Andika jumla ya pembe zote za pembetatu.

Mathematics

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10

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Judith, katika pembetatu, jumla ya pembe zote ni 180∘180^\circ. Tunapewa pembe tatu: 2x∘2x^\circ, 2x∘2x^\circ, na (2x2−60)∘(2x^2 - 60)^\circ.

  1. Andika jumla ya pembe zote za pembetatu.

    Jumla ya pembe za pembetatu ni 180∘180^\circ.

  2. Tengeneza mlinganyo kwa kutumia pembe ulizopewa.

    2x+2x+(2x2−60)=1802x + 2x + (2x^2 - 60) = 180
  3. Rahisisha mlinganyo.

    2x2+4x−60=1802x^2 + 4x - 60 = 180
  4. Panga upya mlinganyo katika mfumo wa quadratic ax2+bx+c=0ax^2 + bx + c = 0.

    2x2+4x−60−180=02x^2 + 4x - 60 - 180 = 0 2x2+4x−240=02x^2 + 4x - 240 = 0
  5. Gawanya mlinganyo mzima kwa 2 ili kurahisisha.

    x2+2x−120=0x^2 + 2x - 120 = 0
  6. Tatua mlinganyo wa quadratic kwa kutumia njia ya kuzidisha (factoring).

    Tunatafuta namba mbili ambazo zinazidishana kutoa −120-120 na kujumlisha kutoa 2. Namba hizo ni 12 na −10-10.

    (x+12)(x−10)=0(x + 12)(x - 10) = 0

    Hii inatoa suluhisho mbili zinazowezekana kwa xx:
    x+12=0  ⟹  x=−12x + 12 = 0 \implies x = -12
    x−10=0  ⟹  x=10x - 10 = 0 \implies x = 10

  7. Angalia suluhisho gani ni halali.

    Pembe za pembetatu haziwezi kuwa hasi.
    Kama x=−12x = -12:
    Pembe ya kwanza: 2x=2(−12)=−24∘2x = 2(-12) = -24^\circ. Hii haiwezekani.
    Kama x=10x = 10:
    Pembe ya kwanza: 2x=2(10)=20∘2x = 2(10) = 20^\circ.
    Pembe ya pili: 2x=2(10)=20∘2x = 2(10) = 20^\circ.
    Pembe ya tatu: 2x2−60=2(10)2−60=2(100)−60=200−60=140∘2x^2 - 60 = 2(10)^2 - 60 = 2(100) - 60 = 200 - 60 = 140^\circ.
    Jumla ya pembe: 20∘+20∘+140∘=180∘20^\circ + 20^\circ + 140^\circ = 180^\circ. Hii ni sahihi.

    Kwa hivyo, thamani halali ya xx ni 10.

    The value of xx is:

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