Solución - Tiger Algebra Solver

$${\left(9B0\right)}_{16}$$

Pasos: $$9·{16}^2$$
+ $$11·{16}^1$$
+ $$0·{16}^0$$ = $$2.480$$.

$$9\cdot {16}^2+11\cdot {16}^1+0\cdot {16}^0$$

Pasos: $$9·{16}^2$$
+ $$11·{16}^1$$
+ $$0·{16}^0$$ = $$2.480$$.

$$2480$$

Pasos: $$9·{16}^2$$
+ $$11·{16}^1$$
+ $$0·{16}^0$$ = $$2,480$$.

$$2480\left(base10\right)$$

$${2.480}_{10}={2.480}_{10}$$.

$$\frac{2480}{10}$$

$${2.480}_{10}={2.480}_{10}$$.

$$2480=10\cdot 248+0$$

$${2.480}_{10}={2.480}_{10}$$.

$$248=10\cdot 24+8$$

$${2.480}_{10}={2.480}_{10}$$.

$$24=10\cdot 2+4$$

$${2.480}_{10}={2.480}_{10}$$.

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

$${2.480}_{10}={2.480}_{10}$$.

$$2480$$

$${2.480}_{10}={2.480}_{10}$$.

$${\left(2480\right)}_{10}$$

$$2,{480}_{10}=2,{480}_{10}$$.