Ufumbuzi - Tiger algebra Suluhishaji

2, 548

2,548

Maelezo kwa hatua

${(9 F 4)}_{16} = b a s e 10$

$9 F 4$ , $16$ , $10$ .

${(9 F 4)}_{16} = 9 \cdot 16^{2} + 15 \cdot 16^{1} + 4 \cdot 16^{0}$

$9 \cdot 16^{2} + 15 \cdot 16^{1} + 4 \cdot 16^{0} = 2548$

Hatua: $9 \cdot 16^{2}$
+ $15 \cdot 16^{1}$
+ $4 \cdot 16^{0}$ = $2, 548$ .

$\frac{2548}{10} = 254 r e m a \in d e r 8$

$\frac{254}{10} = 25 r e m a \in d e r 4$

$\frac{25}{10} = 2 r e m a \in d e r 5$

$\frac{2}{10} = 0 r e m a \in d e r 2$

$8 \to 4 \to 5 \to 2 = 2548$

$2548 (b a s e 10) = {(2548)}_{10}$

$2, 548_{10} = 2, 548_{10}$ .

Je, maelezo haya yalikuwa na manufaa?

Tulifanyaje?

Jifunze zaidi na Tiger

Ubadilishaji wa msingi ni wa msingi kwa sayansi ya kompyuta, mifumo ya kidijitali, na nadharia ya namba.