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Factorize completely: x2 - 16

Mathematics

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Haya, tuendelee na maswali mengine.

7. Find the reciprocal of 0.0045 and express your answer in standard form to 4 significant figures.

Reciprocal ya nambari ni 1 kugawanya nambari hiyo.

  1. Tafuta reciprocal ya 0.0045.

    10.0045=222.2222...\frac{1}{0.0045} = 222.2222...
  2. Andika jibu katika standard form (mfumo sanifu).

    Ili kuandika 222.2222... katika standard form, tunahamisha desimali mara 2 kushoto.

    2.2222...×1022.2222... \times 10^2
  3. Punguza jibu kwa tarakimu muhimu 4 (4 significant figures).

    Tarakimu ya tano ni 2, ambayo ni chini ya 5, kwa hivyo hatubadilishi tarakimu ya nne.

    2.222×1022.222 \times 10^2

    Jibu ni 2.222×102\boxed{2.222 \times 10^2}.

    8. The temperature in a town was recorded as 18.6∘C18.6^\circ C in the morning and 24.2∘C24.2^\circ C in the afternoon. Calculate the mean temperature.

    Mean temperature (wastani wa joto) hupatikana kwa kujumlisha joto zote na kugawanya kwa idadi ya vipimo.

  4. Jumuisha joto zote mbili.

    18.6∘C+24.2∘C=42.8∘C18.6^\circ C + 24.2^\circ C = 42.8^\circ C
  5. Gawanya jumla kwa idadi ya vipimo (2).

    42.8∘C2=21.4∘C\frac{42.8^\circ C}{2} = 21.4^\circ C

    Jibu ni 21.4∘C\boxed{21.4^\circ C}.

    9. Estimate the height of a classroom wall in kilometres.

    Urefu wa ukuta wa darasa kwa kawaida hupimwa kwa mita.

  6. Kadiria urefu wa ukuta wa darasa kwa mita.

    Urefu wa kawaida wa ukuta wa darasa ni takriban mita 3.

  7. Badilisha mita kuwa kilomita.

    Tunajua kuwa kilomita 1 ina mita 1000.

    3 m1000 m/km=0.003 km\frac{3 \text{ m}}{1000 \text{ m/km}} = 0.003 \text{ km}

    Jibu ni 0.003 km\boxed{0.003 \text{ km}}.

    10. Express 243 as a power of a prime number.

    Nambari prime ni nambari inayoweza kugawanyika na 1 na yenyewe tu.

  8. Tafuta vigawo prime vya 243.

    Anza kugawanya 243 na nambari ndogo zaidi prime.

    243÷3=81243 \div 3 = 81 81÷3=2781 \div 3 = 27 27÷3=927 \div 3 = 9 9÷3=39 \div 3 = 3 3÷3=13 \div 3 = 1
  9. Andika 243 kama nguvu ya nambari prime.

    Kwa kuwa 3 imejirudia mara 5, tunaweza kuandika 243 kama 353^5.

    243=3×3×3×3×3=35243 = 3 \times 3 \times 3 \times 3 \times 3 = 3^5

    Jibu ni 35\boxed{3^5}.

    11. Simplify using the laws of indices: (52)12(5^2)^{\frac{1}{2}}

    Tunatumia sheria ya indices inayosema (am)n=am×n(a^m)^n = a^{m \times n}.

  10. Tumia sheria ya indices.

    (52)12=52×12(5^2)^{\frac{1}{2}} = 5^{2 \times \frac{1}{2}}
  11. Rahisisha kipeo.

    52×12=515^{2 \times \frac{1}{2}} = 5^1 51=55^1 = 5

    Jibu ni 5\boxed{5}.

    12. Solve for x: 2x+1=322^{x+1} = 32

    Ili kutatua mlinganyo huu, tunahitaji kufanya besi ziwe sawa.

  12. Andika 32 kama nguvu ya 2.

    32=2×2×2×2×2=2532 = 2 \times 2 \times 2 \times 2 \times 2 = 2^5
  13. Linganisha besi na kisha linganisha vipeo.

    2x+1=252^{x+1} = 2^5

    Kwa kuwa besi ni sawa, vipeo lazima viwe sawa:

    x+1=5x+1 = 5
  14. Tatua kwa xx.

    x=5−1x = 5 - 1 x=4x = 4

    Jibu ni x=4\boxed{x=4}.

    13. The width of a rectangle is xx cm and its length is (x+4)(x+4) cm. Form an algebraic expression for its area.

    Eneo la mstatili (rectangle) hupatikana kwa kuzidisha urefu na upana.

  15. Andika formula ya eneo la mstatili.

    Area=length×width\text{Area} = \text{length} \times \text{width}
  16. Badilisha urefu na upana uliopewa kwenye formula.

    Area=(x+4) cm×x cm\text{Area} = (x+4) \text{ cm} \times x \text{ cm}
  17. Rahisisha usemi.

    Area=x(x+4) cm2\text{Area} = x(x+4) \text{ cm}^2 Area=(x2+4x) cm2\text{Area} = (x^2 + 4x) \text{ cm}^2

    Jibu ni x2+4x\boxed{x^2 + 4x}.

    14. Factorize completely: x2−16x^2 - 16

    Huu ni mfumo wa tofauti ya miraba miwili (difference of two squares), ambayo inaweza kuandikwa kama a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b).

  18. Tambua aa na bb.

    Katika x2−16x^2 - 16, a2=x2a^2 = x^2, kwa hivyo a=xa = x.
    Na b2=16b^2 = 16, kwa hivyo b=16=4b = \sqrt{16} = 4.

  19. Tumia formula ya tofauti ya miraba miwili.

    x2−42=(x−4)(x+4)x^2 - 4^2 = (x-4)(x+4)

    Jibu ni (x−4)(x+4)\boxed{(x-4)(x+4)}.

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