Ufumbuzi - Tiger algebra Suluhishaji

2, 211

2,211

Maelezo kwa hatua

${(8 A 3)}_{16} = b a s e 10$

$8 A 3$ , $16$ , $10$ .

${(8 A 3)}_{16} = 8 \cdot 16^{2} + 10 \cdot 16^{1} + 3 \cdot 16^{0}$

$8 \cdot 16^{2} + 10 \cdot 16^{1} + 3 \cdot 16^{0} = 2211$

Hatua: $8 \cdot 16^{2}$
+ $10 \cdot 16^{1}$
+ $3 \cdot 16^{0}$ = $2, 211$ .

$\frac{2211}{10} = 221 r e m a \in d e r 1$

$\frac{221}{10} = 22 r e m a \in d e r 1$

$\frac{22}{10} = 2 r e m a \in d e r 2$

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

$1 \to 1 \to 2 \to 2 = 2211$

$2211 (b a s e 10) = {(2211)}_{10}$

$2, 211_{10} = 2, 211_{10}$ .

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