Question
Simplify : $a^{\frac{4}{3}} \div a^{-\frac{2}{3}}$
✓
Answer
$a^2$
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Solve for $x : 3^{4x + 1}= (27)^{x + 1}$
→Simplify : $(\sqrt3 - \sqrt2 )^2$
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→Rationalise the denominators of : $\frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}+\sqrt{5}}$
→The area enclosed between two concentric circles is 770$cm ^2$. If the radius of the outer circle is 21 cm, calculate the radius of the inner circle.
[Hint. Let the radius of the inner circle be (21 - x) cm. Then
$\begin{array}{l}\pi(21)^2-\pi(21-x)^2=770 \Rightarrow x^2-42 x+245=0 \Rightarrow(x-35)(x-7)=0 . \\ \text { But, } x \neq 35 . \text { So, } x=7]\end{array}$
→[Hint. Let the radius of the inner circle be (21 - x) cm. Then
$\begin{array}{l}\pi(21)^2-\pi(21-x)^2=770 \Rightarrow x^2-42 x+245=0 \Rightarrow(x-35)(x-7)=0 . \\ \text { But, } x \neq 35 . \text { So, } x=7]\end{array}$
A rectangular field is $112 \ m$ long and $62 \ m$ broad. A cubical tank of edge $6 \ m$ is dug at each of the four corners of the field and the earth so removed is evenly spread on the remaining field. Find the rise in level.
→If two altitudes of a triangle are equal, prove that it is an isosceles triangle.
Represent each of the following on the real line.
(i) $\sqrt{3}$
(ii) $\sqrt{5}$
(iii) $\sqrt{6}$
(iv) $\sqrt{10}$
→(i) $\sqrt{3}$
(ii) $\sqrt{5}$
(iii) $\sqrt{6}$
(iv) $\sqrt{10}$
Factorise:
$2 a^3+16 b^3-3 a-6 b$
→$2 a^3+16 b^3-3 a-6 b$
Simplify:
$\frac{5^{2(n+6)} \times(25)^{-7+2 n}}{(125)^{2 n}}$
→$\frac{5^{2(n+6)} \times(25)^{-7+2 n}}{(125)^{2 n}}$