I'm unsure where EAX is in the emulator. Thanks in advance.

What will be the final value of EAX in this example? mov eax, 0 mov ecx, 10;outer loop counter L1: mov eax, 3 mov ecx, 5; inner loop counter L2: add eax, 5 loop L2;repeat inner loop loop L1;repeat outer loop Verify your solution by running the program using the 8086 emulator. You must include the screen shot of your program to receive credit