Solución - Tiger Algebra Solver

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

Pasos: $$10·{16}^2$$
+ $$12·{16}^1$$
+ $$15·{16}^0$$ = $$2.767$$.

$$10\cdot {16}^2+12\cdot {16}^1+15\cdot {16}^0$$

Pasos: $$10·{16}^2$$
+ $$12·{16}^1$$
+ $$15·{16}^0$$ = $$2.767$$.

$$2767$$

Pasos: $$10·{16}^2$$
+ $$12·{16}^1$$
+ $$15·{16}^0$$ = $$2,767$$.

$$2767\left(base10\right)$$

$${2.767}_{10}={2.767}_{10}$$.

$$\frac{2767}{10}$$

$${2.767}_{10}={2.767}_{10}$$.

$$2767=10\cdot 276+7$$

$${2.767}_{10}={2.767}_{10}$$.

$$276=10\cdot 27+6$$

$${2.767}_{10}={2.767}_{10}$$.

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

$${2.767}_{10}={2.767}_{10}$$.

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

$${2.767}_{10}={2.767}_{10}$$.

$$2767$$

$${2.767}_{10}={2.767}_{10}$$.

$${\left(2767\right)}_{10}$$

$$2,{767}_{10}=2,{767}_{10}$$.