berapa pasangan elektron bebas (PEB) dan pasangan elektron ikatan (PEI) PADA SENYAWA H2SO4????

− 10 x 2 − 41 x − 21 = − ( 10 x 2 + 41 x + 21 ) = − ( 10 x 2 + 35 x + 6 x + 21 ) = = − [( 10 x 2 + 35 x ) + ( 6 x + 21 )] = − [ 5 x ( 2 x + 7 ) + 3 ( 2 x + 7 )] = − ( 2 x + 7 ) ( 5 x + 3 )

Answered by Lilith | 2024-06-10

− 10 x 2 − 41 x − 21 = − 10 x 2 − 35 x − 6 x − 21 = − 5 x ( 2 x + 7 ) − 3 ( 2 x + 7 ) = − ( 5 x + 3 ) ( 2 x + 7 )

Answered by konrad509 | 2024-06-10

The expression − 10 x 2 − 41 x − 21 can be factored as − ( 2 x + 7 ) ( 5 x + 3 ) by using grouping after rearranging and finding coefficients. First, we factor out a negative, then find two numbers that multiply to 210 and add to 41 . The final step involves grouping and factoring out common terms.
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Answered by Lilith | 2024-12-24

PEI = 4 pasang. PEB =. 2 pasang

Answered by lrntrn17 | 2025-07-16