Yechilishi
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2 TA MASALA
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1-masala: Mototsikil M shahardan N shaharga jo’nadi. Agar u 35 km/soat tezlik bilan yursa 2 soat kechikadi, 50km/soat tezlik bilan yursa 1 soat oldin keladi. M dan N gacha bo’lgan masofani toping. YECHILISHI: Mototsiklchining M nuqtadan N nuqtaga v tezlikda borish vaqtini t deb olsak, MN=S deb olsak: { 𝑆 = 35(𝑡 + 2) 𝑆 = 50(𝑡 − 1) tenglamalarni tuzamiz. 35(𝑡 + 2) = 50(𝑡 − 1) 7(𝑡 + 2) = 10(𝑡 − 1) 7𝑡 + 14 = 10𝑡 − 10 3𝑡 = 24 t=8 soat 𝑆 = 35(8 + 2) = 350𝑘𝑚 2-masala: Yuk poyezdi A shahardan B shaharga qarab jo’nadi. Oradan 3 soat o’tgach, A shahardan B shaharga qarab yo’lovchi poyezdi yo’lga chiqdi va oradan 15 soat o’tgach yuk poyezdidan 300 km o’zib ketdi. Agar yo’lovchi poyezdning tezligi yuk poyezdnikidan 30 km/soat ortiq bo’lsa, yuk poyezdning tezligini toping.
tezlik vaqt masofa
Yuk poyezdi x 18 soat 3x+S Yo’lovchi poyezdi x+30 15 soat 3x+S+300
{ 3𝑥 + 𝑆 = 18𝑥 3𝑥 + 𝑆 + 300 = 15(𝑥 + 30) tenglamalar sistemasini tuzib olamiz: 18𝑥 + 300 = 15(𝑥 + 30) 3𝑥 = 450 − 300 𝑥 = 50 𝑘𝑚/𝑠𝑜𝑎𝑡 Download 467.58 Kb. Do'stlaringiz bilan baham: |
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