 15d = 175 (Mlinganyo 2) Step 4: Badilisha Mlinganyo 1 (a=2d) kwenye Mlinganyo 2 na utatue kwa d. 6(2d) + 15d = 175 12d + 15d = 175 27d = 175 d = (175)/(27) Step 5: Tumia thamani ya d kupata a kutoka Mlinganyo 1. a = 2d a = 2 ((175)/(27)) a = (350)/(27) Neno la kwanza ni a = (350)/(27) na tofauti ya kawaida ni d = (175)/(27). 3 done, 2 left today. You're making progress.