0-000729-3-4- 0-09-3-4 - 9-3X- find X

The correct option is D 10^6 We have, ( 0.000729 )^ 3/4×( 0.09 )^ 3/4 =( 729×10^ 6)^ 3/4×( 9×10^ 2)^ 3/4 =( 9^3×10^ 6)^ 3/4×( 9×10^ 2)^ 3/4 ...

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Byju's Answer

Standard XII

Physics

Order of Magnitude

0.000729-3/4×...

Question

$(0.000729)^{- 3 / 4} \times (0.09)^{- 3 / 4}$ = $9^{- 3} X$ , find $X$

10^{3}

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10^{5}

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10^{2}

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10^{6}

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Solution

The correct option is D $10^{6}$
We have,

${(0.000729)}^{- 3 / 4} \times {(0.09)}^{- 3 / 4}$
$= {(729 \times 10^{- 6})}^{- 3 / 4} \times {(9 \times 10^{- 2})}^{- 3 / 4}$
$= {(9^{3} \times 10^{- 6})}^{- 3 / 4} \times {(9 \times 10^{- 2})}^{- 3 / 4}$
$= (9^{- 9 / 4} \times 10^{18 / 4}) \times (9^{- 3 / 4} \times 10^{6 / 4})$
$= 9^{- 3} \times 10^{6}$
$∴ X = 10^{6}$

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Similar questions

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) $8 \times 10^{5} + 6 \times 10^{4} + 4 \times 10^{3} + 2 \times 10^{2} + 9 \times 10^{1} + 6 \times 10^{0}$

Express the following standard form.

(a) 5 × 10⁶ + 4 × 10⁵ + 2 × 10³ + 9 × 10² + 6 × 10⁰

(b) 4 × 10⁵ + 8 ×10³ + 7 × 10¹

Q. Write the number :

4 \times 10^{6} + 8 \times 10^{5} + 6 \times 10^{4} + 4 \times 10^{3} + 4 \times 10^{2} + 6 \times 10 + 9 \times 10^{0}

Q. Find the number from each of the following expanded forms:
(i) 7 × 10⁴ + 6 × 10³ + 0 × 10² + 4 × 10¹ + 5 × 10⁰
(ii) 5 × 10⁵ + 4 × 10⁴ + 2 × 10³ + 3 × 10⁰
(iii) 9 × 10⁵ + 5 × 10² + 3 × 10¹
(iv) 3 × 10⁴+ 4 × 10² + 5 × 10⁰

4 \times 10^{5} + 5 \times 10^{3} + 3 \times 10^{2} + 2 \times 10^{2} =

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(0.000729)−3/4×(0.09)−3/4 = 9−3X, find X

The correct option is D 106We have,(0.000729)−3/4×(0.09)−3/4=(729×10−6)−3/4×(9×10−2)−3/4=(93×10−6)−3/4×(9×10−2)−3/4=(9−9/4×1018/4)×(9−3/4×106/4)=9−3×106∴X=106

$(0.000729)^{- 3 / 4} \times (0.09)^{- 3 / 4}$ = $9^{- 3} X$ , find $X$