Little Alexander
06/02/2024 · Primary School
6. Tenglamani yeching. \( \begin{array}{llll}\text { 7. a parametming qanday qiymatlarida } 9+3 a, 5-2 a \text { va } 15-5 a \text { uzunlikdagi kesmalardan uchburch: } \\ \text { yashash mumkin? A) }(0 ; 2,5) & \text { B) }\left(\frac{1}{6} ; 1,5\right) & \text { C) }\left(\frac{1}{6} ; 1,1\right) & \text { D) }\left(\frac{1}{3} ; 1,2\right)\end{array} \)
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Step-by-step Solution
Uchburchakning mavjudligi uchun, har bir ikki tomonning yig'indisi uchinchi tomondan katta bo'lishi kerak. Berilgan kesmalar uzunliklari:
1. \( 9 + 3a \)
2. \( 5 - 2a \)
3. \( 15 - 5a \)
Uchburchakning mavjudligi shartlarini quyidagicha yozamiz:
1. \( (9 + 3a) + (5 - 2a) > (15 - 5a) \)
2. \( (9 + 3a) + (15 - 5a) > (5 - 2a) \)
3. \( (5 - 2a) + (15 - 5a) > (9 + 3a) \)
Endi har bir shartni yechamiz.
1. \( (9 + 3a) + (5 - 2a) > (15 - 5a) \)
\[
14 + a > 15 - 5a
\]
\[
6a > 1 \implies a > \frac{1}{6}
\]
2. \( (9 + 3a) + (15 - 5a) > (5 - 2a) \)
\[
24 - 2a > 5 - 2a
\]
Bu shart har qanday \( a \) uchun to'g'ri, shuning uchun bu shartni hisobga olish shart emas.
3. \( (5 - 2a) + (15 - 5a) > (9 + 3a) \)
\[
20 - 7a > 9 + 3a
\]
\[
20 - 9 > 10a \implies 11 > 10a \implies a < 1.1
\]
Shunday qilib, bizning shartlarimiz:
\[
\frac{1}{6} < a < 1.1
\]
Bu shartlar \( \left(\frac{1}{6}; 1.1\right) \) intervalida joylashgan. Berilgan variantlar orasida to'g'ri javob:
**C) \( \left(\frac{1}{6}; 1.1\right) \)**.
Quick Answer
Uchburchak mavjudligi uchun \( \frac{1}{6} < a < 1.1 \). Javob: C) \( \left(\frac{1}{6}; 1.1\right) \).
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