y 2 − 6 y + 6 = 0 a = 1 , b = − 6 , c = 6 Δ = b 2 − 4 a c = ( − 6 ) 2 − 4 ⋅ 1 ⋅ 6 = 36 − 24 = 12 Δ = 12 = 4 ⋅ 3 = 2 3 x 1 = 2 a − b − Δ = 2 6 − 2 3 = 2 2 ( 3 − 3 ) = 3 − 3
x 2 = 2 a − b + Δ = 2 6 + 2 3 = 2 2 ( 3 + 3 ) = 3 + 3
y 2 − 6 y + 6 = 0 y 2 − 6 y + 9 − 3 = 0 ( y − 3 ) 2 = 3 ∣ y − 3∣ = 3 y − 3 = 3 ∨ y − 3 = − 3 y = 3 + 3 ∨ y = 3 − 3
The solutions for the equation y 2 − 6 y + 6 = 0 are y = 3 − 3 and y = 3 + 3 . We used the quadratic formula to find these values since factoring was not straightforward. This process involved calculating the discriminant and applying it in the quadratic formula.
;
Jawaban:Tentu! Mari kita hitung nilai R berdasarkan rumus yang diberikan dan nilai-nilai yang diketahui:Rumus:R = \frac{\rho}{2 \times \pi \times L} \times \left( \ln \left( \frac{4 \times L}{a} \right) - 1 \right)Diketahui: * \rho = 10 * \pi \approx 3,14 * L = 2,9 * a = 0,008Langkah 1: Hitung nilai di dalam kurung \ln\frac{4 \times L}{a} = \frac{4 \times 2,9}{0,008} = \frac{11,6}{0,008} = 1450Langkah 2: Hitung nilai \ln(1450)\ln(1450) \approx 7,2793Langkah 3: Hitung nilai di dalam kurung besar\ln \left( \frac{4 \times L}{a} \right) - 1 = 7,2793 - 1 = 6,2793Langkah 4: Hitung penyebut dari pecahan2 \times \pi \times L = 2 \times 3,14 \times 2,9 = 6,28 \times 2,9 = 18,212Langkah 5: Hitung nilai RR = \frac{\rho}{2 \times \pi \times L} \times \left( \ln \left( \frac{4 \times L}{a} \right) - 1 \right)R = \frac{10}{18,212} \times 6,2793R \approx 0,5490 \times 6,2793R \approx 3,4476Jadi, nilai R adalah sekitar 3,4476.