$(2.3\times 10^4) \times (3 \times 10^3)$
$\Rightarrow (2.3 \times 3) \times(10^{3+4})$

$\dfrac{6.3\times 10^{3}}{3\times 10^7}$

We have,

The average distance in KM will be $\frac { 2.3\times { 10 }^{ 14 } }{ 3.9\times { 10 }^{ 4 } } =0.59\times { 10 }^{ 10 }=5.9\times { 10 }^{ 9 }\quad km$

$\dfrac{1.8\times 10^{11}}{2\times 10^4}$

${ 3000 }^{ 2 }\times { 20 }^{ 3 }=(9\times { 10 }^{ 6 })\times (8\times { 10 }^{ 3 })\ =72\times { 10 }^{ 9 }=7.2\times { 10 }^{ 10 }$

Standard form of $900000000 + 800000 + 50000 + 3000 + 20 + 3 = 90,08,53,023$
$\therefore$ Option C is correct.

Since $\displaystyle 1 \leq 1.26<10$  

Standard form of $900000000 + 3000+7=900003007$

We get the answer by multiplying and dividing the number by ${ 10 }^{ 9}=1000000000$

We get the answer by multiplying and dividing the number by 100000000 = ${ 10 }^{ 8 }$

Numeral for sixty million and sixty six is 60,000,066.

Commas are inserted in a number after each period.
Example 39,54, 452

International System: 9,849,475,825

In Indian and international system in the both cases ones period has 3 places
= ones, tens, hundreds

10 crores (or 1000 lakhs) is equal to 100 million (0.1 billion)

There are seven basic symbols,

$I, V, X, L, C, D$ and $M$.

Hence, this is the answer. 

We know that

1 is the factor of every number .

$10+{ 10 }^{ 3 }=10+1000=1010=1.01\times { 10 }^{ 3 }$

$1000000+200000+70000=1270000$

$4.5\times 10^5=450000$

$0.008712=8.712\times 10^{-3}$

We need to write it in a standard form:

$8.73\times 10^7=87300000$

$(200)^3=2^3\times 10^6$

$(400000)^4=4^4\times 10^{20}$

$923.4 = 9.234 \times 100$

$0.00036 = \dfrac{36}{10000} = 36\times10^{-5}$

$2.57 \times 10^{3} = 2.57 \times 1000 = 2570$

$3.506 \times 10^{-2} = \dfrac{3.506}{100} = 0.03506$

Take example $7.495$

Let the digits be $ x $ and $ y $
Given, "The digit in the ten's place of a two-digit number is three times that in the one's places "
$ => x = 3y $ 

$2.53\times { 10 }^{ -4 }=\frac { 2.53 }{ { 10 }^{ 4 } } =0.000253$

On adding, we get

$6.634\times 10^{-3}$ can be written as 

The value of $3200\times 1200$ is 

Option A: For rounding to the nearest hundred, If tens digit is 0,1,2,3,4

$670,000,000+700,000,000=1,370,000,000$

We have,

$ Given\quad that\quad x\quad and\quad y\quad are\quad positive\quad number\quad and\quad x>y.\ We\quad may\quad calculate\ (a)\quad on\quad the\quad positive\quad side\quad of\quad number\quad line\quad x\quad will\quad be\ \quad \quad to\quad the\quad right\quad of\quad y\quad and\quad \ (b)\quad on\quad the\quad negative\quad side\quad of\quad the\quad number\quad line\quad x\quad will\ \quad \quad be\quad to\quad the\quad left\quad of\quad y.\ Statement\quad A\longrightarrow -x>-y\quad i.e\quad -x\quad to\quad the\quad right\quad side\quad of\quad -y\ which\quad is\quad not\quad complying\quad with\quad (b).\quad So\quad it\quad is\quad not\quad true.\ Statement\quad B\longrightarrow -x<-y\quad i.e\quad -x\quad lies\quad to\quad the\quad left\quad of\quad -y.\ This\quad complies\quad with\quad (b).\quad So\quad the\quad statement\quad is\quad true.\ Statement\quad C\longrightarrow \frac { 1 }{ x } >\frac { 1 }{ y } .\quad Here\quad the\quad numerators\quad of\quad the\ two\quad fractions\quad are\quad equal.\quad But\quad the\quad denominator\quad of\quad \frac { 1 }{ x } \ is\quad greater\quad than\quad that\quad of\quad \frac { 1 }{ y } .\ So\quad this\quad statement\quad is\quad not\quad true.\ Statement\quad D\longrightarrow -\frac { 1 }{ x } >\frac { 1 }{ y } ,\ -\frac { 1 }{ x } is\quad a\quad negative\quad number\quad so\quad it\quad will\quad always\quad be\quad smaller\ than\quad a\quad positive\quad number.\ \therefore \quad -\frac { 1 }{ x } \ngtr \frac { 1 }{ y } .\quad So\quad this\quad statement\quad is\quad not\quad true.\ Answer-\quad Statement\quad B\quad is\quad correct.\  $

Every integer can be expressed as a rational number by dividing itself by 1.
Now, it becomes a rational number with the numerator the number itself and denominator 1.

$\begin{array}{l} x=0.566\ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _  \ 10x=5.666\ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ (i) \ 100x=56.6666\ _ \ _ \ _ \ _ \ _ \ _ (ii) \ (ii)-(i) \ 90x=51 \ x=\frac { { 51 } }{ { 90 } }  \end{array}$

Here,$ 2^{111}$ = $2^{3\times27}$
which is divisible by $2^{3} = 8$
Hence, the digit in the unit place is 0