Rešenje - Tiger Algebra Rešavač

$${\left(2B1\right)}_{16}$$

Koraci: $$2·{16}^2$$
+ $$11·{16}^1$$
+ $$1·{16}^0$$ = $$689$$.

$$2\cdot {16}^2+11\cdot {16}^1+1\cdot {16}^0$$

Koraci: $$2·{16}^2$$
+ $$11·{16}^1$$
+ $$1·{16}^0$$ = $$689$$.

$$689$$

Koraci: $$2·{16}^2$$
+ $$11·{16}^1$$
+ $$1·{16}^0$$ = $$689$$.

$$689\left(base10\right)$$

$${689}_{10}={689}_{10}$$.

$$\frac{689}{10}$$

$${689}_{10}={689}_{10}$$.

$$689=10\cdot 68+9$$

$${689}_{10}={689}_{10}$$.

$$68=10\cdot 6+8$$

$${689}_{10}={689}_{10}$$.

$$6=10\cdot 0+6$$

$${689}_{10}={689}_{10}$$.

$$689$$

$${689}_{10}={689}_{10}$$.

$${\left(689\right)}_{10}$$

$${689}_{10}={689}_{10}$$.