Ufumbuzi - Tiger algebra Suluhishaji

4, 045

4,045

Maelezo kwa hatua

${(F C D)}_{16} = b a s e 10$

$F C D$ , $16$ , $10$ .

${(F C D)}_{16} = 15 \cdot 16^{2} + 12 \cdot 16^{1} + 13 \cdot 16^{0}$

$15 \cdot 16^{2} + 12 \cdot 16^{1} + 13 \cdot 16^{0} = 4045$

Hatua: $15 \cdot 16^{2}$
+ $12 \cdot 16^{1}$
+ $13 \cdot 16^{0}$ = $4, 045$ .

$\frac{4045}{10} = 404 r e m a \in d e r 5$

$\frac{404}{10} = 40 r e m a \in d e r 4$

$\frac{40}{10} = 4 r e m a \in d e r 0$

$\frac{4}{10} = 0 r e m a \in d e r 4$

$5 \to 4 \to 0 \to 4 = 4045$

$4045 (b a s e 10) = {(4045)}_{10}$

$4, 045_{10} = 4, 045_{10}$ .

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.