Question
Prove that:
$\sqrt{\frac{1}{4}}+(0.01)^{-\frac{1}{2}}-(27)^{\frac{2}{3}}=\frac{3}{2}$
$\sqrt{\frac{1}{4}}+(0.01)^{-\frac{1}{2}}-(27)^{\frac{2}{3}}=\frac{3}{2}$
✓
Answer
We have,
$\sqrt{\frac{1}{4}}+(0.01)^{-\frac{1}{2}}-(27)^{\frac{2}{3}}=\frac{3}{2}$
$=\frac{1}{2}+\frac{1}{(0.001)^{\frac{1}{2}}}-(3^3)^{\frac{2}{3}}$
$=\frac{1}{2}+\frac{1}{(0.1)^{2\times\frac{1}{2}}}-3^{3\times\frac{2}{3}}$
$=\frac{1}{2}+\frac{1}{0.1}-3^2$
$=\frac{1}{2}+10-9$
$=\frac{1}{2}+1=\frac{3}{2}$
$\Rightarrow\sqrt{\frac{1}{4}}(0.001)^{-\frac{1}{2}}-(27)^{\frac{2}{3}}=\frac{3}{2}$
$\sqrt{\frac{1}{4}}+(0.01)^{-\frac{1}{2}}-(27)^{\frac{2}{3}}=\frac{3}{2}$
$=\frac{1}{2}+\frac{1}{(0.001)^{\frac{1}{2}}}-(3^3)^{\frac{2}{3}}$
$=\frac{1}{2}+\frac{1}{(0.1)^{2\times\frac{1}{2}}}-3^{3\times\frac{2}{3}}$
$=\frac{1}{2}+\frac{1}{0.1}-3^2$
$=\frac{1}{2}+10-9$
$=\frac{1}{2}+1=\frac{3}{2}$
$\Rightarrow\sqrt{\frac{1}{4}}(0.001)^{-\frac{1}{2}}-(27)^{\frac{2}{3}}=\frac{3}{2}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Solve the equation 2y - 1 = y + 1 and represent it graphically on the coordinate plane.
In the given figure, AB || CD and O is the modpoint of AD.
Show that:
→Show that:
- $\triangle\text{AOB}\cong\triangle\text{DOC}.$
- O is the midpoint of BC.
If O is the centre of the circle, find the value of x in the following figure:
Multiply:
(9x2 + 25y2 + 15xy + 12x - 20y + 16) by (3x - 5y + 4)
(9x2 + 25y2 + 15xy + 12x - 20y + 16) by (3x - 5y + 4)
If a + b + c = 0, then write the value of $\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}.$
→Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where,
p(x) = 2x3 + 3x2 - 11x - 3, $\text{g}(\text{x})=\Big(\text{x}+\frac{1}{2}\Big).$
→p(x) = 2x3 + 3x2 - 11x - 3, $\text{g}(\text{x})=\Big(\text{x}+\frac{1}{2}\Big).$
The diameter of the moon is approximately one-fourth the diameter of the earth. What fraction is the volume of the moon of the volume of the earth?
ABC is an isosceles triangle with AB = AC and D is a point on BC such that $\text{AD}\perp\text{BC}$ (see figure). To prove that $\angle\text{BAD} = \angle\text{CAD},$ a student proceeded as follows:
In $\triangle\text{ABD}$ and $\triangle\text{ACD},$
$\text{AB}=\text{AC}$
$\angle\text{B}=\angle\text{C}$
$\angle\text{ADM}=\angle\text{ADC}$
$\therefore\triangle\text{ABD}\cong\triangle\text{ADC}$
$\angle\text{BAD}=\angle\text{CAD}$
What is the defect in the above arguments?
→In $\triangle\text{ABD}$ and $\triangle\text{ACD},$$\text{AB}=\text{AC}$
$\angle\text{B}=\angle\text{C}$
$\angle\text{ADM}=\angle\text{ADC}$
$\therefore\triangle\text{ABD}\cong\triangle\text{ADC}$
$\angle\text{BAD}=\angle\text{CAD}$
What is the defect in the above arguments?
Find the values of a and b in the following:
$\frac{\sqrt{2}+\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}=2-\text{b}\sqrt{6}$
→$\frac{\sqrt{2}+\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}=2-\text{b}\sqrt{6}$
Express the following in the form $\frac{\text{p}}{\text{q}},$ where p and q are integers and $\text{q}\neq0$:
$0.\overline{134}$
→$0.\overline{134}$