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A) Expand completely (x - 0.2)5 b) Use your expansion to find the exact value of (9.8)5 Make b the subject of the formula a = bd / sqrt(b2 + d) The eleventh term of an A.P is four times the second term. If the A.P is 175, find the first term and the common difference.

Mathematics

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Haya Chacha, hapa kuna suluhisho la maswali yako:

7. a) Panua kabisa (x−0.2)5(x - 0.2)^5

Tunatumia Binomial Theorem, ambayo inasema (a+b)n=∑k=0n(nk)an−kbk(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k. Hapa, a=xa=x, b=−0.2b=-0.2, na n=5n=5.

  1. Andika maneno ya upanuzi kwa kutumia Binomial Theorem.

    (x−0.2)5=(50)x5(−0.2)0+(51)x4(−0.2)1+(52)x3(−0.2)2+(53)x2(−0.2)3+(54)x1(−0.2)4+(55)x0(−0.2)5(x - 0.2)^5 = \binom{5}{0}x^5(-0.2)^0 + \binom{5}{1}x^4(-0.2)^1 + \binom{5}{2}x^3(-0.2)^2 + \binom{5}{3}x^2(-0.2)^3 + \binom{5}{4}x^1(-0.2)^4 + \binom{5}{5}x^0(-0.2)^5
  2. Kokotoa coefficients za binomial na nguvu za −0.2-0.2.

    Coefficients ni: (50)=1\binom{5}{0}=1, (51)=5\binom{5}{1}=5, (52)=10\binom{5}{2}=10, (53)=10\binom{5}{3}=10, (54)=5\binom{5}{4}=5, (55)=1\binom{5}{5}=1.
    Nguvu za −0.2-0.2 ni: (−0.2)0=1(-0.2)^0=1, (−0.2)1=−0.2(-0.2)^1=-0.2, (−0.2)2=0.04(-0.2)^2=0.04, (−0.2)3=−0.008(-0.2)^3=-0.008, (−0.2)4=0.0016(-0.2)^4=0.0016, (−0.2)5=−0.00032(-0.2)^5=-0.00032.

  3. Badilisha maadili haya kwenye upanuzi na kurahisisha.

    (x−0.2)5=1⋅x5⋅1+5⋅x4⋅(−0.2)+10⋅x3⋅(0.04)+10⋅x2⋅(−0.008)+5⋅x⋅(0.0016)+1⋅1⋅(−0.00032)(x - 0.2)^5 = 1 \cdot x^5 \cdot 1 + 5 \cdot x^4 \cdot (-0.2) + 10 \cdot x^3 \cdot (0.04) + 10 \cdot x^2 \cdot (-0.008) + 5 \cdot x \cdot (0.0016) + 1 \cdot 1 \cdot (-0.00032) (x−0.2)5=x5−1.0x4+0.4x3−0.08x2+0.008x−0.00032(x - 0.2)^5 = x^5 - 1.0x^4 + 0.4x^3 - 0.08x^2 + 0.008x - 0.00032

    Upanuzi kamili ni x5−x4+0.4x3−0.08x2+0.008x−0.00032\boxed{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(9.8)^5

  4. Linganisha (9.8)5(9.8)^5 na (x−0.2)5(x - 0.2)^5 ili kupata thamani ya xx.

    Ikiwa x−0.2=9.8x - 0.2 = 9.8, basi x=9.8+0.2=10x = 9.8 + 0.2 = 10.

  5. Badilisha x=10x=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(9.8)^5 = (10)^5 - (10)^4 + 0.4(10)^3 - 0.08(10)^2 + 0.008(10) - 0.00032
  6. 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 + 0.4(1000) - 0.08(100) + 0.08 - 0.00032 (9.8)5=100000−10000+400−8+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 = 90000 + 400 - 8 + 0.08 - 0.00032 (9.8)5=90392+0.08−0.00032(9.8)^5 = 90392 + 0.08 - 0.00032 (9.8)5=90392.07968(9.8)^5 = 90392.07968

    Thamani kamili ya (9.8)5(9.8)^5 ni 90392.07968\boxed{90392.07968}.

    8. Fanya bb kuwa somo la fomula a=bdb2+da = \frac{bd}{\sqrt{b^2 + d}}

  7. Ondoa mzizi wa mraba kwa kupandisha pande zote mbili kwa mraba baada ya kuhamisha b2+d\sqrt{b^2+d} upande wa kushoto.

    ab2+d=bda\sqrt{b^2 + d} = bd

    Pandisha pande zote mbili kwa mraba:

    (ab2+d)2=(bd)2(a\sqrt{b^2 + d})^2 = (bd)^2 a2(b2+d)=b2d2a^2(b^2 + d) = b^2d^2
  8. Panua upande wa kushoto.

    a2b2+a2d=b2d2a^2b^2 + a^2d = b^2d^2
  9. Kusanya maneno yenye b2b^2 upande mmoja wa mlinganyo.

    a2d=b2d2−a2b2a^2d = b^2d^2 - a^2b^2
  10. Toa b2b^2 kama kigezo cha kawaida kutoka upande wa kulia.

    a2d=b2(d2−a2)a^2d = b^2(d^2 - a^2)
  11. Fanya b2b^2 kuwa somo la fomula.

    b2=a2dd2−a2b^2 = \frac{a^2d}{d^2 - a^2}
  12. Chukua mzizi wa mraba wa pande zote mbili ili kufanya bb kuwa somo.

    b=±a2dd2−a2b = \pm\sqrt{\frac{a^2d}{d^2 - a^2}}

    Unaweza pia kuandika aa nje ya mzizi wa mraba:

    b=±add2−a2b = \pm a\sqrt{\frac{d}{d^2 - a^2}}

    bb kama somo la fomula ni b=±add2−a2\boxed{b = \pm a\sqrt{\frac{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 aa na tofauti ya kawaida iwe dd.
    Fomula ya neno la nn-th ni Tn=a+(n−1)dT_n = a + (n-1)d.
    Fomula ya jumla ya maneno nn ya kwanza ni Sn=n2(2a+(n−1)d)S_n = \frac{n}{2}(2a + (n-1)d).

  13. Andika milinganyo miwili kulingana na taarifa iliyotolewa.

    Kutoka "Neno la kumi na moja la A.P ni mara nne ya neno la pili":

    T11=4⋅T2T_{11} = 4 \cdot T_2 a+(11−1)d=4(a+(2−1)d)a + (11-1)d = 4(a + (2-1)d) a+10d=4(a+d)a + 10d = 4(a + d) a+10d=4a+4da + 10d = 4a + 4d
  14. Rahisisha mlinganyo wa kwanza ili kueleza aa kwa kutumia dd.

    10d−4d=4a−a10d - 4d = 4a - a 6d=3a6d = 3a a=2d(Mlinganyo 1)a = 2d \quad \text{(Mlinganyo 1)}
  15. Tumia taarifa ya pili "jumla ya maneno sita ya kwanza ya A.P ni 175" kuunda mlinganyo wa pili.

    S6=175S_6 = 175 62(2a+(6−1)d)=175\frac{6}{2}(2a + (6-1)d) = 175 3(2a+5d)=1753(2a + 5d) = 175 6a+15d=175(Mlinganyo 2)6a + 15d = 175 \quad \text{(Mlinganyo 2)}
  16. Badilisha Mlinganyo 1 (a=2da=2d) kwenye Mlinganyo 2 na utatue kwa dd.

    6(2d)+15d=1756(2d) + 15d = 175 12d+15d=17512d + 15d = 175 27d=17527d = 175 d=17527d = \frac{175}{27}
  17. Tumia thamani ya dd kupata aa kutoka Mlinganyo 1.

    a=2da = 2d a=2(17527)a = 2 \left(\frac{175}{27}\right) a=35027a = \frac{350}{27}

    Neno la kwanza ni a=35027\boxed{a = \frac{350}{27}} na tofauti ya kawaida ni d=17527\boxed{d = \frac{175}{27}}.

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