tolong jawabannya soal vektor ​

∣ SM ∣ = 2 1 ​ ∣ S U ∣ ⇒ 4 1 ​ ∣ S U ∣ = x + 15 an d 4 3 ​ ∣ S U ∣ = 4 x − 45 ∣ S U ∣ = 4 ( x + 15 ) an d ∣ S U ∣ = 3 4 ​ ( 4 x − 45 ) 4 ( x + 15 ) = 3 4 ​ ( 4 x − 45 ) / ⋅ 4 3 ​ 3 ( x + 15 ) = 4 x − 45 3 x + 45 = 4 x − 45 3 x − 4 x = − 45 − 45 − x = − 90 x = 90 ∣ S U ∣ = 4 ( x + 15 ) = 4 ⋅ ( 90 + 15 ) = 4 ⋅ 105 = 420

Answered by kate200468 | 2024-06-10

I f T bi sec t s U V t h e n ∣ U T ∣ = ∣ V T ∣ an d ∣ U T ∣ + ∣ V T ∣ = ∣ U V ∣. ∣ U T ∣ = 2 8 7 ​ → ∣ V T ∣ = 2 8 7 ​ ∣ U V ∣ = 2 8 7 ​ + 2 8 7 ​ = 4 8 14 ​ = 4 4 7 ​ = 5 4 3 ​ A n s w er : ∣ U V ∣ = 5 4 3 ​ ​

Answered by Anonymous | 2024-06-10

The length of line segment uv is 5 4 3 ​ . This is found by using the fact that point t bisects line uv and that u t = 2 8 7 ​ . By converting u t into an improper fraction and adding the lengths of u t and t v , we can determine uv .
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Answered by kate200468 | 2024-10-09

Jawaban:Ringkasan soal Masalah:Terdapat tiga gaya yang bekerja pada suatu titik: 20 N ke arah kiri, 6√2 N pada sudut 45° ke atas, dan 6√2 N pada sudut 45° ke bawah. Kita perlu mencari resultan gaya tersebut.Diketahui:- F1 = 20 N (ke kiri)- F2 = 6√2 N (45° ke atas)- F3 = 6√2 N (45° ke bawah)Ditanya:Resultan gaya (total gaya yang bekerja).=?caranya:1. Uraikan gaya F2 dan F3 ke komponen x dan y:- F2x = F2 × cos(45°) = 6√2 × (√2 / 2) = 6 N- F2y = F2 × sin(45°) = 6√2 × (√2 / 2) = 6 N- F3x = F3 × cos(45°) = 6√2 × (√2 / 2) = 6 N- F3y = -F3 × sin(45°) = -6√2 × (√2 / 2) = -6 N2. Hitung total gaya pada arah x (Fx):- Fx = -20 N (dari F1) + 6 N (dari F2x) + 6 N (dari F3x)- Fx = -20 + 6 + 6 = -8 N3. Hitung total gaya pada arah y (Fy):- Fy = 6 N (dari F2y) - 6 N (dari F3y)- Fy = 6 - 6 = 0 NKarna Fy = 0, resultan gaya hanya berada pada arah x, yaitu -8 N.maka, resultan gaya yaitu 8 N ke arah kiri.

Answered by ara1412 | 2025-08-24