A good answer might be:

lui $13, 0x0004
lw  $12, 10($13)

After the lui instruction $13 contains 0x00040000

Filling in the bottom Half

By using the lui instruction, the base register can be loaded with multiples of 0x00010000. But often you want a more specific address in the base register. Use the ori instruction to fill the bottom 16 bits. Recall that:

 ori t,s,imm 

zero-extends imm to 32 bits then does a bitwise OR of that with the contents of register $s. The result goes into register $t.

Say that memory is as at right. The lw instruction (below) loads the word at 0x0060500C into $12.

lui $13, 0x________
ori $13, $13, 0x________
lw  $12, 0xC($13)

QUESTION 10:

Complete the instruction sequence.