Find a particular solution of the indicated linear system that satisfies the initial conditions x₁ (0)=5, x2(0)= -5. 14 The particular solution is x₁ (t) = and x2(t) =

The break-even point occurs where the graphs of C and R intersect. Thus, we find this point by solving the system (x) = 500,000 + 400x |R(x) = 600x [y = 500,000 + 400x or ly = 600x. Using the substitution method, we can substitute 600x for y in the first equation. 600x = 500,000 + 400x Substitute 600x for y in y= 500,000 + 400x. 200x 500,000 Subtract 400x from both sides. x = 2500 Divide both sides by 200. Back-substituting 2500 for x in either of the system's equations (or functions), we obtain R(2500) = 600(2500) = 1,500,000. We used R[x) = 60ox.

Use EROS and backward substitution to solve the system Ax= b, where (三) (6) 21 1 and 1. -3 -1 What is the rank of A?

Find a particular solution of the indicated linear system that satisfies the initial conditions x₁ (0) = 2, x₂ (0) = 3, and x3 (0) = 7. X' = - 27 - 28 2 25 26 - 25 - 25 - 2t - 2 x; x₁ = e 3 The particular solution is x₁ (t) = 6 -5 5 X₂ (t) = X₂ =e³t| nd x3 (t) = and 1 1 X3 1