Solución - Tiger Algebra Solver

$${\left(ABC\right)}_{16}$$

Pasos: $$10·{16}^2$$
+ $$11·{16}^1$$
+ $$12·{16}^0$$ = $$2.748$$.

$$10\cdot {16}^2+11\cdot {16}^1+12\cdot {16}^0$$

Pasos: $$10·{16}^2$$
+ $$11·{16}^1$$
+ $$12·{16}^0$$ = $$2.748$$.

$$2748$$

Pasos: $$10·{16}^2$$
+ $$11·{16}^1$$
+ $$12·{16}^0$$ = $$2,748$$.

$$2748\left(base10\right)$$

$${2.748}_{10}={2.748}_{10}$$.

$$\frac{2748}{10}$$

$${2.748}_{10}={2.748}_{10}$$.

$$2748=10\cdot 274+8$$

$${2.748}_{10}={2.748}_{10}$$.

$$274=10\cdot 27+4$$

$${2.748}_{10}={2.748}_{10}$$.

$$27=10\cdot 2+7$$

$${2.748}_{10}={2.748}_{10}$$.

$$2=10\cdot 0+2$$

$${2.748}_{10}={2.748}_{10}$$.

$$2748$$

$${2.748}_{10}={2.748}_{10}$$.

$${\left(2748\right)}_{10}$$

$$2,{748}_{10}=2,{748}_{10}$$.