Mix Examples - Number Systems Questions in English

(N/A) To add $0. \overline{3}$ and $0.4 \overline{7}$,first convert them into fractions.
$0. \overline{3} = \frac{3}{9} = \frac{1}{3}$
$0.4 \overline{7} = \frac{47 - 4}{90} = \frac{43}{90}$
Now,add the two fractions:
$\frac{1}{3} + \frac{43}{90} = \frac{30}{90} + \frac{43}{90} = \frac{73}{90}$
Converting $\frac{73}{90}$ back to decimal form:
$73 \div 90 = 0.8111... = 0.8 \overline{1}$

(N/A) Step $1$: Convert $0.\overline{52}$ to a fraction.
Let $x = 0.525252...$ $(i)$
$100x = 52.525252...$ (ii)
Subtracting $(i)$ from (ii): $99x = 52$,so $x = \frac{52}{99}$.
Step $2$: Convert $0.4\overline{6}$ to a fraction.
Let $y = 0.4666...$ (iii)
$10y = 4.666...$ (iv)
$100y = 46.666...$ $(v)$
Subtracting (iv) from $(v)$: $90y = 42$,so $y = \frac{42}{90} = \frac{7}{15}$.
Step $3$: Subtract the fractions.
$\frac{52}{99} - \frac{7}{15} = \frac{52 \times 5 - 7 \times 33}{495} = \frac{260 - 231}{495} = \frac{29}{495}$.
Step $4$: Convert to decimal.
$\frac{29}{495} = 0.0585858... = 0.0\overline{58}$.

(N/A) Let $x = 0.1\overline{6}$. This can be written as $x = 0.1666...$ (Equation $1$).
Multiply both sides by $10$ to shift the decimal point:
$10x = 1.6666...$ (Equation $2$).
Subtract Equation $1$ from Equation $2$:
$10x - x = (1.6666...) - (0.1666...)$
$9x = 1.5$
$x = \frac{1.5}{9}$
To remove the decimal,multiply the numerator and denominator by $10$:
$x = \frac{15}{90}$
Simplify the fraction by dividing both by $15$:
$x = \frac{1}{6}$.
Thus,$0.1\overline{6} = \frac{1}{6}$.

(N/A) Let $x = 0.142857142857...$ (Equation $1$).
Since there are $6$ repeating digits,multiply both sides of Equation $1$ by $10^6 = 1,000,000$:
$1,000,000x = 142857.142857142857...$ (Equation $2$).
Subtract Equation $1$ from Equation $2$:
$1,000,000x - x = 142857.142857... - 0.142857...$
$999,999x = 142857$.
$x = \frac{142857}{999999}$.
Dividing both numerator and denominator by $142857$,we get:
$x = \frac{1}{7}$.
Thus,$0.\overline{142857} = \frac{1}{7}$.

(N/A) Let $x = 0.\overline{076923}$.
This can be written as $x = 0.076923076923...$ (Equation $1$).
Since there are $6$ repeating digits, multiply both sides by $10^6 = 1,000,000$:
$1,000,000x = 76923.076923076923...$ (Equation $2$).
Subtract Equation $1$ from Equation $2$:
$1,000,000x - x = 76923.076923... - 0.076923...$
$999,999x = 76923$.
$x = \frac{76923}{999999}$.
Dividing both numerator and denominator by $76923$:
$x = \frac{76923 \div 76923}{999999 \div 76923} = \frac{1}{13}$.
Thus, $0.\overline{076923} = \frac{1}{13}$.

(N/A) To visualize $2.64444$ on the number line,we use the process of successive magnification:
$1$. Since $2.64444$ lies between $2$ and $3$,we divide the distance between $2$ and $3$ into $10$ equal parts and locate $2.6$ and $2.7$.
$2$. Now,$2.64444$ lies between $2.6$ and $2.7$. We magnify this region and divide it into $10$ equal parts to locate $2.64$ and $2.65$.
$3$. Next,$2.64444$ lies between $2.64$ and $2.65$. We magnify this region and divide it into $10$ equal parts to locate $2.644$ and $2.645$.
$4$. Continuing this process,$2.64444$ lies between $2.644$ and $2.645$. We magnify this region to locate $2.6444$ and $2.6445$.
$5$. Finally,we magnify the region between $2.6444$ and $2.6445$ to locate the point $2.64444$ on the number line.

(N/A) Step $1$: Locate $2.3$ and $2.4$ on the number line. Divide the distance between $2.3$ and $2.4$ into $10$ equal parts. The first mark represents $2.31$,the second $2.32$,and so on.
Step $2$: Locate $2.36$ and $2.37$ on the number line. Divide the distance between $2.36$ and $2.37$ into $10$ equal parts.
Step $3$: The $5$th mark between $2.36$ and $2.37$ represents $2.365$.

(N/A) To visualise $-4.126$ on the number line using successive magnification,we follow these steps:
$1$. Since $-4.126$ lies between $-4$ and $-5$,we divide the segment between $-4$ and $-5$ into $10$ equal parts and locate $-4.1$ and $-4.2$.
$2$. Now,$-4.126$ lies between $-4.1$ and $-4.2$. We magnify this segment and divide it into $10$ equal parts to locate $-4.12$ and $-4.13$.
$3$. Finally,$-4.126$ lies between $-4.12$ and $-4.13$. We magnify this segment and divide it into $10$ equal parts to locate the point representing $-4.126$ exactly.

To represent $3.\overline{42}$ up to $4$ decimal places, we need to locate $3.4242$ on the number line.
Step $1$: $3.4242$ lies between $3$ and $4$. Divide the interval $[3, 4]$ into $10$ equal parts and magnify the interval $[3.4, 3.5]$.
Step $2$: $3.4242$ lies between $3.4$ and $3.5$. Divide the interval $[3.4, 3.5]$ into $10$ equal parts and magnify the interval $[3.42, 3.43]$.
Step $3$: $3.4242$ lies between $3.42$ and $3.43$. Divide the interval $[3.42, 3.43]$ into $10$ equal parts and magnify the interval $[3.424, 3.425]$.
Step $4$: $3.4242$ lies between $3.424$ and $3.425$. Divide the interval $[3.424, 3.425]$ into $10$ equal parts. The point $3.4242$ is the $2$nd mark after $3.424$.

(D) To represent $5.4949$ on the number line using successive magnification:
$1$. Locate $5.4949$ between $5$ and $6$. Divide the interval into $10$ equal parts to find $5.4$ and $5.5$.
$2$. Magnify the interval between $5.4$ and $5.5$. Divide it into $10$ equal parts to locate $5.49$ and $5.50$.
$3$. Magnify the interval between $5.49$ and $5.50$. Divide it into $10$ equal parts to locate $5.494$ and $5.495$.
$4$. Magnify the interval between $5.494$ and $5.495$. Divide it into $10$ equal parts. The point $5.4949$ lies between $5.494$ and $5.495$,specifically at the $9^{th}$ subdivision.

(A) To add the given expressions,we group the like terms involving $\sqrt{3}$ and $\sqrt{5}$ together.
$(4 \sqrt{3}+2 \sqrt{5})+(6 \sqrt{3}-4 \sqrt{5})$
$= (4 \sqrt{3}+6 \sqrt{3}) + (2 \sqrt{5}-4 \sqrt{5})$
$= (4+6) \sqrt{3} + (2-4) \sqrt{5}$
$= 10 \sqrt{3} - 2 \sqrt{5}$

(B) To multiply $3 \sqrt{7}$ and $5 \sqrt{7}$,we follow these steps:
Step $1$: Multiply the coefficients (the numbers outside the square roots) together: $3 \times 5 = 15$.
Step $2$: Multiply the square root terms together: $\sqrt{7} \times \sqrt{7} = 7$.
Step $3$: Multiply the results from Step $1$ and Step $2$: $15 \times 7 = 105$.
Therefore,$3 \sqrt{7} \times 5 \sqrt{7} = 105$.

(A) To divide $12 \sqrt{30}$ by $3 \sqrt{5}$,we write it as a fraction:
$\frac{12 \sqrt{30}}{3 \sqrt{5}}$
We can simplify the coefficients: $\frac{12}{3} = 4$.
Next,we simplify the square roots using the property $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$:
$\sqrt{\frac{30}{5}} = \sqrt{6}$
Combining these,we get $4 \times \sqrt{6} = 4 \sqrt{6}$.

(A) To simplify the expression $(3+\sqrt{5})(4-\sqrt{11})$,we use the distributive property of multiplication over addition ($FOIL$ method):
$(3+\sqrt{5})(4-\sqrt{11}) = 3(4-\sqrt{11}) + \sqrt{5}(4-\sqrt{11})$
$= (3 \times 4) - (3 \times \sqrt{11}) + (\sqrt{5} \times 4) - (\sqrt{5} \times \sqrt{11})$
$= 12 - 3\sqrt{11} + 4\sqrt{5} - \sqrt{5 \times 11}$
$= 12 - 3\sqrt{11} + 4\sqrt{5} - \sqrt{55}$

(A) To simplify the expression $(\sqrt{15}+\sqrt{7})(\sqrt{15}-\sqrt{7})$,we use the algebraic identity $(a+b)(a-b) = a^2 - b^2$.
Here,$a = \sqrt{15}$ and $b = \sqrt{7}$.
Applying the identity:
$(\sqrt{15})^2 - (\sqrt{7})^2 = 15 - 7 = 8$.

(A) To simplify the expression $(\sqrt{11}-\sqrt{3})^{2}$,we use the algebraic identity $(a-b)^{2} = a^{2} - 2ab + b^{2}$.
Here,$a = \sqrt{11}$ and $b = \sqrt{3}$.
Substituting these values into the identity:
$(\sqrt{11}-\sqrt{3})^{2} = (\sqrt{11})^{2} - 2(\sqrt{11})(\sqrt{3}) + (\sqrt{3})^{2}$
$= 11 - 2\sqrt{11 \times 3} + 3$
$= 11 - 2\sqrt{33} + 3$
$= 14 - 2\sqrt{33}$

(N/A) To rationalise the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $(5 \sqrt{3} + 3 \sqrt{5})$.
$\frac{30}{5 \sqrt{3}-3 \sqrt{5}} = \frac{30}{5 \sqrt{3}-3 \sqrt{5}} \times \frac{5 \sqrt{3}+3 \sqrt{5}}{5 \sqrt{3}+3 \sqrt{5}}$
$= \frac{30(5 \sqrt{3}+3 \sqrt{5})}{(5 \sqrt{3})^{2}-(3 \sqrt{5})^{2}}$
$= \frac{30(5 \sqrt{3}+3 \sqrt{5})}{75-45}$
$= \frac{30(5 \sqrt{3}+3 \sqrt{5})}{30}$
$= 5 \sqrt{3}+3 \sqrt{5}$

(A) To rationalise the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $(5-2 \sqrt{3})$.
$\frac{1}{5+2 \sqrt{3}} = \frac{1}{5+2 \sqrt{3}} \times \frac{5-2 \sqrt{3}}{5-2 \sqrt{3}}$
Using the identity $(a+b)(a-b) = a^2 - b^2$ in the denominator:
$= \frac{5-2 \sqrt{3}}{(5)^2 - (2 \sqrt{3})^2}$
$= \frac{5-2 \sqrt{3}}{25 - (4 \times 3)}$
$= \frac{5-2 \sqrt{3}}{25 - 12}$
$= \frac{5-2 \sqrt{3}}{13}$

To rationalize the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $\sqrt{m^{2}+n^{2}}-m$.
$\frac{n^{2}}{\sqrt{m^{2}+n^{2}}+m} = \frac{n^{2}}{\sqrt{m^{2}+n^{2}}+m} \times \frac{\sqrt{m^{2}+n^{2}}-m}{\sqrt{m^{2}+n^{2}}-m}$
Using the identity $(a+b)(a-b) = a^{2}-b^{2}$ in the denominator:
$= \frac{n^{2}(\sqrt{m^{2}+n^{2}}-m)}{(\sqrt{m^{2}+n^{2}})^{2} - (m)^{2}}$
$= \frac{n^{2}(\sqrt{m^{2}+n^{2}}-m)}{m^{2}+n^{2}-m^{2}}$
$= \frac{n^{2}(\sqrt{m^{2}+n^{2}}-m)}{n^{2}}$
$= \sqrt{m^{2}+n^{2}}-m$

(A) Given: $\frac{2 \sqrt{6}-\sqrt{5}}{3 \sqrt{5}-2 \sqrt{6}}=a+b \sqrt{30}$
To rationalize the denominator,multiply the numerator and denominator by $(3 \sqrt{5}+2 \sqrt{6})$:
$\frac{2 \sqrt{6}-\sqrt{5}}{3 \sqrt{5}-2 \sqrt{6}} \times \frac{3 \sqrt{5}+2 \sqrt{6}}{3 \sqrt{5}+2 \sqrt{6}}=a+b \sqrt{30}$
Using the identity $(x-y)(x+y) = x^2 - y^2$ in the denominator:
$\frac{(2 \sqrt{6}-\sqrt{5})(3 \sqrt{5}+2 \sqrt{6})}{(3 \sqrt{5})^{2}-(2 \sqrt{6})^{2}}=a+b \sqrt{30}$
Expanding the numerator:
$(2 \sqrt{6})(3 \sqrt{5}) + (2 \sqrt{6})(2 \sqrt{6}) - (\sqrt{5})(3 \sqrt{5}) - (\sqrt{5})(2 \sqrt{6}) = 6 \sqrt{30} + 24 - 15 - 2 \sqrt{30} = 9 + 4 \sqrt{30}$
Calculating the denominator:
$(3 \sqrt{5})^2 - (2 \sqrt{6})^2 = 45 - 24 = 21$
So,$\frac{9+4 \sqrt{30}}{21} = a+b \sqrt{30}$
$\frac{9}{21} + \frac{4 \sqrt{30}}{21} = a+b \sqrt{30}$
$\frac{3}{7} + \frac{4}{21} \sqrt{30} = a+b \sqrt{30}$
Comparing the rational and irrational parts,we get $a=\frac{3}{7}$ and $b=\frac{4}{21}$.

(N/A) Given: $x = 5 + 2\sqrt{6}$.
First,find $\frac{1}{x}$:
$\frac{1}{x} = \frac{1}{5 + 2\sqrt{6}} = \frac{1}{5 + 2\sqrt{6}} \times \frac{5 - 2\sqrt{6}}{5 - 2\sqrt{6}} = \frac{5 - 2\sqrt{6}}{25 - 24} = 5 - 2\sqrt{6}$.
Now,calculate $x + \frac{1}{x}$:
$x + \frac{1}{x} = (5 + 2\sqrt{6}) + (5 - 2\sqrt{6}) = 10$.
To find $x^{2} + \frac{1}{x^{2}}$:
$x^{2} + \frac{1}{x^{2}} = (x + \frac{1}{x})^{2} - 2 = (10)^{2} - 2 = 100 - 2 = 98$.
To find $x^{3} + \frac{1}{x^{3}}$:
$x^{3} + \frac{1}{x^{3}} = (x + \frac{1}{x})^{3} - 3(x + \frac{1}{x}) = (10)^{3} - 3(10) = 1000 - 30 = 970$.

(N/A) To add the given expressions,we group the like terms:
$(4 \sqrt{6}-8 \sqrt{10}) + (3 \sqrt{6}+12 \sqrt{10})$
$= (4 \sqrt{6} + 3 \sqrt{6}) + (-8 \sqrt{10} + 12 \sqrt{10})$
$= (4 + 3) \sqrt{6} + (-8 + 12) \sqrt{10}$
$= 7 \sqrt{6} + 4 \sqrt{10}$

(A) To multiply $5 \sqrt{3}$ and $4 \sqrt{12}$,we follow these steps:
$1$. Write the expression as $(5 \sqrt{3}) \times (4 \sqrt{12})$.
$2$. Multiply the coefficients: $5 \times 4 = 20$.
$3$. Multiply the radicals: $\sqrt{3} \times \sqrt{12} = \sqrt{3 \times 12} = \sqrt{36}$.
$4$. Calculate the square root: $\sqrt{36} = 6$.
$5$. Multiply the results: $20 \times 6 = 120$.
Therefore,the product is $120$.

(B) To simplify the expression $5 \sqrt{2} + 2 \sqrt{8} - 3 \sqrt{32} + 4 \sqrt{128}$, we first simplify each radical term:
$1$. $5 \sqrt{2} = 5 \sqrt{2}$
$2$. $2 \sqrt{8} = 2 \sqrt{4 \times 2} = 2 \times 2 \sqrt{2} = 4 \sqrt{2}$
$3$. $3 \sqrt{32} = 3 \sqrt{16 \times 2} = 3 \times 4 \sqrt{2} = 12 \sqrt{2}$
$4$. $4 \sqrt{128} = 4 \sqrt{64 \times 2} = 4 \times 8 \sqrt{2} = 32 \sqrt{2}$
Now, substitute these values back into the expression:
$5 \sqrt{2} + 4 \sqrt{2} - 12 \sqrt{2} + 32 \sqrt{2}$
Combine the coefficients:
$(5 + 4 - 12 + 32) \sqrt{2} = 29 \sqrt{2}$

(A) To rationalise the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $(\sqrt{5} + \sqrt{3})$.
$\frac{1}{\sqrt{5}-\sqrt{3}} \times \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}$
$= \frac{\sqrt{5}+\sqrt{3}}{(\sqrt{5})^2 - (\sqrt{3})^2}$
$= \frac{\sqrt{5}+\sqrt{3}}{5 - 3}$
$= \frac{\sqrt{5}+\sqrt{3}}{2}$

(A) To rationalise the denominator of $\frac{4}{\sqrt{10}+\sqrt{6}}$,multiply the numerator and the denominator by the conjugate of the denominator,which is $(\sqrt{10}-\sqrt{6})$.
$\frac{4}{\sqrt{10}+\sqrt{6}} \times \frac{\sqrt{10}-\sqrt{6}}{\sqrt{10}-\sqrt{6}}$
$= \frac{4(\sqrt{10}-\sqrt{6})}{(\sqrt{10})^2 - (\sqrt{6})^2}$
$= \frac{4(\sqrt{10}-\sqrt{6})}{10-6}$
$= \frac{4(\sqrt{10}-\sqrt{6})}{4}$
$= \sqrt{10}-\sqrt{6}$

(N/A) To rationalise the denominator,we multiply the numerator and the denominator by the conjugate of the denominator,which is $(3 \sqrt{2} + 2 \sqrt{3})$.
$\frac{18}{3 \sqrt{2}-2 \sqrt{3}} \times \frac{3 \sqrt{2}+2 \sqrt{3}}{3 \sqrt{2}+2 \sqrt{3}}$
$= \frac{18(3 \sqrt{2}+2 \sqrt{3})}{(3 \sqrt{2})^2 - (2 \sqrt{3})^2}$
$= \frac{18(3 \sqrt{2}+2 \sqrt{3})}{(9 \times 2) - (4 \times 3)}$
$= \frac{18(3 \sqrt{2}+2 \sqrt{3})}{18 - 12}$
$= \frac{18(3 \sqrt{2}+2 \sqrt{3})}{6}$
$= 3(3 \sqrt{2}+2 \sqrt{3})$

(A) To rationalise the denominator of $\frac{1}{7-4 \sqrt{3}}$,we multiply the numerator and the denominator by the conjugate of the denominator,which is $7+4 \sqrt{3}$.
$\frac{1}{7-4 \sqrt{3}} \times \frac{7+4 \sqrt{3}}{7+4 \sqrt{3}} = \frac{7+4 \sqrt{3}}{(7)^2 - (4 \sqrt{3})^2}$
$= \frac{7+4 \sqrt{3}}{49 - (16 \times 3)}$
$= \frac{7+4 \sqrt{3}}{49 - 48}$
$= \frac{7+4 \sqrt{3}}{1}$
$= 7+4 \sqrt{3}$

$(17+12 \sqrt{2})$ To rationalise the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $(3+2 \sqrt{2})$.
$\frac{3+2 \sqrt{2}}{3-2 \sqrt{2}} \times \frac{3+2 \sqrt{2}}{3+2 \sqrt{2}}$
$= \frac{(3+2 \sqrt{2})^2}{(3)^2 - (2 \sqrt{2})^2}$
$= \frac{3^2 + (2 \sqrt{2})^2 + 2(3)(2 \sqrt{2})}{9 - (4 \times 2)}$
$= \frac{9 + 8 + 12 \sqrt{2}}{9 - 8}$
$= \frac{17 + 12 \sqrt{2}}{1}$
$= 17 + 12 \sqrt{2}$

(A) To rationalise the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $(5-2 \sqrt{6})$.
$\frac{5-2 \sqrt{6}}{5+2 \sqrt{6}} \times \frac{5-2 \sqrt{6}}{5-2 \sqrt{6}}$
$= \frac{(5-2 \sqrt{6})^2}{(5)^2 - (2 \sqrt{6})^2}$
$= \frac{5^2 - 2(5)(2 \sqrt{6}) + (2 \sqrt{6})^2}{25 - (4 \times 6)}$
$= \frac{25 - 20 \sqrt{6} + 24}{25 - 24}$
$= \frac{49 - 20 \sqrt{6}}{1}$
$= 49 - 20 \sqrt{6}$

(A) Given expression is $\frac{\sqrt{5}}{\sqrt{10}}$.
We can simplify this as $\sqrt{\frac{5}{10}} = \sqrt{\frac{1}{2}}$.
This is equal to $\frac{1}{\sqrt{2}}$.
Rationalizing the denominator,we get $\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$.
Substituting the value of $\sqrt{2} = 1.4142$,we get $\frac{1.4142}{2} = 0.7071$.

(B) Given that $\sqrt{5} = 2.236$.
We need to evaluate $\frac{4 - \sqrt{5}}{\sqrt{5}}$.
This expression can be simplified as $\frac{4}{\sqrt{5}} - 1$.
Substituting the value of $\sqrt{5}$: $\frac{4}{2.236} - 1$.
Calculating $\frac{4}{2.236} \approx 1.788895$.
Subtracting $1$ from this value gives $1.788895 - 1 = 0.788895$.
Rounding to four decimal places,we get $0.7889$.
However,checking the calculation $\frac{4 - 2.236}{2.236} = \frac{1.764}{2.236} \approx 0.788886$.
Rounding to four decimal places,we get $0.7889$.
Given the options provided,$0.7888$ is the closest match.

(N/A) Given $x = 3 + 2\sqrt{2}$.
First,find $\frac{1}{x}$:
$\frac{1}{x} = \frac{1}{3 + 2\sqrt{2}} = \frac{1}{3 + 2\sqrt{2}} \times \frac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} = \frac{3 - 2\sqrt{2}}{9 - 8} = 3 - 2\sqrt{2}$.
Now,calculate $x + \frac{1}{x}$:
$x + \frac{1}{x} = (3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) = 6$.
To find $x^{2} + \frac{1}{x^{2}}$,use the identity $(x + \frac{1}{x})^{2} = x^{2} + \frac{1}{x^{2}} + 2$:
$x^{2} + \frac{1}{x^{2}} = (x + \frac{1}{x})^{2} - 2 = (6)^{2} - 2 = 36 - 2 = 34$.
To find $x^{3} + \frac{1}{x^{3}}$,use the identity $(x + \frac{1}{x})^{3} = x^{3} + \frac{1}{x^{3}} + 3(x + \frac{1}{x})$:
$x^{3} + \frac{1}{x^{3}} = (x + \frac{1}{x})^{3} - 3(x + \frac{1}{x}) = (6)^{3} - 3(6) = 216 - 18 = 198$.
Thus,the values are $34$ and $198$.

(D) Given $x = 7 - 4\sqrt{3}$.
First,find $\frac{1}{x}$:
$\frac{1}{x} = \frac{1}{7 - 4\sqrt{3}}$.
Rationalizing the denominator:
$\frac{1}{x} = \frac{1}{7 - 4\sqrt{3}} \times \frac{7 + 4\sqrt{3}}{7 + 4\sqrt{3}} = \frac{7 + 4\sqrt{3}}{49 - (16 \times 3)} = \frac{7 + 4\sqrt{3}}{49 - 48} = 7 + 4\sqrt{3}$.
Now,find $x + \frac{1}{x}$:
$x + \frac{1}{x} = (7 - 4\sqrt{3}) + (7 + 4\sqrt{3}) = 14$.
We know that $(x + \frac{1}{x})^{2} = x^{2} + \frac{1}{x^{2}} + 2$.
Therefore,$x^{2} + \frac{1}{x^{2}} = (x + \frac{1}{x})^{2} - 2$.
Substituting the value:
$x^{2} + \frac{1}{x^{2}} = (14)^{2} - 2 = 196 - 2 = 194$.

(A) To solve $\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}=a+b \sqrt{35}$,we rationalize the denominator by multiplying the numerator and denominator by the conjugate $(\sqrt{7}+\sqrt{5})$.
$\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}} \times \frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}+\sqrt{5}} = \frac{(\sqrt{7}+\sqrt{5})^2}{(\sqrt{7})^2-(\sqrt{5})^2}$
$= \frac{7+5+2\sqrt{35}}{7-5}$
$= \frac{12+2\sqrt{35}}{2}$
$= 6+\sqrt{35}$
Comparing $6+\sqrt{35}$ with $a+b\sqrt{35}$,we get $a=6$ and $b=1$.

(N/A) To solve this,we rationalize each term in the expression.
Step $1$: Rationalize the first term: $\frac{1}{1+\sqrt{2}} = \frac{1}{\sqrt{2}+1} \times \frac{\sqrt{2}-1}{\sqrt{2}-1} = \frac{\sqrt{2}-1}{2-1} = \sqrt{2}-1$.
Step $2$: Rationalize the second term: $\frac{1}{\sqrt{2}+\sqrt{3}} = \frac{1}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} = \frac{\sqrt{3}-\sqrt{2}}{3-2} = \sqrt{3}-\sqrt{2}$.
Step $3$: Rationalize the third term: $\frac{1}{\sqrt{3}+\sqrt{4}} = \frac{1}{\sqrt{4}+\sqrt{3}} \times \frac{\sqrt{4}-\sqrt{3}}{\sqrt{4}-\sqrt{3}} = \frac{\sqrt{4}-\sqrt{3}}{4-3} = \sqrt{4}-\sqrt{3} = 2-\sqrt{3}$.
Step $4$: Add the results: $(\sqrt{2}-1) + (\sqrt{3}-\sqrt{2}) + (2-\sqrt{3})$.
Step $5$: Simplify the expression: $\sqrt{2} - 1 + \sqrt{3} - \sqrt{2} + 2 - \sqrt{3} = -1 + 2 = 1$.
Thus,the expression equals $1$.

(A) To find the values of $a$ and $b$,we rationalize the denominator of the given expression:
$\frac{5+3 \sqrt{3}}{7+4 \sqrt{3}} \times \frac{7-4 \sqrt{3}}{7-4 \sqrt{3}}$
$= \frac{(5+3 \sqrt{3})(7-4 \sqrt{3})}{(7)^2 - (4 \sqrt{3})^2}$
$= \frac{35 - 20 \sqrt{3} + 21 \sqrt{3} - 12(3)}{49 - 16(3)}$
$= \frac{35 + \sqrt{3} - 36}{49 - 48}$
$= \frac{-1 + \sqrt{3}}{1}$
$= -1 + 1 \sqrt{3}$
Comparing this with $a+b \sqrt{3}$,we get $a = -1$ and $b = 1$.

(N/A) Given $x = 2 + \sqrt{3}.$
First,find $\frac{1}{x} = \frac{1}{2 + \sqrt{3}}.$
Rationalizing the denominator: $\frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}.$
Now,$x + \frac{1}{x} = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4.$
To find $x^{2} + \frac{1}{x^{2}},$ use the identity $(x + \frac{1}{x})^{2} = x^{2} + \frac{1}{x^{2}} + 2.$
$4^{2} = x^{2} + \frac{1}{x^{2}} + 2 \implies 16 = x^{2} + \frac{1}{x^{2}} + 2 \implies x^{2} + \frac{1}{x^{2}} = 14.$
To find $x^{3} + \frac{1}{x^{3}},$ use the identity $(x + \frac{1}{x})^{3} = x^{3} + \frac{1}{x^{3}} + 3(x + \frac{1}{x}).$
$4^{3} = x^{3} + \frac{1}{x^{3}} + 3(4) \implies 64 = x^{3} + \frac{1}{x^{3}} + 12 \implies x^{3} + \frac{1}{x^{3}} = 52.$
Thus,the values are $14$ and $52.$

(N/A) To represent $\sqrt{8.2}$ on the number line,follow these steps:
$1$. Draw a line segment $AB = 8.2 \text{ units}$ on the number line.
$2$. From point $B$,mark a point $C$ such that $BC = 1 \text{ unit}$. Now,$AC = 9.2 \text{ units}$.
$3$. Find the midpoint $O$ of $AC$. The distance $AO = OC = 4.6 \text{ units}$.
$4$. Draw a semicircle with center $O$ and radius $OA$ (or $OC$).
$5$. Draw a perpendicular line at point $B$ to the line $AC$,intersecting the semicircle at point $D$.
$6$. The length $BD$ is equal to $\sqrt{8.2}$.
$7$. With $B$ as the center and $BD$ as the radius,draw an arc on the number line. The point where the arc intersects the number line represents $\sqrt{8.2}$.

(N/A) To represent $\sqrt{5.6}$ on the number line,follow these steps:
$1$. Draw a line segment $AB = 5.6 \text{ units}$ on the number line.
$2$. From point $B$,mark a point $C$ such that $BC = 1 \text{ unit}$. Now,$AC = 6.6 \text{ units}$.
$3$. Find the midpoint $O$ of $AC$. The distance $AO = OC = 3.3 \text{ units}$.
$4$. Draw a semicircle with center $O$ and radius $3.3 \text{ units}$ passing through $A$ and $C$.
$5$. Draw a perpendicular line at point $B$ to $AC$,which intersects the semicircle at point $D$.
$6$. The length $BD$ is equal to $\sqrt{5.6}$.
$7$. With $B$ as the center and $BD$ as the radius,draw an arc on the number line. The point where the arc intersects the number line represents $\sqrt{5.6}$.

(A) To represent $\sqrt{7.5}$ on the number line,follow these steps:
$1$. Draw a line segment $AB$ of length $7.5 \text{ units}$.
$2$. Extend the line segment $AB$ by $1 \text{ unit}$ to point $C$,such that $BC = 1 \text{ unit}$. Now,$AC = 8.5 \text{ units}$.
$3$. Find the midpoint $O$ of $AC$. The distance $AO = OC = 4.25 \text{ units}$.
$4$. With $O$ as the center and $OA$ as the radius,draw a semicircle.
$5$. Draw a perpendicular line at point $B$ to the line $AC$,which intersects the semicircle at point $D$.
$6$. The length $BD$ is equal to $\sqrt{7.5}$.
$7$. Using $B$ as the center and $BD$ as the radius,draw an arc on the number line to mark the point representing $\sqrt{7.5}$.

(D) To evaluate $\frac{\sqrt{2}}{2+\sqrt{2}}$,we first rationalize the denominator by multiplying the numerator and the denominator by the conjugate $(2-\sqrt{2})$:
$\frac{\sqrt{2}}{2+\sqrt{2}} \times \frac{2-\sqrt{2}}{2-\sqrt{2}} = \frac{2\sqrt{2} - (\sqrt{2})^2}{2^2 - (\sqrt{2})^2}$
$= \frac{2\sqrt{2} - 2}{4 - 2} = \frac{2(\sqrt{2} - 1)}{2} = \sqrt{2} - 1$
Given $\sqrt{2} = 1.414$,we substitute this value:
$1.414 - 1 = 0.414$
Thus,the correct value is $0.414$.

(A) To find the value of $\frac{1}{\sqrt{3} + \sqrt{2}}$,we first rationalize the denominator by multiplying the numerator and the denominator by the conjugate $(\sqrt{3} - \sqrt{2})$.
$\frac{1}{\sqrt{3} + \sqrt{2}} \times \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} = \frac{\sqrt{3} - \sqrt{2}}{(\sqrt{3})^2 - (\sqrt{2})^2}$
$= \frac{\sqrt{3} - \sqrt{2}}{3 - 2} = \frac{\sqrt{3} - \sqrt{2}}{1} = \sqrt{3} - \sqrt{2}$
Given $\sqrt{3} = 1.732$ and $\sqrt{2} = 1.414$,we substitute these values:
$1.732 - 1.414 = 0.318$
Thus,the value is $0.318$.

(B) Given the expression: $\frac{\sqrt{10} - \sqrt{5}}{2}$
We can rewrite $\sqrt{10}$ as $\sqrt{2} \times \sqrt{5}$.
So,the expression becomes: $\frac{\sqrt{2} \times \sqrt{5} - \sqrt{5}}{2}$
Factor out $\sqrt{5}$ from the numerator: $\frac{\sqrt{5}(\sqrt{2} - 1)}{2}$
Substitute the given values $\sqrt{5} = 2.236$ and $\sqrt{2} = 1.414$:
$= \frac{2.236(1.414 - 1)}{2}$
$= \frac{2.236(0.414)}{2}$
$= 1.118 \times 0.414$
$= 0.462852$
Rounding to three decimal places,we get $0.463$.

(C) Given expression: $\frac{4}{3 \sqrt{3}-2 \sqrt{2}} + \frac{3}{3 \sqrt{3}+2 \sqrt{2}}$
Rationalize the first term: $\frac{4(3 \sqrt{3}+2 \sqrt{2})}{(3 \sqrt{3}-2 \sqrt{2})(3 \sqrt{3}+2 \sqrt{2})} = \frac{12 \sqrt{3}+8 \sqrt{2}}{(3 \sqrt{3})^2 - (2 \sqrt{2})^2} = \frac{12 \sqrt{3}+8 \sqrt{2}}{27-8} = \frac{12 \sqrt{3}+8 \sqrt{2}}{19}$
Rationalize the second term: $\frac{3(3 \sqrt{3}-2 \sqrt{2})}{(3 \sqrt{3}+2 \sqrt{2})(3 \sqrt{3}-2 \sqrt{2})} = \frac{9 \sqrt{3}-6 \sqrt{2}}{27-8} = \frac{9 \sqrt{3}-6 \sqrt{2}}{19}$
Adding both terms: $\frac{12 \sqrt{3}+8 \sqrt{2} + 9 \sqrt{3}-6 \sqrt{2}}{19} = \frac{21 \sqrt{3}+2 \sqrt{2}}{19}$
Substitute the values $\sqrt{3} = 1.732$ and $\sqrt{2} = 1.414$:
$= \frac{21(1.732) + 2(1.414)}{19} = \frac{36.372 + 2.828}{19} = \frac{39.2}{19} \approx 2.06315...$
Rounding to three decimal places,we get $2.063$.

(D) To simplify the expression $3^{\frac{2}{3}} \cdot 3^{\frac{4}{3}}$,we use the law of exponents: $a^m \cdot a^n = a^{m+n}$.
Applying this rule,we get:
$3^{\frac{2}{3}} \cdot 3^{\frac{4}{3}} = 3^{\frac{2}{3} + \frac{4}{3}}$
$= 3^{\frac{2+4}{3}}$
$= 3^{\frac{6}{3}}$
$= 3^2$
$= 9$

(A) To simplify the expression $\left(4^{\frac{1}{5}}\right)^{3}$,we use the law of exponents: $(a^m)^n = a^{m \times n}$.
Applying this rule:
$\left(4^{\frac{1}{5}}\right)^{3} = 4^{\frac{1}{5} \times 3}$
$= 4^{\frac{3}{5}}$
Thus,the simplified form is $4^{\frac{3}{5}}$.

(A) To simplify the expression $\frac{11^{\frac{1}{3}}}{11^{\frac{1}{5}}}$,we use the law of exponents: $\frac{a^m}{a^n} = a^{m-n}$.
Applying this rule,we get $11^{\frac{1}{3} - \frac{1}{5}}$.
To subtract the fractions in the exponent,we find a common denominator,which is $15$.
So,$\frac{1}{3} - \frac{1}{5} = \frac{5}{15} - \frac{3}{15} = \frac{2}{15}$.
Therefore,the simplified expression is $11^{\frac{2}{15}}$.

(A) Using the law of exponents,$a^m \cdot b^m = (a \cdot b)^m$.
Given expression: $7^{\frac{1}{4}} \cdot 12^{\frac{1}{4}}$.
Applying the rule: $(7 \times 12)^{\frac{1}{4}}$.
Calculating the product: $84^{\frac{1}{4}}$.

(D) To find the value of $64^{\frac{2}{3}}$,we first express $64$ as a power of $2$ or $4$.
Since $64 = 4^3$,we can write the expression as:
$64^{\frac{2}{3}} = (4^3)^{\frac{2}{3}}$.
Using the exponent rule $(a^m)^n = a^{m \times n}$,we get:
$4^{3 \times \frac{2}{3}} = 4^2$.
Calculating the value,$4^2 = 16$.