Simplify the following - - 0-0081

The correct option is C 0.09 0.09 : Since (0.09)(0.09)=0.0081 , #160; √(0.0081)=0.09 . #160;You can also rewrite 0.0081 as 81 ...

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Standard XII

Mathematics

Algebra of Limits

Simplify the ...

Question

Simplify the following : $\sqrt{0.0081}$

0.02

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0.09

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0.06

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0.05

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Solution

The correct option is C $0.09$
$0.09$ : Since $(0.09) (0.09) = 0.0081$ ,

$\sqrt{0.0081} = 0.09$ .

You can also rewrite $0.0081$ as $81 \times 10^{- 4}$

$\sqrt{81 \times 10^{- 4}} = \sqrt{81} \times \sqrt{10^{- 4}} = 9 \times {(10^{- 4})}^{\frac{1}{2}} = 9 \times 10^{- 2} = 0.09$

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Similar questions

Q. Simplify:

\frac{0.0081}{0.09} (1000)

Q. Evaluate:

\frac{0.06 \times 0.06 \times 0.06 - 0.05 \times 0.05 \times 0.05}{0.06 \times 0.06 + 0.06 \times 0.05 + 0.05 \times 0.05}

Q. Question 5 (ii)
A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows:

\begin{matrix} 0.03 & 0.08 & 0.08 & 0.09 & 0.04 & 0.17 0.16 & 0.05 & 0.02 & 0.06 & 0.18 & 0.20 0.11 & 0.08 & 0.12 & 0.13 & 0.22 & 0.07 0.08 & 0.01 & 0.10 & 0.06 & 0.09 & 0.18 0.11 & 0.07 & 0.05 & 0.07 & 0.01 & 0.04 \end{matrix}

(ii) For how many days, was the concentration of sulphur dioxide more than 0.11 parts per million?

A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows: $\begin{matrix} 0.03 & 0.08 & 0.08 & 0.09 & 0.04 & 0.17 0.16 & 0.05 & 0.02 & 0.06 & 0.18 & 0.20 0.11 & 0.08 & 0.12 & 0.13 & 0.22 & 0.07 0.08 & 0.01 & 0.10 & 0.06 & 0.09 & 0.18 0.11 & 0.07 & 0.05 & 0.07 & 0.01 & 0.04 \end{matrix}$
The concentration of sulphur dioxide was more than 0.11 parts per million for 8 days.

Q. Question 5 (ii)
A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows:

\begin{matrix} 0.03 & 0.08 & 0.08 & 0.09 & 0.04 & 0.17 0.16 & 0.05 & 0.02 & 0.06 & 0.18 & 0.20 0.11 & 0.08 & 0.12 & 0.13 & 0.22 & 0.07 0.08 & 0.01 & 0.10 & 0.06 & 0.09 & 0.18 0.11 & 0.07 & 0.05 & 0.07 & 0.01 & 0.04 \end{matrix}

(ii) For how many days, was the concentration of sulphur dioxide more than 0.11 parts per million?

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Simplify the following : √0.0081

The correct option is C 0.090.09: Since (0.09)(0.09)=0.0081, √0.0081=0.09. You can also rewrite 0.0081 as 81×10−4√81×10−4=√81×√10−4=9×(10−4)12=9×10−2=0.09

Simplify the following : $\sqrt{0.0081}$

The correct option is C $0.09$
$0.09$ : Since $(0.09) (0.09) = 0.0081$ ,

$\sqrt{0.0081} = 0.09$ .

You can also rewrite $0.0081$ as $81 \times 10^{- 4}$

$\sqrt{81 \times 10^{- 4}} = \sqrt{81} \times \sqrt{10^{- 4}} = 9 \times {(10^{- 4})}^{\frac{1}{2}} = 9 \times 10^{- 2} = 0.09$