36 32 28 ....... 20 16 12
Answer (3)
If there are no duplicates among the rational numbers and no duplicates among the irrational numbers, then there must be exactly 6 points.
None of the rational numbers can also be irrational, and none of the irrational numbers can also be rational.
A set of six numbers can be graphed with a minimum of 2 distinct points if there are duplicates among the numbers. If all numbers are unique, then you would need a maximum of 6 distinct points. The exact number depends on duplicates among the rational and irrational numbers.
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[tex]U_{4} = 24[/tex]PEMBAHASAN :Pola barisan aritmetika = 36, 32, 28, .... , 20, 16, 12.Suku pertama (a) = [tex]U_{1} = 36[/tex]Beda (b) = [tex]U_{2} - U_{1} = 32 - 36 = - 4[/tex]Rumus suku ke-n [tex](U_{n})[/tex] adalah[tex]U_{n} = a + (n - 1)b[/tex][tex]U_{n} = 36 + (n - 1)( - 4)[/tex][tex]U_{n} = 36 - 4n + 4[/tex][tex]U_{n} = 36 + 4 - 4n[/tex][tex]U_{n} = 40 - 4n[/tex]Jadi, nilai dari suku ke-4 [tex](U_{4})[/tex] adalah[tex]U_{n} = 40 - 4n[/tex][tex]U_{4} = 40 - 4(4)[/tex][tex]U_{4} = 40 - 16[/tex][tex]U_{4} = 24[/tex]
