Solución - Tiger Algebra Solver

$${\left(5B4\right)}_{16}$$

Pasos: $$5·{16}^2$$
+ $$11·{16}^1$$
+ $$4·{16}^0$$ = $$1.460$$.

$$5\cdot {16}^2+11\cdot {16}^1+4\cdot {16}^0$$

Pasos: $$5·{16}^2$$
+ $$11·{16}^1$$
+ $$4·{16}^0$$ = $$1.460$$.

$$1460$$

Pasos: $$5·{16}^2$$
+ $$11·{16}^1$$
+ $$4·{16}^0$$ = $$1,460$$.

$$1460\left(base10\right)$$

$${1.460}_{10}={1.460}_{10}$$.

$$\frac{1460}{10}$$

$${1.460}_{10}={1.460}_{10}$$.

$$1460=10\cdot 146+0$$

$${1.460}_{10}={1.460}_{10}$$.

$$146=10\cdot 14+6$$

$${1.460}_{10}={1.460}_{10}$$.

$$14=10\cdot 1+4$$

$${1.460}_{10}={1.460}_{10}$$.

$$1=10\cdot 0+1$$

$${1.460}_{10}={1.460}_{10}$$.

$$1460$$

$${1.460}_{10}={1.460}_{10}$$.

$${\left(1460\right)}_{10}$$

$$1,{460}_{10}=1,{460}_{10}$$.