The correct option is A (9 × 10^0) nbsp;+ (2 × 10^1) + (3 × 10^2) nbsp;+ (4 × 10^3) We know that a^0=1 for any non zero integer a. 4329 = 4000 + 300 + 20 + ...

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Byju's Answer

Standard VII

Mathematics

Expanding a Rational Number Using Powers

4329=-

Question

4329 = ___

$(9 \times 10^{0}) + (2 \times 10^{1}) + (3 \times 10^{2}) + (4 \times 10^{3})$

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$(4 \times 10^{- 3}) + (3 \times 10^{1}) + (2 \times 10^{2}) + (9 \times 10^{3})$

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$(4 \times 10^{- 3}) + (3 \times 10^{- 2}) + (2 \times 10^{- 1}) + (9 \times 10^{0})$

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$(9 \times 10^{- 3}) + (2 \times 10^{- 2}) + (3 \times 10^{- 1}) + (4 \times 10^{3})$

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Solution

The correct option is A
$(9 \times 10^{0}) + (2 \times 10^{1}) + (3 \times 10^{2}) + (4 \times 10^{3})$

We know that $a^{0} = 1$ for any non-zero integer $a .$
$4329 = 4000 + 300 + 20 + 9$
$= (4 \times 10^{3}) + (3 \times 10^{2}) + (2 \times 10^{1}) + (9 \times 10^{0})$

Suggest Corrections

4329 = ___

The correct option is A (9×100)+(2×101)+(3×102)+(4×103) We know that a0=1 for any non-zero integer a. 4329=4000+300+20+9 =(4×103)+(3×102)+(2×101)+(9×100)

The correct option is A
$(9 \times 10^{0}) + (2 \times 10^{1}) + (3 \times 10^{2}) + (4 \times 10^{3})$

We know that $a^{0} = 1$ for any non-zero integer $a .$
$4329 = 4000 + 300 + 20 + 9$
$= (4 \times 10^{3}) + (3 \times 10^{2}) + (2 \times 10^{1}) + (9 \times 10^{0})$