\frac{\left(\frac{K_t\cdot \:n_G\cdot \:K_a\cdot \:K_b\cdot \:K_m\cdot \:K_n}{s^2\cdot \left(n^2_G\cdot \:J\cdot \:L\right)+s\cdot \left(n^2_G\cdot \:J\cdot \:R+n^2_G\cdot \:b_2\cdot \:L+n^2_G\cdot \:b_3\cdot \:L-b_1\cdot \:L+\left(R\cdot \:n^2_G\cdot \:b_2+R\cdot \:n^2_G\cdot \:b_3-R\cdot \:b_1\right)\right)}\right)}{n_G-\left(K_b-K_a\cdot \:K_b\cdot \:n_G\right)\cdot \left(\frac{K_t\cdot \:n_G\cdot \:\:K_a\cdot \:\:K_b\cdot \:\:K_m\cdot \:\:K_n}{s^2\cdot \left(n^2_G\cdot \:\:J\cdot \:\:L\right)+s\cdot \left(n^2_G\cdot \:\:J\cdot \:\:R+n^2_G\cdot \:\:b_2\cdot \:\:L+n^2_G\cdot \:\:b_3\cdot \:\:L-b_1\cdot \:\:L+\left(R\cdot \:\:n^2_G\cdot \:\:b_2+R\cdot \:\:n^2_G\cdot \:\:b_3-R\cdot \:\:b_1\right)\right)}\right)}

asked by guest
on Mar 25, 2025 at 8:16 pm



Mathbot Says...

I wasn't able to parse your question, but the HE.NET team is hard at work making me smarter.