Ufumbuzi - Tiger algebra Suluhishaji

3, 247

3,247

Maelezo kwa hatua

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

$C A F$ , $16$ , $10$ .

${(C A F)}_{16} = 12 \cdot 16^{2} + 10 \cdot 16^{1} + 15 \cdot 16^{0}$

$12 \cdot 16^{2} + 10 \cdot 16^{1} + 15 \cdot 16^{0} = 3247$

Hatua: $12 \cdot 16^{2}$
+ $10 \cdot 16^{1}$
+ $15 \cdot 16^{0}$ = $3, 247$ .

$\frac{3247}{10} = 324 r e m a \in d e r 7$

$\frac{324}{10} = 32 r e m a \in d e r 4$

$\frac{32}{10} = 3 r e m a \in d e r 2$

$\frac{3}{10} = 0 r e m a \in d e r 3$

$7 \to 4 \to 2 \to 3 = 3247$

$3247 (b a s e 10) = {(3247)}_{10}$

$3, 247_{10} = 3, 247_{10}$ .

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