0 157 1 65 0 0108 0 0028 Z 0 157 1 65 0 0108 0 0028 1 1

Last updated: 11/28/2023

0 157 1 65 0 0108 0 0028 Z 0 157 1 65 0 0108 0 0028 1 1 solving for T The equation 1 is in quadratic form solving for by evaluating the roots of the quadratic equation Rewring the equatin 1 3 Solving for root Z 1 2 concentration Discarding the negative root 0 06 0 0 0556 0 05 0 04 0 03 0 02 0 82572 0 157 0 008 0 0 04315 0 01 ax bx c 0 T 0 root 0 root T 0 03 b b 4 a c 2 a 0 15 0 152 4 0 825 0 008 2 0 825 0 15 0 0225 0 0264 2 0 825 0 5 0 15 0 2211 2 0 825 1 0 4311 Distance Z 1 5 y 0 0128x 0 0557 R 0 9998 2 2 5