Carilah golongan dan periode nya beserta kulit dan subkulitnya 1.12 mg 2.20ca 3.17ci

x + 1 x ​ < x − 1 x ​ x + 1 x ​ − x − 1 x ​ < 0 x ( x + 1 1 ​ − x + 1 1 ​ ) < 0 x 2 − 1 2 x ​ < 0 Consider the function and define the domain of the function y = x 2 − 1 2 x ​ x 2 − 1  = 0 x 2  = 1 x  = ± 1 Function is zero x 2 − 1 2 x ​ = 0 2 x = 0 x = 0
Answer: ( − 1 ; 0 ) ∪ ( 1 ; + ∞ )

Answered by optimistic | 2024-06-10

( x + 5 ) / ( x + 2 ) < 0 <=> ( x + 5 < 0 and x + 2 > 0 ) or ( x + 5 > 0 or x + 2 < 0 ) <=> ( x < -5 and x > - 2 ) or ( x > -5 and x < -2 ) <=> ( x∈Ф ) or ( - 5 < x < - 2 ) <=> - 5 < x < - 2.

The answer is the fourth choice.

Answered by crisforp | 2024-06-10

To solve x + 1 x ​ < x − 1 x ​ , we combine fractions and find critical points at x = − 1 , 0 , 1 . The solution is ( − 1 , 0 ) ∪ ( 1 , + ∞ ) with a domain excluding x = − 1 and x = 1 .
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Answered by optimistic | 2024-12-26

1. Elemen: Magnesium (Mg)   - Nomor Atom: 12   - Konfigurasi Elektron: 1s² 2s² 2p⁶ 3s²   - Golongan: 2 (Golongan Logam Alkalina Tanah)   - Periode: 3   - Kulit dan Subkulit:     - Kulit pertama (K): 2 elektron (subkulit 1s²)     - Kulit kedua (L): 8 elektron (subkulit 2s² 2p⁶)     - Kulit ketiga (M): 2 elektron (subkulit 3s²)2. Elemen: Kalsium (Ca)   - Nomor Atom: 20   - Konfigurasi Elektron: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²   - Golongan: 2 (Golongan Logam Alkalina Tanah)   - Periode: 4   - Kulit dan Subkulit:     - Kulit pertama (K): 2 elektron (subkulit 1s²)     - Kulit kedua (L): 8 elektron (subkulit 2s² 2p⁶)     - Kulit ketiga (M): 8 elektron (subkulit 3s² 3p⁶)     - Kulit keempat (N): 2 elektron (subkulit 4s²)3. Elemen: Klorin (Cl)   - Nomor Atom: 17   - Konfigurasi Elektron: 1s² 2s² 2p⁶ 3s² 3p⁵   - Golongan: 17 (Golongan Halogen)   - Periode: 3   - Kulit dan Subkulit:     - Kulit pertama (K): 2 elektron (subkulit 1s²)     - Kulit kedua (L): 8 elektron (subkulit 2s² 2p⁶)     - Kulit ketiga (M): 7 elektron (subkulit 3s² 3p⁵)

Answered by pupusdwisuryoanggono | 2025-07-18