Square And Qube

$$ = \sqrt {0.000441} $$
$$ = \sqrt {\frac{{441}}{{{{10}^6}}}} $$
$$ = \frac{{\sqrt {441} }}{{\sqrt {{{10}^6}} }}$$
$$ = \frac{{21}}{{{{10}^3}}}$$
$$ = \frac{{21}}{{1000}}$$
$$ = 0.021$$

$$ = {1.5^2} \times \sqrt {0.0225} $$
$$ = {1.5^2} \times \sqrt {\frac{{225}}{{10000}}} $$
$$ = 2.25 \times \frac{{15}}{{100}}$$
$$ = 2.25 \times 0.15$$
$$ = 0.3375$$

Given expression,
$$ = \sqrt {\frac{1}{{100}}} + \sqrt {\frac{{81}}{{100}}} + \sqrt {\frac{{121}}{{100}}} + $$      $$\sqrt {\frac{9}{{10000}}} $$
$$ = \frac{1}{{10}} + \frac{9}{{10}} + \frac{{11}}{{10}} + \frac{3}{{100}}$$
$$ = 0.1 + 0.9 + 1.1 + 0.03$$
$$ = 2.13$$

$${\text{Given expression,}}$$
$$\sqrt {1.30} + \sqrt {1300} + \sqrt {0.0130} $$
$$ = \sqrt {\frac{{130}}{{100}}} + \sqrt {13 \times 100} + \sqrt {\frac{{130}}{{10000}}} $$
$$ = \frac{{\sqrt {130} }}{{10}} + \sqrt {13} \times 10 + \frac{{\sqrt {130} }}{{100}}$$
$$ = \frac{{11.40}}{{10}} + 3.605 \times 10 + \frac{{11.40}}{{100}}$$
$$ = 1.14 + 36.05 + 0.114$$
$$ = 37.304$$

T

Clearly, the woman was born between 1800 and 1850 is 1849.
$${\text{And, }}1849 = {\left( {43} \right)^2}$$
So, the woman was 43 years old in 1849
$${\text{Thus, he was born in }}$$
$$ = \left( {1849 - 43} \right)$$
$$ = 1806$$

$$ ⇒ {\text{Q}} = \sqrt {\frac{{{{\left( {8{\text{R}}} \right)}^2}}}{4}} $$
$$ ⇒ {\text{Q}} = \frac{{\sqrt {{{\left( {8{\text{R}}} \right)}^2}} }}{{\sqrt 4 }}$$
$$ ⇒ {\text{Q}} = \frac{{8{\text{R}}}}{2}$$
$$ ⇒ {\text{Q}} = 4{\text{R}}$$

The smallest such number is 1444 $$\left[ {1444 = {{\left( {38} \right)}^2}} \right]$$
It lies between 1000 and 2000.

$${\text{We have,}}$$
$$1 \times 2 \times 3 \times 4 = 24$$
$${\text{And }}24 + 1 = 25\left[ {25 = {5^2}} \right]$$
$$2 \times 3 \times 4 \times 5 = 120{\text{ }}$$
$${\text{And 1}}20 + 1 = 121\left[ {121 = {{11}^2}} \right]$$
$$3 \times 4 \times 5 \times 6 = 360{\text{ }}$$
$${\text{And }}360 + 1 = 361\left[ {361 = {{19}^2}} \right]$$
$$4 \times 5 \times 6 \times 7 = 840{\text{ }}$$
$${\text{And }}840 + 1 = 841\left[ {841 = {{29}^2}} \right]$$
$$\therefore p = 1$$