1 - 103-2 - 102-3 - 101-4 - 100-2 - 10-1-3 - 10-2-

The correct option is A 1234.23 nbsp;Using a^0=1 and a^ m=1/a^m for a non zero integer a and any integer m, we have (1× 10^3)+(2× 10^2)+(3× 10^1)+(4× 10^0)+( ...

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Byju's Answer

Standard VIII

Mathematics

Expanded Form of a Number

1 × 103+2 × 1...

Question

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

1234.23

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1230.23

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1234.32

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Solution

The correct option is A $1234.23$

Using $a^{0} = 1$ and $a^{- m} = \frac{1}{a^{m}}$ for a non-zero integer $a$ and any integer $m,$ we have
$(1 \times 10^{3}) + (2 \times 10^{2}) + (3 \times 10^{1}) + (4 \times 10^{0}) + (2 \times 10^{- 1}) + (3 \times 10^{- 2})$
$= 1000 + 200 + 30 + 4 + (2 \times \frac{1}{10}) + (3 \times \frac{1}{100})$
$= 1234 + 0.2 + 0.03$
$= 1234.23$

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$(2 \times 10^{4}) + (3 \times 10^{3}) + (1 \times 10^{2}) + (6 \times 10^{1}) + (7 \times 10^{0}) + (4 \times 10^{- 1}) + (2 \times 10^{- 2}) + (8 \times 10^{- 3})$

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

is the expanded form of the number ___.

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

Write the numeral whose expanded form is given below:
(i) $6 \times 10^{4} + 3 \times 10^{3} + 0 \times 10^{2} + 7 \times 10^{1} + 8 \times 10^{0}$
(ii) $9 \times 10^{6} + 7 \times 10^{5} + 0 \times 10^{4} + 3 \times 10^{3} + 4 \times 10^{2} + 6 \times 10^{1} + 2 \times 10^{0}$
(iii) $8 \times 10^{5} + 6 \times 10^{4} + 4 \times 10^{3} + 2 \times 10^{2} + 9 \times 10^{1} + 6 \times 10^{0}$

Find the number from each of the following expanded forms:

(a) 8 × 10⁴ + 6 × 10³ + 0 × 10² + 4 × 10¹ + 5 × 10⁰

(b) 4 × 10⁵+ 5 × 10³+ 3 × 10²+ 2 × 10⁰

(d) 9 × 10⁵ + 2 × 10² + 3 × 10¹

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(1×103)+(2×102)+(3×101)+(4×100)+(2×10−1)+(3×10−2) =

The correct option is A 1234.23 Using a0=1 and a−m=1am for a non-zero integer a and any integer m, we have (1×103)+(2×102)+(3×101)+(4×100)+(2×10−1)+(3×10−2) =1000+200+30+4+(2×110)+(3×1100) =1234+0.2+0.03 =1234.23

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