Mix Examples - Number Systems Questions in English

(A) To find the value of $625^{\frac{3}{4}}$,we first express $625$ as a power of $5$.
Since $5 \times 5 \times 5 \times 5 = 625$,we can write $625 = 5^{4}$.
Now,substitute this into the expression: $625^{\frac{3}{4}} = (5^{4})^{\frac{3}{4}}$.
Using the power of a power rule $(a^{m})^{n} = a^{m \times n}$,we get $5^{4 \times \frac{3}{4}}$.
Simplifying the exponent,$4 \times \frac{3}{4} = 3$.
Therefore,the expression becomes $5^{3}$.
Calculating $5^{3} = 5 \times 5 \times 5 = 125$.

(B) To find the value of $64^{\frac{5}{6}}$,we first express $64$ as a power of $2$.
We know that $64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^{6}$.
Substituting this into the expression,we get:
$64^{\frac{5}{6}} = (2^{6})^{\frac{5}{6}}$.
Using the exponent rule $(a^{m})^{n} = a^{m \times n}$,we multiply the exponents:
$2^{6 \times \frac{5}{6}} = 2^{5}$.
Finally,calculating $2^{5}$:
$2^{5} = 2 \times 2 \times 2 \times 2 \times 2 = 32$.

(A) Given expression: $\left(\frac{125}{64}\right)^{-\frac{2}{3}}$
First,we use the property $(a/b)^{-n} = (b/a)^n$ to remove the negative exponent:
$\left(\frac{64}{125}\right)^{\frac{2}{3}}$
Next,express $64$ as $4^3$ and $125$ as $5^3$:
$\left(\frac{4^3}{5^3}\right)^{\frac{2}{3}} = \left(\left(\frac{4}{5}\right)^3\right)^{\frac{2}{3}}$
Using the power rule $(x^a)^b = x^{a \cdot b}$:
$\left(\frac{4}{5}\right)^{3 \cdot \frac{2}{3}} = \left(\frac{4}{5}\right)^2$
Finally,calculate the square:
$\frac{4^2}{5^2} = \frac{16}{25}$

(A) Given expression: $(343)^{-\frac{2}{3}}$
We know that $343 = 7^3$.
Substituting this into the expression: $(7^3)^{-\frac{2}{3}}$
Using the power rule $(a^m)^n = a^{m \times n}$,we get: $7^{3 \times (-\frac{2}{3})}$
$= 7^{-2}$
$= \frac{1}{7^2}$
$= \frac{1}{49}$

(A) Given expression: $(625)^{-\frac{3}{4}}$
We know that $625 = 5^4$.
Substituting this into the expression: $(5^4)^{-\frac{3}{4}}$
Using the exponent rule $(a^m)^n = a^{m \times n}$,we get: $5^{4 \times (-\frac{3}{4})}$
$= 5^{-3}$
$= \frac{1}{5^3}$
$= \frac{1}{125}$

(A) Given expression is $\sqrt[5]{(243)^{-3}}$.
We know that $243 = 3^5$.
Substituting this into the expression: $\sqrt[5]{(3^5)^{-3}}$.
Using the power rule $(a^m)^n = a^{m \times n}$,we get $(3^5)^{-3} = 3^{-15}$.
Now,the expression becomes $\sqrt[5]{3^{-15}}$.
Using the radical rule $\sqrt[n]{x} = x^{1/n}$,we get $(3^{-15})^{1/5} = 3^{-15 \times 1/5} = 3^{-3}$.
$3^{-3} = \frac{1}{3^3} = \frac{1}{27}$.

(B) Given expression: $\frac{(25)^{\frac{3}{2}} \times (243)^{\frac{3}{5}}}{(16)^{\frac{5}{4}} \times (8)^{\frac{4}{3}}}$
Step $1$: Express each base as a power of its prime factors.
$25 = 5^2$,$243 = 3^5$,$16 = 2^4$,$8 = 2^3$.
Step $2$: Substitute these into the expression:
$= \frac{(5^2)^{\frac{3}{2}} \times (3^5)^{\frac{3}{5}}}{(2^4)^{\frac{5}{4}} \times (2^3)^{\frac{4}{3}}}$
Step $3$: Apply the power rule $(a^m)^n = a^{m \times n}$:
$= \frac{5^{2 \times \frac{3}{2}} \times 3^{5 \times \frac{3}{5}}}{2^{4 \times \frac{5}{4}} \times 2^{3 \times \frac{4}{3}}}$
$= \frac{5^3 \times 3^3}{2^5 \times 2^4}$
Step $4$: Simplify the powers:
$= \frac{125 \times 27}{2^{5+4}}$
$= \frac{3375}{2^9}$
$= \frac{3375}{512}$

(D) Step $1$: Simplify each term individually.
$2^{-3} = \frac{1}{2^3} = \frac{1}{8} = 0.125$
Step $2$: Simplify $(0.01)^{-\frac{1}{2}}$.
$(0.01)^{-\frac{1}{2}} = (\frac{1}{100})^{-\frac{1}{2}} = (100)^{\frac{1}{2}} = \sqrt{100} = 10$
Step $3$: Simplify $(27)^{\frac{2}{3}}$.
$(27)^{\frac{2}{3}} = (3^3)^{\frac{2}{3}} = 3^{3 \times \frac{2}{3}} = 3^2 = 9$
Step $4$: Combine the results.
$0.125 + 10 - 9 = 0.125 + 1 = 1.125$
Thus,the simplified value is $1.125$.

(C) Given the equation: $125^{x} = \frac{25}{5^{x}}$
Express all terms with base $5$:
$(5^{3})^{x} = \frac{5^{2}}{5^{x}}$
Using the power of a power rule $(a^{m})^{n} = a^{mn}$:
$5^{3x} = 5^{2} \cdot 5^{-x}$
Using the product rule $a^{m} \cdot a^{n} = a^{m+n}$:
$5^{3x} = 5^{2-x}$
Since the bases are equal,equate the exponents:
$3x = 2 - x$
$3x + x = 2$
$4x = 2$
$x = \frac{2}{4} = \frac{1}{2}$

(A) Given that $a = \frac{\sqrt{5}}{8}$.
We are given the equation $\frac{8}{a} = b \sqrt{5}$.
Substitute the value of $a$ into the equation:
$\frac{8}{\frac{\sqrt{5}}{8}} = b \sqrt{5}$
$8 \times \frac{8}{\sqrt{5}} = b \sqrt{5}$
$\frac{64}{\sqrt{5}} = b \sqrt{5}$
To solve for $b$,multiply both sides by $\frac{1}{\sqrt{5}}$:
$b = \frac{64}{\sqrt{5} \times \sqrt{5}}$
$b = \frac{64}{5}$.

(C) Given equation: $2^{x-2} \cdot 3^{2x-6} = 36$
We know that $36 = 6^2 = (2 \cdot 3)^2 = 2^2 \cdot 3^2$.
So,the equation becomes: $2^{x-2} \cdot 3^{2x-6} = 2^2 \cdot 3^2$.
By comparing the powers of $2$ and $3$ on both sides:
For base $2$: $x - 2 = 2 \implies x = 4$.
For base $3$: $2x - 6 = 2 \implies 2x = 8 \implies x = 4$.
Since both conditions yield $x = 4$,the value of $x$ is $4$.

(D) Given equation: $(\frac{2}{5})^{5} \times (\frac{25}{4})^{3} = (\frac{5}{2})^{3x-2}$
First,express all bases in terms of $(\frac{5}{2})$:
$(\frac{2}{5})^{5} = (\frac{5}{2})^{-5}$
$(\frac{25}{4})^{3} = ((\frac{5}{2})^{2})^{3} = (\frac{5}{2})^{6}$
Now substitute these into the equation:
$(\frac{5}{2})^{-5} \times (\frac{5}{2})^{6} = (\frac{5}{2})^{3x-2}$
Using the law of exponents $a^m \times a^n = a^{m+n}$:
$(\frac{5}{2})^{-5+6} = (\frac{5}{2})^{3x-2}$
$(\frac{5}{2})^{1} = (\frac{5}{2})^{3x-2}$
Since the bases are equal,equate the exponents:
$1 = 3x - 2$
$3x = 3$
$x = 1$

(B) To compare the numbers $\sqrt{3}, \sqrt[3]{4}, \sqrt[4]{10}$,we express them as powers with a common index.
$1$. The indices are $2, 3, 4$. The least common multiple $(LCM)$ of $2, 3, 4$ is $12$.
$2$. Convert each number to the $12^{th}$ root:
$\sqrt{3} = 3^{1/2} = 3^{6/12} = \sqrt[12]{3^6} = \sqrt[12]{729}$
$\sqrt[3]{4} = 4^{1/3} = 4^{4/12} = \sqrt[12]{4^4} = \sqrt[12]{256}$
$\sqrt[4]{10} = 10^{1/4} = 10^{3/12} = \sqrt[12]{10^3} = \sqrt[12]{1000}$
$3$. Comparing the values inside the $12^{th}$ root: $256 < 729 < 1000$.
$4$. Therefore,the ascending order is $\sqrt[12]{256} < \sqrt[12]{729} < \sqrt[12]{1000}$,which corresponds to $\sqrt[3]{4} < \sqrt{3} < \sqrt[4]{10}$.

(N/A) Using the laws of exponents,specifically $\frac{x^m}{x^n} = x^{m-n}$ and $(x^m)^n = x^{m \times n}$:
Step $1$: Simplify each term.
$\left(\frac{x^a}{x^b}\right)^{a+b} = (x^{a-b})^{a+b} = x^{(a-b)(a+b)} = x^{a^2-b^2}$.
$\left(\frac{x^b}{x^c}\right)^{b+c} = (x^{b-c})^{b+c} = x^{(b-c)(b+c)} = x^{b^2-c^2}$.
$\left(\frac{x^c}{x^a}\right)^{c+a} = (x^{c-a})^{c+a} = x^{(c-a)(c+a)} = x^{c^2-a^2}$.
Step $2$: Multiply the simplified terms using $x^m \times x^n = x^{m+n}$.
$x^{a^2-b^2} \times x^{b^2-c^2} \times x^{c^2-a^2} = x^{(a^2-b^2) + (b^2-c^2) + (c^2-a^2)}$.
Step $3$: Simplify the exponent.
$a^2 - b^2 + b^2 - c^2 + c^2 - a^2 = 0$.
Therefore,$x^0 = 1$. Hence,the expression equals $1$.

(N/A) Step $1$: Calculate the sum inside the parentheses on the Left Hand Side $(LHS)$:
$1^{3} + 2^{3} + 3^{3} + 4^{3} + 5^{3} = 1 + 8 + 27 + 64 + 125 = 225$.
Step $2$: Evaluate the $LHS$:
$(225)^{\frac{1}{2}} = \sqrt{225} = 15$.
Step $3$: Calculate the sum inside the first parentheses on the Right Hand Side $(RHS)$:
$1^{3} + 2^{3} + 3^{3} + 4^{3} = 1 + 8 + 27 + 64 = 100$.
Step $4$: Evaluate the $RHS$:
$(100)^{\frac{1}{2}} + (5^{3})^{\frac{1}{3}} = \sqrt{100} + 5^{(3 \times \frac{1}{3})} = 10 + 5^{1} = 10 + 5 = 15$.
Step $5$: Since $LHS = 15$ and $RHS = 15$,the equation is proven.

(D) Given expression: $\left(\frac{1}{27}\right)^{-\frac{2}{3}}$
First,express $27$ as a power of $3$: $27 = 3^3$.
So,$\frac{1}{27} = \frac{1}{3^3} = 3^{-3}$.
Substitute this into the expression: $(3^{-3})^{-\frac{2}{3}}$.
Using the power rule $(a^m)^n = a^{m \times n}$,we get: $3^{-3 \times -\frac{2}{3}}$.
Simplify the exponent: $-3 \times -\frac{2}{3} = 2$.
Therefore,the expression becomes $3^2 = 9$.

(A) Given expression: $64^{-\frac{1}{3}}\left(64^{\frac{1}{3}}-64^{\frac{2}{3}}\right)$
We know that $64 = 4^3$.
Substituting this into the expression:
$= (4^3)^{-\frac{1}{3}} \left((4^3)^{\frac{1}{3}} - (4^3)^{\frac{2}{3}}\right)$
$= 4^{3 \times -\frac{1}{3}} \left(4^{3 \times \frac{1}{3}} - 4^{3 \times \frac{2}{3}}\right)$
$= 4^{-1} \left(4^1 - 4^2\right)$
$= \frac{1}{4} \left(4 - 16\right)$
$= \frac{1}{4} \left(-12\right)$
$= -3$

(B) To find the value of $\frac{8^{1/3} \times 16^{1/3}}{32^{-1/3}}$,we express each base as a power of $2$:
$8 = 2^3$,$16 = 2^4$,and $32 = 2^5$.
Substituting these into the expression:
$\frac{(2^3)^{1/3} \times (2^4)^{1/3}}{(2^5)^{-1/3}} = \frac{2^{3 \times 1/3} \times 2^{4 \times 1/3}}{2^{5 \times -1/3}} = \frac{2^1 \times 2^{4/3}}{2^{-5/3}}$.
Using the laws of exponents $a^m \times a^n = a^{m+n}$ and $\frac{a^m}{a^n} = a^{m-n}$:
$= 2^{1 + 4/3 - (-5/3)} = 2^{1 + 4/3 + 5/3} = 2^{1 + 9/3} = 2^{1 + 3} = 2^4$.
$2^4 = 16$.

(C) Given expression: $\frac{4}{(216)^{-\frac{2}{3}}} + \frac{1}{(256)^{-\frac{3}{4}}} + \frac{2}{(243)^{-\frac{1}{5}}}$
Step $1$: Simplify each term using the property $a^{-n} = \frac{1}{a^n}$ or $\frac{1}{a^{-n}} = a^n$.
Term $1$: $\frac{4}{(216)^{-\frac{2}{3}}} = 4 \times (216)^{\frac{2}{3}} = 4 \times ((6^3)^{\frac{2}{3}}) = 4 \times 6^2 = 4 \times 36 = 144$.
Term $2$: $\frac{1}{(256)^{-\frac{3}{4}}} = 1 \times (256)^{\frac{3}{4}} = (4^4)^{\frac{3}{4}} = 4^3 = 64$.
Term $3$: $\frac{2}{(243)^{-\frac{1}{5}}} = 2 \times (243)^{\frac{1}{5}} = 2 \times ((3^5)^{\frac{1}{5}}) = 2 \times 3^1 = 6$.
Step $2$: Add the results together:
$144 + 64 + 6 = 214$.
Thus,the correct option is $C$.

(FALSE) The statement is False.
Whole numbers are the set of numbers ${0, 1, 2, 3, ...}$.
Integers are the set of numbers ${..., -3, -2, -1, 0, 1, 2, 3, ...}$.
Since negative numbers (e.g.,$-1, -2, -3$) are integers but are not whole numbers,the statement that 'each integer is a whole number' is incorrect.

(B) The statement is False.
Calculation: $(-5)^{2} = (-5) \times (-5) = 25$.
Since $25 \neq -25$,the given statement is false.

(A) The statement is True.
When a negative number is raised to an odd power,the result is negative.
Since $11$ is an odd number,$(-1)^{11} = -1 \times -1 \times ... \times -1$ ($11$ times) $= -1$.

(B) The statement is False.
According to the laws of radicals,for any positive real numbers $a$ and $b$,$\sqrt{a} \times \sqrt{b} = \sqrt{a \times b}$.
Applying this rule: $\sqrt{3} \times \sqrt{5} = \sqrt{3 \times 5} = \sqrt{15}$.
Since $\sqrt{15} \neq \sqrt{8}$,the given statement is false.

(B) The statement is False.
Explanation:
We know that $\sqrt{49} = 7$.
Since $7$ can be expressed in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$ (e.g.,$\frac{7}{1}$),it is a rational number.
Therefore,the statement that $\sqrt{49}$ is an irrational number is false.

(A) Let $x = 0.\overline{3} = 0.333\dots$ (Equation $1$).
Multiply both sides by $10$:
$10x = 3.333\dots$ (Equation $2$).
Subtract Equation $1$ from Equation $2$:
$10x - x = 3.333\dots - 0.333\dots$
$9x = 3$
$x = \frac{3}{9} = \frac{1}{3}$.
Therefore,$0.\overline{3} = \frac{1}{3}$.

(B) Let $x = 4.185185185...$ (Equation $1$)
Since there are $3$ repeating digits,multiply by $1000$:
$1000x = 4185.185185...$ (Equation $2$)
Subtract Equation $1$ from Equation $2$:
$1000x - x = 4185.185185... - 4.185185...$
$999x = 4181$
$x = \frac{4181}{999}$
Now,convert to a mixed fraction:
$x = 4 + \frac{181}{999}$
Wait,checking the repeating decimal $4.185185...$ as a fraction:
$x = 4 + \frac{185}{999}$
Simplify $\frac{185}{999}$ by dividing by $37$:
$185 \div 37 = 5$
$999 \div 37 = 27$
So,$x = 4 \frac{5}{27}$.

(C) To solve the expression $(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})$,we use the algebraic identity $(a+b)(a-b) = a^2 - b^2$.
Here,$a = \sqrt{5}$ and $b = \sqrt{3}$.
Substituting these values into the identity:
$(\sqrt{5})^2 - (\sqrt{3})^2 = 5 - 3 = 2$.
Therefore,the correct option is $C$.

(D) To determine the nature of the number $\frac{\sqrt{50}}{\sqrt{98}}$,we simplify the expression:
$\frac{\sqrt{50}}{\sqrt{98}} = \sqrt{\frac{50}{98}}$
Divide both the numerator and the denominator by $2$:
$\sqrt{\frac{50 \div 2}{98 \div 2}} = \sqrt{\frac{25}{49}}$
Since $\sqrt{25} = 5$ and $\sqrt{49} = 7$,the expression simplifies to $\frac{5}{7}$.
$A$ number that can be expressed in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$,is a rational number.
Since $\frac{5}{7}$ fits this definition,the number is rational.

(A) Using the law of exponents,$(a^{m})^{n} = a^{m \times n}$.
Applying this to the given expression: $(5^{-2})^{3} = 5^{-2 \times 3} = 5^{-6}$.
Since $a^{-n} = \frac{1}{a^{n}}$,we have $5^{-6} = \frac{1}{5^{6}}$.
Therefore,the correct option is $A$.

(B) To find the rationalising factor of an expression of the form $a - \sqrt{b}$,we multiply it by its conjugate $a + \sqrt{b}$ to eliminate the square root.
Here,the given expression is $4 - \sqrt{5}$.
Multiplying $4 - \sqrt{5}$ by $4 + \sqrt{5}$ gives $(4)^2 - (\sqrt{5})^2 = 16 - 5 = 11$,which is a rational number.
Therefore,the rationalising factor of $4 - \sqrt{5}$ is $4 + \sqrt{5}$.

(C) Let $x = 0.3\overline{6} = 0.3666...$ (Equation $1$)
Multiply both sides by $10$: $10x = 3.666...$ (Equation $2$)
Multiply Equation $1$ by $100$: $100x = 36.666...$ (Equation $3$)
Subtract Equation $2$ from Equation $3$:
$100x - 10x = 36.666... - 3.666...$
$90x = 33$
$x = \frac{33}{90}$
Simplify the fraction by dividing both numerator and denominator by $3$:
$x = \frac{11}{30}$

(D) To determine the nature of the number $\sqrt{5^{2}+12^{2}}$,we first calculate the value inside the square root.
Step $1$: Calculate the squares: $5^{2} = 25$ and $12^{2} = 144$.
Step $2$: Add the values: $25 + 144 = 169$.
Step $3$: Find the square root: $\sqrt{169} = 13$.
Since $13$ is a positive integer,it belongs to the set of natural numbers. Therefore,the correct option is $D$.

(A) Given expression: $\sqrt{5} + \sqrt{5} = 2\sqrt{5}$.
Since $\sqrt{5}$ is an irrational number and the product of a non-zero rational number $(2)$ and an irrational number $(\sqrt{5})$ is always an irrational number,therefore,$2\sqrt{5}$ is an irrational number.

(B) To determine the nature of the number $(\sqrt{5}+3)^{2}$,we expand it using the algebraic identity $(a+b)^{2} = a^{2} + 2ab + b^{2}$.
Here,$a = \sqrt{5}$ and $b = 3$.
$(\sqrt{5}+3)^{2} = (\sqrt{5})^{2} + 2(\sqrt{5})(3) + (3)^{2}$
$= 5 + 6\sqrt{5} + 9$
$= 14 + 6\sqrt{5}$.
Since $\sqrt{5}$ is an irrational number,$6\sqrt{5}$ is also irrational. The sum of a rational number $(14)$ and an irrational number $(6\sqrt{5})$ is always an irrational number.
Therefore,$(\sqrt{5}+3)^{2}$ is an irrational number.

(C) The given expression is $\sqrt{5} \times \sqrt{5}$.
By the property of square roots,$\sqrt{a} \times \sqrt{a} = a$ for any $a \ge 0$.
Therefore,$\sqrt{5} \times \sqrt{5} = 5$.
The number $5$ is a positive integer,which belongs to the set of natural numbers $\{1, 2, 3, \dots\}$.
Thus,$\sqrt{5} \times \sqrt{5}$ is a natural number.

(D) To find a rational number between $7$ and $8$,we evaluate each option:
$A) \frac{33}{5} = 6.6$ (This is less than $7$)
$B) \frac{51}{6} = 8.5$ (This is greater than $8$)
$C) \frac{60}{7} \approx 8.57$ (This is greater than $8$)
$D) \frac{61}{8} = 7.625$ (Since $7 < 7.625 < 8$,this is the correct rational number between $7$ and $8$).

(A) Given the expression: $(\sqrt{5}+3)^{2} = a+b\sqrt{5}$.
Using the algebraic identity $(x+y)^{2} = x^{2} + 2xy + y^{2}$,where $x = \sqrt{5}$ and $y = 3$:
$(\sqrt{5})^{2} + 2(\sqrt{5})(3) + (3)^{2} = a+b\sqrt{5}$.
$5 + 6\sqrt{5} + 9 = a+b\sqrt{5}$.
$14 + 6\sqrt{5} = a+b\sqrt{5}$.
By comparing the rational and irrational parts on both sides,we get $a = 14$ and $b = 6$.

(B) To determine the decimal expansion of the rational number $-\frac{11}{4}$,we perform the division:
$11 \div 4 = 2.75$.
Since the division results in a finite number of decimal places,the decimal expansion is terminating.
Therefore,$-\frac{11}{4} = -2.75$,which is a terminating decimal.

(C) Using the law of exponents $(a^m)^n = a^{m \times n}$,we have:
$(5^{\frac{3}{4}})^{\frac{4}{3}} = 5^{\frac{3}{4} \times \frac{4}{3}}$
$= 5^{\frac{12}{12}}$
$= 5^1$
$= 5$

(IRRATIONAL) number is called irrational if it cannot be written in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$. Since $7$ is not a perfect square,its square root,$\sqrt{7}$,is an irrational number.

(A) The square root of a number $x$ is a value $y$ such that $y^2 = x$.
Since $7^2 = 7 \times 7 = 49$,the principal square root of $49$ is $7$.
Therefore,$\sqrt{49} = 7$.

(N/A) The square root of a number $x$ is a value that,when multiplied by itself,gives $x$.
Since $11 \times 11 = 121$ and $(-11) \times (-11) = 121$,the square root of $121$ is $11$ or $-11$.

(1 1/12) First,convert the mixed fraction into an improper fraction:
$1 \frac{25}{144} = \frac{1 \times 144 + 25}{144} = \frac{144 + 25}{144} = \frac{169}{144}$
Now,take the square root of the improper fraction:
$\sqrt{\frac{169}{144}} = \frac{\sqrt{169}}{\sqrt{144}} = \frac{13}{12}$
Finally,convert the improper fraction back into a mixed fraction:
$\frac{13}{12} = 1 \frac{1}{12}$

(1/2) To solve $(64)^{-\frac{1}{6}}$,we first express $64$ as a power of $2$.
We know that $64 = 2^6$.
Substituting this into the expression,we get:
$(2^6)^{-\frac{1}{6}}$
Using the exponent rule $(a^m)^n = a^{m \times n}$,we multiply the exponents:
$2^{6 \times (-\frac{1}{6})} = 2^{-1}$
Since $a^{-n} = \frac{1}{a^n}$,we have:
$2^{-1} = \frac{1}{2^1} = \frac{1}{2}$

(B) To find the decimal expansion of $\frac{2}{3}$, we perform long division by dividing $2$ by $3$.
When we divide $2$ by $3$, we get $0.666...$ which can be written as $0.\overline{6}$.
Since the remainder never becomes $0$ and the digit $6$ repeats infinitely, the decimal expansion is non-terminating and recurring.

(B) The $n$-th root of $1$ is defined as the number $x$ such that $x^n = 1$.
For any positive integer $n$,$1^n = 1 \times 1 \times \dots \times 1 = 1$.
Therefore,$\sqrt[11]{1} = 1$.

(C) To solve $(729)^{\frac{1}{3}}$,we find the prime factorization of $729$.
$729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^6$.
Alternatively,$729 = 9 \times 9 \times 9 = 9^3$.
Therefore,$(729)^{\frac{1}{3}} = (9^3)^{\frac{1}{3}}$.
Using the exponent rule $(a^m)^n = a^{m \times n}$,we get $9^{3 \times \frac{1}{3}} = 9^1 = 9$.

(TERMINATING) To determine the type of decimal expansion of $\frac{47}{50}$,we first simplify the fraction or examine the denominator.
Given fraction: $\frac{47}{50}$.
The denominator is $50 = 2^1 \times 5^2$.
Since the denominator is of the form $2^n \times 5^m$,where $n$ and $m$ are non-negative integers,the decimal expansion is terminating.
Calculation: $\frac{47}{50} = \frac{47 \times 2}{50 \times 2} = \frac{94}{100} = 0.94$.
Therefore,the decimal expansion is terminating.

(A) To solve the expression $\sqrt{2} \cdot \sqrt{3} \cdot \sqrt{6}$,we use the property of radicals $\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}$.
First,multiply the terms inside the square roots: $\sqrt{2} \cdot \sqrt{3} = \sqrt{2 \cdot 3} = \sqrt{6}$.
Now,substitute this back into the original expression: $\sqrt{6} \cdot \sqrt{6}$.
Using the property $\sqrt{a} \cdot \sqrt{a} = a$,we get $\sqrt{6} \cdot \sqrt{6} = 6$.
Therefore,the correct answer is $6$.

(B) The number $\pi$ is defined as the ratio of the circumference of a circle to its diameter. Its decimal expansion is non-terminating and non-recurring. Therefore,$\pi$ is an irrational number.