162-bet.
11-vazifa. Berilgan rasmdan foydalanib katta aylananing diametrini toping. CB = 4 cm.
Ishlanishi π:
CB=4 cm
OC=CB=4 cm
r=OB=OC+CB=8 cm
d=2Β·r=2Β·8=16 cm
Javob: katta aylanani diametri 16 cm.
https://t.me/matematika6uzb
Umrzoq G'aniyev: https://t.me/GUA_math
11-vazifa. Berilgan rasmdan foydalanib katta aylananing diametrini toping. CB = 4 cm.
Ishlanishi π:
CB=4 cm
OC=CB=4 cm
r=OB=OC+CB=8 cm
d=2Β·r=2Β·8=16 cm
Javob: katta aylanani diametri 16 cm.
https://t.me/matematika6uzb
Umrzoq G'aniyev: https://t.me/GUA_math
π144π€―20π―19π16π14π₯10π€5
163-bet.
2-vazifa. Radiusi 5 m boΚ»lgan favvora atrofi panjara bilan oΚ»raldi. Taxminan necha metr panjara ishlatilgan? (Ο = 3,14 deb oling).
Ishlanishi:
Favvora aylana shaklida bo'lgani uchun panjarani uzunligi aylana uzunligiga teng.
C=2Οr=2Β·3,14Β·5=3,14Β·10=31,4 m.
Javob: 31,4 metr.
https://t.me/matematika6uzb
Umrzoq G'aniyev: https://t.me/GUA_math
2-vazifa. Radiusi 5 m boΚ»lgan favvora atrofi panjara bilan oΚ»raldi. Taxminan necha metr panjara ishlatilgan? (Ο = 3,14 deb oling).
Ishlanishi:
Favvora aylana shaklida bo'lgani uchun panjarani uzunligi aylana uzunligiga teng.
C=2Οr=2Β·3,14Β·5=3,14Β·10=31,4 m.
Javob: 31,4 metr.
https://t.me/matematika6uzb
Umrzoq G'aniyev: https://t.me/GUA_math
Telegram
i-MATEMATIKA 6-SINF YECHIMLARI (2024)
'6-sinf matematika' 2022 yil darsligi yechimlari.
Murojaat uchun:
https://t.me/GUA_math
Muhokama guruhiπ
https://t.me/+-3E8MbEVyjk0NDcy
Murojaat uchun:
https://t.me/GUA_math
Muhokama guruhiπ
https://t.me/+-3E8MbEVyjk0NDcy
π167π―41π€36π28π22π₯21π€―21
166-bet.
16-vazifani ishlanishi ππ½
a) S=ΟrΒ²=3,14Β·3Β²=3,14Β·9=28,26 cmΒ². C=2Οr=2Β·3,14Β·3=18,84 cm;
b) S=ΟrΒ²=3,14Β·1Β²=3,14Β·1=3,14 cmΒ². C=2Οr=2Β·3,14Β·1=6,28 cm;
c) S=ΟrΒ²=3,14Β·4Β²=3,14Β·16=50,24 cmΒ². C=2Οr=2Β·3,14Β·4=25,12 cm;
d) S=ΟrΒ²=3,14Β·2,5Β²=3,14Β·6,25=19,625 cmΒ². C=2Οr=2Β·3,14Β·2,25= 14,13 cm;
e) S=ΟrΒ²=3,14Β·1,3Β²=3,14Β·1,69=5,3066 mΒ². C=2Οr=2Β·3,14Β·1,3= 8,164 m;
f) S=ΟrΒ²=3,14Β·125Β²=3,14Β·15 625=49062,5 mmΒ². C=2Οr=2Β·3,14Β·125=785 mm.
https://t.me/matematika6uzb
Umrzoq G'aniyev: https://t.me/GUA_math
16-vazifani ishlanishi ππ½
a) S=ΟrΒ²=3,14Β·3Β²=3,14Β·9=28,26 cmΒ². C=2Οr=2Β·3,14Β·3=18,84 cm;
b) S=ΟrΒ²=3,14Β·1Β²=3,14Β·1=3,14 cmΒ². C=2Οr=2Β·3,14Β·1=6,28 cm;
c) S=ΟrΒ²=3,14Β·4Β²=3,14Β·16=50,24 cmΒ². C=2Οr=2Β·3,14Β·4=25,12 cm;
d) S=ΟrΒ²=3,14Β·2,5Β²=3,14Β·6,25=19,625 cmΒ². C=2Οr=2Β·3,14Β·2,25= 14,13 cm;
e) S=ΟrΒ²=3,14Β·1,3Β²=3,14Β·1,69=5,3066 mΒ². C=2Οr=2Β·3,14Β·1,3= 8,164 m;
f) S=ΟrΒ²=3,14Β·125Β²=3,14Β·15 625=49062,5 mmΒ². C=2Οr=2Β·3,14Β·125=785 mm.
https://t.me/matematika6uzb
Umrzoq G'aniyev: https://t.me/GUA_math
π255π―46π€―32π27π€21π21π₯11
166-bet.
18-vazifani ishlanishi ππ½
Javob: 88 cm.
https://t.me/matematika6uzb
Umrzoq G'aniyev: https://t.me/GUA_math
18-vazifani ishlanishi ππ½
Javob: 88 cm.
https://t.me/matematika6uzb
Umrzoq G'aniyev: https://t.me/GUA_math
π158π45π―24π€―18π€17π16π₯13
166-bet.
19-vazifani ishlanishi ππ½
Javob: 50 000 marta.
https://t.me/matematika6uzb
Umrzoq G'aniyev: https://t.me/GUA_math
19-vazifani ishlanishi ππ½
Javob: 50 000 marta.
https://t.me/matematika6uzb
Umrzoq G'aniyev: https://t.me/GUA_math
π170π―85π34π€―31π€25π16π₯15
Forwarded from Π£ΠΌΡΠ·ΠΎΠΊ ΠΠ°Π½ΠΈΠ΅Π²
193-bet.
13-masalani ishlanishi ππ½
https://t.me/matematika6sinfuzb
Umrzoq G'aniyev: https://t.me/GUA_math
13-masalani ishlanishi ππ½
https://t.me/matematika6sinfuzb
Umrzoq G'aniyev: https://t.me/GUA_math
π166π―27π19π15π€―15π₯11
194-bet.
π8-masala.
8 ta olma 16 000 soΚ»m turadi. Xuddi shunday 18 ta olma necha pul boΚ»ladi? 64 000 soΚ»mga nechta olma sotib olish mumkinβ
Ishlanishi:
1) Proporsiya tuzib olamiz:
β 8 ta olma β 16 000 so'm β
β 18 ta olma β x so'm β
β
8:18=16 000:x
8x=18Β·16 000
x=(18Β·16 000):8=18Β·2 000=36 000
Javob: 36 000 so'm.
2) Proporsiya tuzib olamiz:
β 8 ta olma β 16 000 so'm β
β x ta olma β 64 000 so'm β
β
8:x=16 000:64 000
8:x=1:4
x=8Β·4
x=32
Javob: 32 ta olma.
https://t.me/matematika6uzb
Umrzoq G'aniyev: https://t.me/GUA_math
π8-masala.
8 ta olma 16 000 soΚ»m turadi. Xuddi shunday 18 ta olma necha pul boΚ»ladi? 64 000 soΚ»mga nechta olma sotib olish mumkinβ
Ishlanishi:
1) Proporsiya tuzib olamiz:
β 8 ta olma β 16 000 so'm β
β 18 ta olma β x so'm β
β
8:18=16 000:x
8x=18Β·16 000
x=(18Β·16 000):8=18Β·2 000=36 000
Javob: 36 000 so'm.
2) Proporsiya tuzib olamiz:
β 8 ta olma β 16 000 so'm β
β x ta olma β 64 000 so'm β
β
8:x=16 000:64 000
8:x=1:4
x=8Β·4
x=32
Javob: 32 ta olma.
https://t.me/matematika6uzb
Umrzoq G'aniyev: https://t.me/GUA_math
π255π35π€―32π29π―28π₯25π€24
201-bet.
10-masala.
Chizmadagi uchburchakning perimetri va yuzini a = 5,5 cm, b = 12 cm va c = 13 cm boΚ»lganda hisoblang.
Ishlanishi:
1) P=a+b+c=5,5+12+13=30,5 cm
2) S=0,5ab=0,5Β·5,5Β·12=0,5Β·2Β·5,5Β·2Β·3= 1Β·11Β·3=33 cmΒ²
Javob: P=30,5 cm va S=33 cmΒ².
https://t.me/matematika6uzb
Umrzoq G'aniyev: https://t.me/GUA_math
10-masala.
Chizmadagi uchburchakning perimetri va yuzini a = 5,5 cm, b = 12 cm va c = 13 cm boΚ»lganda hisoblang.
Ishlanishi:
1) P=a+b+c=5,5+12+13=30,5 cm
2) S=0,5ab=0,5Β·5,5Β·12=0,5Β·2Β·5,5Β·2Β·3= 1Β·11Β·3=33 cmΒ²
Javob: P=30,5 cm va S=33 cmΒ².
https://t.me/matematika6uzb
Umrzoq G'aniyev: https://t.me/GUA_math
π191π39π―32π€28π€―23π18π₯17
212-bet.
7-vazifa.
Traktorchi 4 soatda 12 ha yerni haydadi. U shunday mehnat unumdorligi bilan ishlasa, 6 soatda qancha yerni haydaydi?
Ishlanishi:
Proporsiya tuzib olamiz:
4 soat β 12 ha
6 soat β x
4x = 12Β·6
4x = 72
x = 72:4
x = 18 (ha).
Javob: 18 ha.
https://t.me/matematika6uzb
PULLIK GURUHGA KIRISH UCHUN MUROJAAT QILINGπ
π https://t.me/GUA_math
P.S.
Pullik guruhda darslik to'liq ishlab tushuntirib berilgan.
7-vazifa.
Traktorchi 4 soatda 12 ha yerni haydadi. U shunday mehnat unumdorligi bilan ishlasa, 6 soatda qancha yerni haydaydi?
Ishlanishi:
Proporsiya tuzib olamiz:
4 soat β 12 ha
6 soat β x
4x = 12Β·6
4x = 72
x = 72:4
x = 18 (ha).
Javob: 18 ha.
https://t.me/matematika6uzb
PULLIK GURUHGA KIRISH UCHUN MUROJAAT QILINGπ
π https://t.me/GUA_math
P.S.
Pullik guruhda darslik to'liq ishlab tushuntirib berilgan.
π250π41π―36π€32π₯25π18π€―17
205-bet.
1-vazifani ishlanishi ππ½
https://t.me/matematika6uzb
PULLIK GURUHGA KIRISH UCHUN MUROJAAT QILINGπ
π https://t.me/GUA_math
P.S.
Pullik guruhda darslik to'liq ishlab tushuntirib berilgan.
1-vazifani ishlanishi ππ½
https://t.me/matematika6uzb
PULLIK GURUHGA KIRISH UCHUN MUROJAAT QILINGπ
π https://t.me/GUA_math
P.S.
Pullik guruhda darslik to'liq ishlab tushuntirib berilgan.
π214π€―34π―34π27π€22π16π₯11
227-bet.
37-masala.
Ota oΚ»gΚ»lidan 27 yoshga katta. Agar ularning yoshi yigΚ»indisi 47 boΚ»lsa, oΚ»gΚ»il va otasining yoshini toping.
Ishlanishi:
O'g'ilning yoshi β x bo'lsin.
U holda otasining yoshi x+27 bo'ladi.
Masala shartiga ko'ra,
x+(x+27)=47
x+x+27=47
2x=47-27
2x=20
x=10
Demak, o'g'il 10 yosh, ota esa 37 yosh.
https://t.me/matematika6uzb
PULLIK GURUHGA KIRISH UCHUN MUROJAAT QILINGπ
π https://t.me/GUA_math
P.S.
Pullik guruhda darslik to'liq ishlab tushuntirib berilgan.
37-masala.
Ota oΚ»gΚ»lidan 27 yoshga katta. Agar ularning yoshi yigΚ»indisi 47 boΚ»lsa, oΚ»gΚ»il va otasining yoshini toping.
Ishlanishi:
O'g'ilning yoshi β x bo'lsin.
U holda otasining yoshi x+27 bo'ladi.
Masala shartiga ko'ra,
x+(x+27)=47
x+x+27=47
2x=47-27
2x=20
x=10
Demak, o'g'il 10 yosh, ota esa 37 yosh.
https://t.me/matematika6uzb
PULLIK GURUHGA KIRISH UCHUN MUROJAAT QILINGπ
π https://t.me/GUA_math
P.S.
Pullik guruhda darslik to'liq ishlab tushuntirib berilgan.
π353π38π30π―29π€―18π₯13π€11
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i-ALGEBRA 7-SINF YECHIMLARI (2024-25) QORAKO'L
Yangi algebra 7 (2022 yil) darsligi yechimlari.
Murojaat uchun: @GUA_math
https://t.me/matematika6uzb
Murojaat uchun: @GUA_math
https://t.me/matematika6uzb
π117π―36π34π32π₯20π€18π€―14
Forwarded from i-ALGEBRA 7-SINF YECHIMLARI (2024-25) QORAKO'L
BRAIN DAQIQAβοΈβ
πShilliqurt 23 metrli daraxt tepasiga chiqmoqchi. U har kuni kunduzi 5 metr ko'tariladi, kechasi esa 2 metrga sirg'anib pastga tushadi. Shilliqurt necha kunda daraxt uchiga chiqadiβ
πShilliqurt 23 metrli daraxt tepasiga chiqmoqchi. U har kuni kunduzi 5 metr ko'tariladi, kechasi esa 2 metrga sirg'anib pastga tushadi. Shilliqurt necha kunda daraxt uchiga chiqadiβ
Anonymous Quiz
33%
9 kun
31%
8 kun
23%
7 kun
12%
6 kun
π654π€―242π€223π―173π112π83π₯77
Telegram
i-ALGEBRA 7-SINF YECHIMLARI (2024-25) QORAKO'L
Yangi algebra 7 (2022 yil) darsligi yechimlari.
Murojaat uchun: @GUA_math
https://t.me/matematika6uzb
Murojaat uchun: @GUA_math
https://t.me/matematika6uzb
π121π38π₯31π―28π€―24π18π€4
i-MATEMATIKA 6-SINF YECHIMLARI (2024)
1-misol, a) ishlanishi ππ½ 8-bet https://t.me/matematika6uzb Umrzoq G'aniyev: https://t.me/GUA_math
8-bet, 1-vazifaning a) misoli ishlanishiπ
π340π98π₯78π€―78π69π―46π€36
Telegram
i-GEOMETRIYA 7-SINF YECHIMLARI (2024-25) QORAKO'L
https://t.me/GUA_math
π438π166π―146π₯127π€―109π€75π73
Forwarded from "AHA" MATEMATIKA-6 (MASHQ DAFTARI-1-2)-2025 πΊπΏ
Avgust oyidan boshlab "Easy math" o'quv markazida 6-7-8-9-sinf o'quvchilari uchun maxsus "Matematika" kursi ochildi. Darslarni o'zim (Umrzoq G'aniyev) o'taman.
Matematikani oddiy va qiziqarli o'rganib, Milliy Sertifikat olgan holda litsey va universitetlarga yuqori ballar bilan kirishni xoxlasangiz, ikkilanmasdan murojaat qiling!
https://t.me/GUA_math
+998935117214
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Umrzoq G'aniyev
π https://t.me/matematika6mashqdaftari π https://t.me/matematika5_2025
π7