Nilai rata² lima orang kelas 10 adalah 82 filza mengikuti ulangan sususlan rata² menjadi 84 maka nilai filza adalah
Answer (3)
x:8=c1 reminder 3 x=M₈ +3
x:10=c2 reminder 3 x=M₁₀ +3
x:12=c3 reminder 3 x=M₁₂ +3
x:30=c4 reminder 3 x=M₃₀ +3
x=[M₈;M₁₀;M₁₂;M₃₀]+3
8=2³ 10=2* 5 12=2² * 3 30=2*3 *5
200 < x < 500
x= 2³ * 5 3 +3 x=8 5*3+3 x=120+3 x=123 it doesn't corespond
x=2⁴ 15+3 x=16 15+3 x=240+3 x=243
x=8 5 9+3 x=40*9+3 x=360+3 x=363
x=8 5 3 4+3 x=40 12+3 x=480+3 x=483
x=8*5² 3+3 x=24 25+3 x=600+3 x=603 it doesn't corespond
x={243;363;483}
The smallest number between 200 and 500 that leaves a remainder of 3 when divided by 8, 10, 12, and 30 is 243. This was determined by finding the least common multiple of these divisors and solving the related equations. Therefore, the answer is 243.
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Jawaban:94Penjelasan dengan langkah-langkah:Total Nilai Awal : 82 x 5 = 410Jumlah Siswa : 5 + 1 = 6Total Nilai Akhir : 84 x 6= 504Nilai Filza : 504 - 410 = 94
