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SECTION A: General Mathematics (Questions 1-9) 1. A perfume bottle contains 1 mL of liquid perfume. Each spray releases 0.125 mL of the perfume. How many sprays can the bottle produce altogether?

Mathematics

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2 parts
a

W2+5W−40=0W^2 + 5W - 40 = 0

b

−5+1852 cm\frac{-5 + \sqrt{185}}{2} \text{ cm}

5+1852 cm\frac{5 + \sqrt{185}}{2} \text{ cm}

Step-by-step solution

4 steps

Huu hapa ni utatuzi wa swali la 16:

Part a· 1 step

Tengeneza mlinganyo wa quadratic.

  1. Fafanua vigezo na uunde mlinganyo kulingana na taarifa uliyopewa.

    Acha upana wa mstatili uwe WW cm.
    Urefu wa mstatili utakuwa L=(W+5)L = (W + 5) cm.
    Eneo la mstatili A=urefu×upana=(W+5)WA = \text{urefu} \times \text{upana} = (W + 5)W.
    Tunapewa kuwa eneo A=40 cm2A = 40 \text{ cm}^2.
    Kwa hiyo, tunaweza kuandika mlinganyo:

    (W+5)W=40(W + 5)W = 40

    Panua mlinganyo:

    W2+5W=40W^2 + 5W = 40

    Panga upya mlinganyo katika fomu ya kawaida ya quadratic ax2+bx+c=0ax^2 + bx + c = 0:

    W2+5W−40=0\boxed{W^2 + 5W - 40 = 0}

Part b

−5+1852 cm\frac{-5 + \sqrt{185}}{2} \text{ cm}

Tafuta vipimo vya mstatili kwa kuzidisha.

  1. Jaribu kuzidisha mlinganyo wa quadratic.

    Mlinganyo ni W2+5W−40=0W^2 + 5W - 40 = 0.
    Tunahitaji namba mbili ambazo bidhaa yake ni 1×(−40)=−401 \times (-40) = -40 na jumla yake ni 5 (kizigeu cha WW).
    Orodha ya vigawo vya 40 ni (1, 40), (2, 20), (4, 10), (5, 8).
    Hakuna jozi ya vigawo hivi, hata tukitumia alama zinazofaa, itatoa jumla ya 5. Kwa mfano, 8×(−5)=−408 \times (-5) = -40, lakini 8+(−5)=3≠58 + (-5) = 3 \neq 5.
    Kwa hivyo, mlinganyo huu wa quadratic hauwezi kuzidishwa katika vigawo vya mstari na vigawo kamili.

  2. Tatua kwa kutumia fomula ya quadratic kwa kuwa kuzidisha kwa namba kamili haiwezekani.

    Fomula ya quadratic kwa ax2+bx+c=0ax^2 + bx + c = 0 ni x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
    Katika mlinganyo wetu, a=1a=1, b=5b=5, na c=−40c=-40.

    W=−5±52−4(1)(−40)2(1)W = \frac{-5 \pm \sqrt{5^2 - 4(1)(-40)}}{2(1)} W=−5±25+1602W = \frac{-5 \pm \sqrt{25 + 160}}{2} W=−5±1852W = \frac{-5 \pm \sqrt{185}}{2}

    Kuna thamani mbili zinazowezekana kwa WW:

    W1=−5+1852W_1 = \frac{-5 + \sqrt{185}}{2} W2=−5−1852W_2 = \frac{-5 - \sqrt{185}}{2}

    Kwa kuwa upana hauwezi kuwa hasi, tunachukua thamani chanya ya WW.

    W=−5+1852 cmW = \frac{-5 + \sqrt{185}}{2} \text{ cm}

    Takriban W≈−5+13.6012≈8.6012≈4.30 cmW \approx \frac{-5 + 13.601}{2} \approx \frac{8.601}{2} \approx 4.30 \text{ cm}.

  3. Kokotoa urefu kwa kutumia upana halali.

    Urefu L=W+5L = W + 5.

    L=(−5+1852)+5L = \left(\frac{-5 + \sqrt{185}}{2}\right) + 5

    Ili kuongeza, tunahitaji kupata denominator ya kawaida:

    L=−5+1852+102L = \frac{-5 + \sqrt{185}}{2} + \frac{10}{2} L=−5+185+102L = \frac{-5 + \sqrt{185} + 10}{2} L=5+1852 cmL = \frac{5 + \sqrt{185}}{2} \text{ cm}

    Takriban L≈5+13.6012≈18.6012≈9.30 cmL \approx \frac{5 + 13.601}{2} \approx \frac{18.601}{2} \approx 9.30 \text{ cm}.

    Kwa hiyo, vipimo vya mstatili ni:
    Upana = −5+1852 cm\boxed{\frac{-5 + \sqrt{185}}{2} \text{ cm}}
    Urefu = 5+1852 cm\boxed{\frac{5 + \sqrt{185}}{2} \text{ cm}}

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