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Use mathematical tables to find the reciprocal of 0.0247, hence evaluate [3]3.0250.0247, correct to 2 decimal places.

Mathematics

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Hapa kuna hatua za kutatua swali la 2:

Sehemu (a): Kuamua kuratibu za T
Tunatumia fomula ya sehemu kwa mgawanyiko wa ndani.
Kuratibu za P ni (x1,y1)=(1,−2)(x_1, y_1) = (1, -2).
Kuratibu za Q ni (x2,y2)=(4,10)(x_2, y_2) = (4, 10).
Uwiano ni m:n=2:1m:n = 2:1.

  1. Tumia fomula ya sehemu kupata kuratibu za T.

    Tx=nx1+mx2m+n=1(1)+2(4)2+1=1+83=93=3T_x = \frac{nx_1 + mx_2}{m+n} = \frac{1(1) + 2(4)}{2+1} = \frac{1 + 8}{3} = \frac{9}{3} = 3 Ty=ny1+my2m+n=1(−2)+2(10)2+1=−2+203=183=6T_y = \frac{ny_1 + my_2}{m+n} = \frac{1(-2) + 2(10)}{2+1} = \frac{-2 + 20}{3} = \frac{18}{3} = 6

    Kuratibu za T ni (3,6)\boxed{(3, 6)}.

    Sehemu (b) (i): Tafuta mteremko wa mstari ulio wima kwa PQ

  2. Tafuta mteremko wa mstari PQ.

    mPQ=y2−y1x2−x1=10−(−2)4−1=10+23=123=4m_{PQ} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{10 - (-2)}{4 - 1} = \frac{10 + 2}{3} = \frac{12}{3} = 4
  3. Tafuta mteremko wa mstari ulio wima kwa PQ.

    Mteremko wa mstari ulio wima ni hasi ya kinyume cha mteremko wa mstari wa asili.

    m⊥=−1mPQ=−14m_{\perp} = -\frac{1}{m_{PQ}} = -\frac{1}{4}

    Mteremko wa mstari ulio wima kwa PQ ni −14\boxed{-\frac{1}{4}}.

    Sehemu (b) (ii): Kuamua mlinganyo wa mstari ulio wima kwa PQ na kupitia T

  4. Tumia mteremko m⊥m_{\perp} na kuratibu za T (3,6)(3, 6) kupata mlinganyo wa mstari.

    Tunatumia fomula ya pointi-mteremko: y−y1=m(x−x1)y - y_1 = m(x - x_1).

    y−6=−14(x−3)y - 6 = -\frac{1}{4}(x - 3)

    Panua pande zote mbili kwa 4:

    4(y−6)=−(x−3)4(y - 6) = -(x - 3) 4y−24=−x+34y - 24 = -x + 3

    Panga upya mlinganyo:

    x+4y−24−3=0x + 4y - 24 - 3 = 0 x+4y−27=0x + 4y - 27 = 0

    Mlinganyo wa mstari ni x+4y−27=0\boxed{x + 4y - 27 = 0}.

    Sehemu (iii): Kokotoa umbali TR

  5. Tafuta kuratibu za R ambapo mstari unakutana na mhimili wa y.

    Kwenye mhimili wa y, x=0x = 0. Badilisha x=0x=0 kwenye mlinganyo wa mstari x+4y−27=0x + 4y - 27 = 0.

    0+4y−27=00 + 4y - 27 = 0 4y=274y = 27 y=274=6.75y = \frac{27}{4} = 6.75

    Kuratibu za R ni (0,6.75)(0, 6.75).

  6. Kokotoa umbali kati ya T $(3, 6)$ na R $(0, 6.75)$ kwa kutumia fomula ya umbali.

    D=(x2−x1)2+(y2−y1)2D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} D=(0−3)2+(6.75−6)2D = \sqrt{(0 - 3)^2 + (6.75 - 6)^2} D=(−3)2+(0.75)2D = \sqrt{(-3)^2 + (0.75)^2} D=9+0.5625D = \sqrt{9 + 0.5625} D=9.5625D = \sqrt{9.5625} D≈3.092329D \approx 3.092329

    Umbali TR, kwa tarakimu tatu muhimu, ni 3.09\boxed{3.09}.

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