Diketahui sebuah lingkaran memiliki pusat di titik
P(3,−2) dan berjari-jari 5 satuan.
Tentukan persamaan lingkaran tersebut!

On or near the Earth's surface, without air resistance, it's 9.81 meters per second². (32.2 feet per second²)

Answered by AL2006 | 2024-06-10

Freely falling acceleration is the acceleration due to gravity 'g'. It's value is 9.8 N, but for convenience it is taken as 10 N.

Answered by sairah | 2024-06-10

The free-fall acceleration on or near the Earth's surface is approximately 9.81 m/s 2 . This value indicates the rate at which an object speeds up as it falls due to gravity, irrespective of its mass. The direction of this acceleration is downward toward the center of the Earth.
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Answered by AL2006 | 2024-12-23

Jawaban:(x-3)² + (y+2)² = 25Penjelasan dengan langkah-langkah:Rumus jari-jari lingkaran yg memiliki pusat di titik P(a,b) adalah (x-a)² + (y-b)² = R²Jika lingkaran memiliki pusat di titik P(3,−2) dan berjari-jari 5 satuan, maka persamaan lingkaran tersebut adalah:(x-3)² + (y-(-2))² = 5²(x-3)² + (y+2)² = 25

Answered by PrasTy0 | 2025-07-02

Penyelesaian[tex]P(h,k)=(3,-2)[/tex][tex]r=5[/tex][tex](x-h)^2 + (y-k)^2 = r^2[/tex][tex](x-3)^2 + (y+2)^2 = 5^2[/tex][tex]ax^2 + by^2 + cx + dy + e = 0[/tex][tex]x^{2}-6x+9 + y^{2}+4y+4 = 25[/tex][tex]x^{2}+y^{2}-6x+4y +13 -25 = 0[/tex][tex]x^{2}+y^{2}-6x+4y -12 = 0[/tex]bentuk baku = [tex]\boxed{(x-3)^2 + (y+2)^2 = 25}[/tex]bentuk umum = [tex]\boxed{x^{2}+y^{2}-6x+4y-12 = 0}[/tex]

Answered by vinganzbeut | 2025-07-02