An investor's retirement account pays an annual nominal interest rate of 4.2%, convertible monthly. On January 1 of year y, the investor's account balance was X. The investor then deposited 100 at the end of every quarter. On May 1 of year (y+ 10), the account balance was 1.9X. Determine which of the following is an equation of value that can be used to solve for X. (A) (1.0105)13121.9X+k=142(1.0151)k1100=X (B) X+k=142(1.0035)(3k1)100=(1.0035)121.9X (C) X+k=141(1.0035)3k100=(1.0035)1241.9X (D) X+k=141(1.0105)k1100=(1.0105)31241.9X (E) X+k=142(1.0105)k1100=(1.0105)31241.9X