The digit in the unit's place of the number represented by $(7^{95} - 7^{58})$ is

Since we only need the unit's digit, patterns in powers of 7 matter:

Unit's digit in powers of 7 follows pattern: 7,9,3,1,7,9,3,1,... and so on (cycles every 4 powers)

For $7^{58}$:

58 mod 4 = 2

So $7^{58}$ ends in same digit as $7^2$ = 49, which is 9

For $7^{95}$:

95 mod 4 = 3

So $7^{95}$ ends in same digit as $7^3$ = 343, which is 3

Therefore, $7^{95} - 7^{58}$ ends in $(..3 - ..9) = ..4$