Question 12 Marks
Write the following in decimal form and say what kind of decimal expansion has.
$\frac{7}{25}$
Answer$\frac{7}{25}=0.28$ By actual division, we have:
It is a terminating decimal expansion. View full question & answer→Question 22 Marks
Solve for $\text{x}\Big(\frac{2}{5}\Big)^{2\text{x}-2}=\frac{32}{3125}.$
Answer$\text{x}\Big(\frac{2}{5}\Big)^{2\text{x}-2}=\frac{32}{3125}$
$\Rightarrow\Big(\frac{2}{5}\Big)^{2\text{x}-2}=\frac{2^5}{5^5}$
$\Rightarrow\Big(\frac{2}{5}\Big)^{2\text{x}-2}=\Big(\frac{2}{5}\Big)^5$
$\Rightarrow2\text{x}-2=5$
$\Rightarrow2\text{x}=7$
$\Rightarrow\text{x}=\frac{7}{2}$
View full question & answer→Question 32 Marks
Find the value of x in the following:
$\sqrt[5]{5\text{x}+2}=2$
Answer$\sqrt[5]{5\text{x}+2}=2$
$\Rightarrow(5\text{x}+2)^\frac{1}{5}=2$
$\Rightarrow\bigg[(5\text{x}+2)^\frac{1}{5}\bigg]=2^5$
$\Rightarrow5\text{x}+2=32$
$\Rightarrow5\text{x}=30$
$\Rightarrow\text{x}=6$
View full question & answer→Question 42 Marks
Simplify:
$\big(-3+\sqrt{5}\big)\big(-3-\sqrt{5}\big)$
Answer$\big(-3+\sqrt{5}\big)\big(-3-\sqrt{5}\big)$
$=(-3)^2-\big(\sqrt{5}\big)^2$
$=9-5$
$=4$
View full question & answer→Question 52 Marks
Write the following in decimal form and say what kind of decimal expansion has.
$\frac{261}{400}$
Answer$\frac{261}{400}=0.6525$
It is a terminating decimal expansion. View full question & answer→Question 62 Marks
Evaluate $\frac{2^{\text{n}}+2^{\text{n}-1}}{2^{\text{n}+1}-2^{\text{n}}}.$
Answer$\frac{2^{\text{n}}+2^{\text{n}-1}}{2^{\text{n}+1}-2^{\text{n}}}$
$=\frac{2^{\text{n}-1}(2+1)}{2^{\text{n}}(2-1)}$
$=\frac{2^{\text{n}-1}\times3}{2^{\text{n}}\times1}$
$=\frac{3}{2^{\text{n}-\text{n}+1}}$
$=\frac{3}{2}$
View full question & answer→Question 72 Marks
Find four rational numbers lying between $\frac{3}{7}$ and $\frac{5}{7}.$
Answern = 4
n + 1 = 4 + 1 = 5
$\frac{3}{7}=\frac{3}{7}\times\frac{5}{5}=\frac{15}{35}$
$\frac{5}{7}=\frac{5}{7}\times\frac{5}{5}=\frac{25}{35}$
Thus, rational numbers between $\frac{3}{7}$ and $\frac{5}{7}$ are $\frac{16}{35},\frac{17}{35},\frac{18}{35},\frac{19}{35}.$
View full question & answer→Question 82 Marks
Examine whether the following numbers are rational or irrational.
$\sqrt{7}-2$
AnswerLet us assume, to the contrary, that $\sqrt{7}-2$ is rational.
Then, $\sqrt{7}-2=\frac{\text{p}}{\text{q}},$ where p and q are coprime and $\text{q}\neq0.$
$\Rightarrow\sqrt{7}=\frac{\text{p}}{\text{q}}+2$
$\Rightarrow\sqrt{7}=\frac{\text{p}+2\text{q}}{\text{q}}$
Since, p and q are are integers.
$\Rightarrow\frac{\text{p}+2\text{q}}{\text{q}}$ is rational.
So, $\sqrt{7}$ is also rational.
But this contradicts the fact that $\sqrt{7}$ is irrational.
This contradiction has arisen because of our incorrect assumption that $\sqrt{7}-2$ is rational.
Hence, $\sqrt{7}-2$ is irrational.
View full question & answer→Question 92 Marks
Is the product of a rational and an irrational number always irrational? Give an example.
AnswerYes, the product of a rational and an irrational number is always an irrational number.
Example: 2 is a rational number and $\sqrt{3}$ is an irrational number.
Now, $2\times\sqrt{3}=2\sqrt{3},$ which is an irrational number.
View full question & answer→Question 102 Marks
Prove that:
$\sqrt{\text{x}^{-1}\text{y}}\times\sqrt{\text{y}^{-1}\text{z}}\times\sqrt{\text{z}^{-1}\text{x}}=1.$
Answer$\text{L.H.S}=\sqrt{\text{x}^{-1}\text{y}}\times\sqrt{\text{y}^{-1}\text{z}}\times\sqrt{\text{z}^{-1}\text{x}}=1.$
$=\sqrt{\frac{\text{y}}{\text{x}}}\times\sqrt{\frac{\text{z}}{\text{y}}}\times\sqrt{\frac{\text{x}}{\text{z}}}$
$=\sqrt{\frac{\text{y}}{\text{x}}\times\frac{\text{z}}{\text{y}}\times\frac{\text{x}}{\text{z}}}$
$=\sqrt1$
$=1$
$=\text{R.H.S}$
View full question & answer→Question 112 Marks
Simplify:
$\Bigg(\frac{15^\frac{1}{4}}{3^\frac{1}{2}}\Bigg)^{-2}$
Answer $\Bigg(\frac{15^\frac{1}{4}}{3^\frac{1}{2}}\Bigg)^{-2}$
$=\Bigg(\frac{3^\frac{1}{2}}{15^\frac{1}{4}}\Bigg)^2$
$=\frac{3^{\frac{1}{2}\times2}}{15^{\frac{1}{4}\times2}}$
$=\frac{3}{15^\frac{1}{2}}$
View full question & answer→Question 122 Marks
Examine whether the following number are rational or irrational:
$\big(5-\sqrt{5}\big)\big(5+\sqrt{5}\big)$
Answer$\big(5-\sqrt{5}\big)\big(5+\sqrt{5}\big)$
$=(5)^2-\big(\sqrt{5}\big)^2$
$=25-5$
$=20$
Thus, the given number is rational.
View full question & answer→Question 132 Marks
Let a be a rational number and b be an irrational number. Is ab necessarily an irrational number? Justify your answer with an example.
Answera be a rational number and b be an irrational number then ab necessarily will be an irrational number.
Example: 6 is a rational number but $\sqrt{5}$ is irrational. And $6\sqrt{5}$ is also an irrational number.
View full question & answer→Question 142 Marks
Examine whether the following numbers are rational or irrational.
$3+\sqrt{3}$
AnswerLet us assume, to the contrary, that $3+\sqrt{3}$ is rational.
Then, $3+\sqrt{3}=\frac{\text{p}}{\text{q}},$ where p and q are coprime and $\text{q}\neq0.$
$\Rightarrow\sqrt{3}=\frac{\text{p}}{\text{q}}-3$
$\Rightarrow\sqrt{3}=\frac{\text{p}-3\text{q}}{\text{q}}$
Since, p and q are are integers.
$\Rightarrow\frac{\text{p}-3\text{q}}{\text{q}}$ is rational.
So, $\sqrt{3}$ is also rational.
But this contradicts the fact that $\sqrt{3}$ is irrational.
This contradiction has arisen because of our incorrect assumption that $3+\sqrt{3}$ is rational.
Hence, $3+\sqrt{3}$ is irrational.
View full question & answer→Question 152 Marks
Multiply:
$\sqrt{10}$ by $\sqrt{40}$
Answer$\sqrt{10}$ by $\sqrt{40}$
$\sqrt{10}\times\sqrt{40}=\sqrt{10\times40}$
$=\sqrt{2\times5\times2\times2\times2\times5}$
$=(2\times2\times5)=20$
View full question & answer→Question 162 Marks
Multiply:
$3\sqrt{8}$ by $3\sqrt{2}$
Answer$3\sqrt{8}$ by $3\sqrt{2}$
$3\sqrt{8}\times3\sqrt{2}=3\times3\times\sqrt{8}\times\sqrt{2}$
$=9\times\sqrt{8\times2}$
$=9\times\sqrt{2\times2\times2\times2}$
$=(9\times2\times2)=36$
View full question & answer→Question 172 Marks
Write the following in decimal form and say what kind of decimal expansion has.
$\frac{5}{8}$
Answer$\frac{5}{8}=0.625$
By actual division, we have:
It is a terminating decimal expansion. View full question & answer→Question 182 Marks
Simplify $\Bigg[\Big\{(256)^{-\frac{1}{2}}\Big\}^{-\frac{1}{4}}\Bigg]^2.$
Answer$\Bigg[\Big\{(256)^{-\frac{1}{2}}\Big\}^{-\frac{1}{4}}\Bigg]^2$
$=\Bigg[\Big\{\big(16^2\big)^{-\frac{1}{2}}\Big\}^{-\frac{1}{4}}\Bigg]^2$
$=\Bigg[\Big\{16^{-1}\Big\}^{-\frac{1}{4}}\Bigg]^2$
$=\Bigg[16^{-1\times\big(-\frac{1}{4}\big)}\Bigg]^2$
$=\Bigg[16^{\frac{1}{4}}\Bigg]^2$
$=\Bigg[2^{4\times\frac{1}{4}}\Bigg]^2$
$=2^2$
$=4$
View full question & answer→Question 192 Marks
Rationalise the denominator of the following:
$\frac{1}{\sqrt{5}-2}$
AnswerIf a and b are integers, then
$\big(\text{a}+\sqrt{\text{b}}\big)$ and $\big(\text{a}-\sqrt{\text{b}}\big)$ are rationalising factor of each other, as $\big(\text{a}+\sqrt{\text{b}}\big)\big(\text{a}-\sqrt{\text{b}}\big)=\big(\text{a}^2-\text{b}\big),$ which is rational.
Therefore, we have,
$=\frac{1}{\big(\sqrt{5}-2\big)}=\frac{1}{\sqrt{5}-2}\times\frac{\sqrt{5}+2}{\sqrt{5}+2}$
$=\frac{\sqrt{5}+2}{\big(\sqrt{5}\big)^2-(2)^2}=\frac{\sqrt{5}+2}{5-4}$
$=\frac{\sqrt{5}+2}{1}=\sqrt{5}+2$
View full question & answer→Question 202 Marks
Examine whether the following number are rational or irrational:
$\sqrt{8}+4\sqrt{32}-6\sqrt{2}$
Answer$\sqrt{8}+4\sqrt{32}-6\sqrt{2}$
$=\sqrt{4\times2}+4\sqrt{16\times2}-6\sqrt{2}$
$=2\sqrt{2}+16\sqrt{2}-6\sqrt{2}$
$=12\sqrt{2}$
Thus, the given number is irrational.
View full question & answer→Question 212 Marks
Examine whether the following numbers are rational or irrational.
$\sqrt{7}\times\sqrt{343}$
AnswerAs, $\sqrt{7}\times\sqrt{343}$
$=\sqrt{7\times343}$
$=\sqrt{2401}$
$=49,$ which is an integer
Hence, $\sqrt{7}\times\sqrt{343}$ is rational.
View full question & answer→Question 222 Marks
Write the following in decimal form and say what kind of decimal expansion has.
$\frac{15}{13}$
Answer$\frac{15}{13}=0.\overline{384615}$
It is a non-terminating recurring decimal. View full question & answer→Question 232 Marks
Multiply:
$3\sqrt{5}$ by $2\sqrt{5}$
Answer$3\sqrt{5}$ by $2\sqrt{5}$
$3\sqrt{5}\times2\sqrt{5}=3\times2\times\sqrt{5}\times\sqrt{5}$
$=(3\times2\times5)=30$
View full question & answer→Question 242 Marks
Examine whether the following number are rational or irrational:
$\frac{2\sqrt{13}}{3\sqrt{52}-4\sqrt{117}}$
Answer$\frac{2\sqrt{13}}{3\sqrt{52}-4\sqrt{117}}$
$=\frac{2\sqrt{13}}{3\sqrt{4\times13}-4\sqrt{9\times13}}$
$=\frac{2\sqrt{13}}{3\times2\sqrt{13}-4\times3\sqrt{13}}$
$=\frac{2\sqrt{13}}{6\sqrt{13}-12\sqrt{13}}$
$=\frac{2\sqrt{13}}{-6\sqrt{13}}$
$=-\frac{1}{3}$
Thus, the given number is rational.
View full question & answer→Question 252 Marks
Give an example of a number $x$ such that $x^2$ is an irrational number and $x^3$ is a rational number.
AnswerThe cube roots of natural numbers which are not perfect cubes are all irrational numbers.
Let $\text{x}=\sqrt[3]{2}=2^{\frac{1}{3}}$
Now,
$\text{x}^2=\big(2^{\frac{1}{3}}\big)^2=2^{\frac{2}{3}}=\big(2^2\big)^{\frac{1}{3}}=4^\frac{1}{3},$ which is an irrational number
Also,
$\text{x}^3=\Big(2^{\frac{1}{3}}\Big)^3=2^{3\times\frac{1}{3}}=2,$ which is a rational number.
View full question & answer→Question 262 Marks
Write the following in decimal form and say what kind of decimal expansion has.
$\frac{3}{11}$
Answer$\frac{3}{11}=0.\overline{27}$
It is a non-terminating recurring decimal. View full question & answer→Question 272 Marks
Simplify the product $\sqrt[3]{2}\times\sqrt[4]{2}\times\sqrt[12]{32}.$
Answer$\sqrt[3]{2}\times\sqrt[4]{2}\times\sqrt[12]{32}$
$=2^\frac{1}{3}\times2^\frac{1}{4}\times32^\frac{1}{12}$
$=2^\frac{1}{3}\times2^\frac{1}{4}\times2^{5\times\frac{1}{12}}$
$=2^\frac{1}{3}\times2^\frac{1}{4}\times2^\frac{5}{12}$
$=2^{\frac{1}{3}+\frac{1}{4}+\frac{5}{12}}$
$=2^{\frac{4+3+5}{12}}$
$=2^\frac{12}{12}$
$=2$
View full question & answer→Question 282 Marks
Find two rational numbers of the form $\frac{\text{p}}{\text{q}}$ between the numbers 0.2121121112... and 0.2020020002...
AnswerThe rational numbers between the numbers 0.2121121112... and 0.2020020002... are:
$0.21=\frac{21}{100}$ and $0.205=\frac{206}{1000}=\frac{41}{200}$
Disclaimer: There are an infinite number of rational numbers between two irrational numbers.
View full question & answer→Question 292 Marks
Simplify:
$\Big(\frac{7776}{243}\Big)^{-\frac{3}{5}}$
Answer$\Big(\frac{7776}{243}\Big)^{-\frac{3}{5}}$
$=\Big(\frac{243}{7776}\Big)^{\frac{3}{5}}$
$=\Big(\frac{3^5}{6^5}\Big)^{\frac{3}{5}}$
$=\frac{3^{5\times\frac{3}{5}}}{6^{5\times\frac{3}{5}}}$
$=\frac{3^3}{6^3}$
$=\frac{3\times3\times3}{6\times6\times6}$
$=\frac{1}{8}$
View full question & answer→Question 302 Marks
Simplify:
$\big(3-\sqrt{3}\big)^2$
Answer$\big(3-\sqrt{3}\big)^2$
$=(3)^2+\big(\sqrt{3}\big)^2-2.3.\sqrt{3}$
$=9+3-6\sqrt{3}$
$=12-6\sqrt{3}$
View full question & answer→Question 312 Marks
Simplify:
$\big(\sqrt{5}-\sqrt{3}\big)^2$
Answer$\big(\sqrt{5}-\sqrt{3}\big)^2$
$=\big(\sqrt{5}\big)^2+\big(\sqrt{3}\big)^2-2\sqrt{5}.\sqrt{3}$
$=5+3-2\sqrt{15}$
$=8-2\sqrt{15}$
View full question & answer→Question 322 Marks
Rationalise the denominator of the following:
$\frac{1}{\sqrt{7}-\sqrt{6}}$
Answer$\frac{1}{\sqrt{7}-\sqrt{6}}$
$=\frac{1}{\sqrt{7}-\sqrt{6}}\times\frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}$
$=\frac{\sqrt{7}+\sqrt{6}}{\big(\sqrt{7}\big)^2-\big(\sqrt{6}\big)^2}$
$=\frac{\sqrt{7}+\sqrt{6}}{7-6}$
$=\sqrt{7}+\sqrt{6}$
View full question & answer→Question 332 Marks
Write the following in decimal form and say what kind of decimal expansion has.
$2\frac{5}{12}$
Answer$2\frac{5}{12}=\frac{29}{12}=2.41\overline{6}$ By actual division, we have:
It is a non-terminating decimal expansion. View full question & answer→Question 342 Marks
Prove that:
$\frac{\text{x}^{\text{a}(\text{b}-\text{c})}}{\text{x}^{\text{b}(\text{a}-\text{c})}}\div\Big(\frac{\text{x}^\text{b}}{\text{x}^\text{a}}\Big)^\text{c}=1$
Answer$\text{L.H.S}=\frac{\text{x}^{\text{a}(\text{b}-\text{c})}}{\text{x}^{\text{b}(\text{a}-\text{c})}}\div\Big(\frac{\text{x}^\text{b}}{\text{x}^\text{a}}\Big)^\text{c}$
$=\frac{\text{X}^{\text{ab}-\text{ac}}}{\text{X}^{\text{ab}-\text{bc}}}\div\frac{\text{X}^\text{bc}}{\text{X}^\text{ac}}$
$=\text{X}^{\text{ab}-\text{ac}-\text{ab}+\text{bc}}\div\text{X}^{\text{bc}-\text{ac}}$
$=\text{X}^{\text{bc}-\text{ac}}\div\text{X}^{\text{bc}-\text{ac}}$
$=1$
$=\text{R.H.S}$
View full question & answer→Question 352 Marks
Simplify:
$\sqrt{72}+\sqrt{800}-\sqrt{18}$
Answer$\sqrt{72}+\sqrt{800}-\sqrt{18}$
$=\sqrt{ 36\times2}+\sqrt{400\times2}-\sqrt{9\times2}$
$=6\sqrt{2}+20\sqrt{2}-3\sqrt{2}$
$=23\sqrt{2}$
View full question & answer→Question 362 Marks
Find a rational number between 1.3 and 1.4
Answer1.3 and 1.4
Let
x = 1.3 and y = 1.4
Rational number lying between x and y.
$\frac{1}{2}(\text{x}+\text{y})=\frac{1}{2}\big(1.3+1.4\big)$
$=\frac{1}{2}(2.7)=1.35$
View full question & answer→Question 372 Marks
Examine whether the following numbers are rational or irrational.
$\sqrt{\frac{13}{117}}$
Answer$\sqrt{\frac{13}{117}}=\sqrt{\frac{1}{9}}=\frac{1}{3},$ which is rational
Hence, $\sqrt{\frac{13}{117}}$ is rational.
View full question & answer→Question 382 Marks
Find the value of x in the following:
$\frac{3^{3\text{x}}\times3^{2\text{x}}}{3^\text{x}}=\sqrt[4]{3^{20}}$
Answer$\frac{3^{3\text{x}}\times3^{2\text{x}}}{3^\text{x}}=\sqrt[4]{3^{20}}$
$\Rightarrow\frac{3^{3\text{x}+2\text{x}}}{3^\text{x}}=3^{20\times\frac{1}{4}}$
$\Rightarrow\frac{3^{5\text{x}}}{3^\text{x}}=3^5$
$\Rightarrow3^{4\text{x}}=3^5$
$\Rightarrow4\text{x}=5$
$\Rightarrow\text{x}=\frac{5}{4}$
View full question & answer→Question 392 Marks
If a = 2, b = 3, find the values of:
$\big(\text{a}^{\text{a}}+\text{b}^{\text{b}}\big)^{-1}$
AnswerGiven, a = 2 and b = 3
$\big(\text{a}^{\text{a}}+\text{b}^{\text{b}}\big)^{-1}=(2^2+3^3)^{-1}$
$=(4+27)^{-1}$
$=(31)^{-1}$
$=\frac{1}{31}$
View full question & answer→Question 402 Marks
Simplify:
$\Bigg(\frac{15^\frac{1}{3}}{9^\frac{1}{4}}\Bigg)^{-6}$
Answer$\Bigg(\frac{15^\frac{1}{3}}{9^\frac{1}{4}}\Bigg)^{-6}$
$=\Bigg(\frac{9^\frac{1}{4}}{15^\frac{1}{3}}\Bigg)^6$
$=\Bigg(\frac{3^{2\times\frac{1}{4}}}{15^\frac{1}{3}}\Bigg)^6$
$=\Bigg(\frac{3^\frac{1}{2}}{15^\frac{1}{3}}\Bigg)^6$
$=\frac{3^{\frac{1}{2}\times6}}{15^{\frac{1}{3}\times6}}$
$=\frac{3^3}{15^2}$
$=\frac{27}{225}$
View full question & answer→Question 412 Marks
Find the value of x in the following:
$\Big(\frac{3}{4}\Big)^3\Big(\frac{4}{3}\Big)^{-7}=\Big(\frac{3}{4}\Big)^{2\text{x}}$
Answer $\Big(\frac{3}{4}\Big)^3\Big(\frac{4}{3}\Big)^{-7}=\Big(\frac{3}{4}\Big)^{2\text{x}}$
$\Rightarrow\Big(\frac{3}{4}\Big)^3\Big(\frac{3}{4}\Big)^7=\Big(\frac{3}{4}\Big)^{2\text{x}}$
$\Rightarrow\Big(\frac{3}{4}\Big)^{3+7}=\Big(\frac{3}{4}\Big)^{2\text{x}}$
$\Rightarrow\Big(\frac{3}{4}\Big)^{10}=\Big(\frac{3}{4}\Big)^{2\text{x}}$
$\Rightarrow2\text{x}=10$
$\Rightarrow\text{x}=5$
View full question & answer→Question 422 Marks
Simplify:
$\Big(\frac{32}{243}\Big)^{-\frac{4}{5}}$
Answer$\Big(\frac{32}{243}\Big)^{-\frac{4}{5}}$
$=\Big(\frac{243}{32}\Big)^{\frac{4}{5}}$
$=\Big(\frac{3^5}{2^5}\Big)^{\frac{4}{5}}$
$=\frac{3^{5\times\frac{4}{5}}}{2^{5\times\frac{4}{5}}}$
$=\frac{3^4}{2^4}$
$=\frac{81}{16}$
View full question & answer→Question 432 Marks
Write the following in decimal form and say what kind of decimal expansion has.
$\frac{231}{625}$
Answer$\frac{231}{625}=0.3696$
It is a terminating decimal expansion. View full question & answer→Question 442 Marks
Rationalise the denominator of the following:
$\frac{4}{\sqrt{11}-\sqrt{7}}$
Answer$\frac{4}{\sqrt{11}-\sqrt{7}}$
$=\frac{4}{\sqrt{11}-\sqrt{7}}\times\frac{\sqrt{11}+\sqrt{7}}{\sqrt{11}+\sqrt{7}}$
$=\frac{4\big(\sqrt{11}+\sqrt{7}\big)}{\big(\sqrt{11}\big)^2-\big(\sqrt{7}\big)^2}$
$=\frac{4\big(\sqrt{11}+\sqrt{7}\big)}{11-7}$
$=\frac{4\big(\sqrt{11}+\sqrt{7}\big)}{4}$
$=\sqrt{11}+\sqrt{7}$
View full question & answer→Question 452 Marks
Find a rational number between -1 and $\frac{1}{2}$
Answer$-1$ and $\frac{1}{2}$
Let:
$\text{x}=-1$ and $\text{y}=\frac{1}{2}$
Rational number lying between x and y.
$\frac{1}{2}(\text{x}+\text{y})=\frac{1}{2}\Big(-1+\frac{1}{2}\Big)$
$=-\frac{1}{4}$
View full question & answer→Question 462 Marks
Multiply:
$3\sqrt{28}$ by $2\sqrt{7}$
Answer$3\sqrt{28}$ by $2\sqrt{7}$
$3\sqrt{28}\times2\sqrt{7}=3\times2\times\sqrt{28}\times\sqrt{7}$
$=6\times\sqrt{28\times7}$
$=6\times\sqrt{2\times2\times7\times7}$
$=(6\times2\times4)=84$
View full question & answer→Question 472 Marks
Multiply:
$2\sqrt{6}$ by $3\sqrt{3}$
Answer$2\sqrt{6}$ by $3\sqrt{3}$
$2\sqrt{6}\times3\sqrt{3}=2\times3\times\sqrt{6}\times\sqrt{3}$
$=6\times\sqrt{6\times3}$
$=6\times\sqrt{2\times3\times3}$
$=6\times3\sqrt{2}=18\sqrt{2}$
View full question & answer→Question 482 Marks
How many irrational numbers lie between $\sqrt{2}$ and $\sqrt{3}?$ Find any three irrational numbers lying between $\sqrt{2}$ and $\sqrt{3}.$
AnswerThere are infinite number of irrational numbers lying between $\sqrt{2}$ and $\sqrt{3}.$
As, $\sqrt{2}=1.414$ and $\sqrt{3}=1.732$
So, the three irrational numbers lying between $\sqrt{2}$ and $\sqrt{3}$ are:
1.420420042000..., 1.505005000... and 1.616116111...
View full question & answer→Question 492 Marks
Write the following in decimal form and say what kind of decimal expansion has.
$\frac{11}{24}$
Answer$\frac{11}{24}=0.458\overline{3}$
By actual division, we have:
It is a non-terminating recurring decimal expansion. View full question & answer→Question 502 Marks
Is the product of two irrationals always irrational? Justify your answer.
AnswerProduct of two irrational numbers is not always an irrational number.
Example: $\sqrt{5}$ is irrational number. And $\sqrt{5}\times\sqrt{5}=5$ is a rational number. But the product of another two irrational numbers $\sqrt{2}$ and $\sqrt{3}$ is $\sqrt{6}$ which is also an irrational numbers.
View full question & answer→