Sum of the angles of a quadrilateral is 360°.
ஃ 4x° + 7x° + 15x° + 10x° = 360°
[Angle sum property of quadrilateral]
⇒ 36x°= 360°
⇒ x=10°
ஃ Smallest angle = 4x° = 40°
Largest angle = 15x°= 150°

\(\sqrt[4]{16}=\sqrt[4]{2\times2\times2\times2}=2\)
\(\sqrt[3]{343}=\sqrt[3]{7\times7\times7}=7\)
\(\sqrt[5]{243}=\sqrt[5]{3\times3\times3\times3\times3}=3\)
\(\sqrt{196}\) = 14
ஃ \(\sqrt[4]{16}-6\sqrt[3]{343}+18\sqrt[5]{243}-\sqrt{196}\)
= 2- 6 x 7 + 18 x 3 - 14
= 2 - 42 + 54 - 14
= 56 - 56 = 0

From l² = r² + h², we have
\(r=\sqrt{l^{2}-h^{2}}=\sqrt{28^{2}-21^{2}} \mathrm{~cm}=7 \sqrt{7} \mathrm{~cm}\) 
So, volume of the cone \(=\frac{1}{3} \pi r^{2} h=\frac{1}{3} \times \frac{22}{7} \times 7 \sqrt{7} \times 7 \sqrt{7} \times 21 \mathrm{~cm}^{3}\)
= 7546 cm³

\(\sqrt { 297 } cm\)

84 cm²

\(\angle\)ACB=1/2 * \(\angle\)AOB
=1/2*90^o
=45^o
In \(\Delta\)ACB, \(\angle\)CAB=180^o-(30^o+45^o)
=105^o
\(\angle\)OAB=\(\angle\)OBA
=45^o
(Angles opp. to equal sides of triangle are equal as OA = OB radius of same circle)
\(\angle\)CAO = 105°- \(\angle\)OAB
= 105°-45°
= 60°

\(128y^3\)

\((a-5)(a+3)(a+2)(a-4)\)

\(-\frac{3}{2}\)

Quotient =\(3x^3-x^3-x-1\)

Remainder=-5

28

5

\(a=2,\quad b=\frac { 5 }{ 6 } \)

\(-8\sqrt { 5 } \)

Diameter = 7 cm
ஃ  Radius (r) = \(\frac{7}{2}\)cm
Height (h) = 4 cm
ஃ  Volume of soup in the cylindrical bowl
=\(\pi r^2h\)
\(=\frac{22}{7}\times (\frac{7}{2})^2\times 4 cm^2\)
= 154 cm²
ஃ  Volume of soup to be prepared daily to serve 250 patients
= 154 x 250 cm³ = 38500 cm³ ( or  38.5l)
Hence, the hospital has to prepare 28500 cm³ (or 38.5l) of soup daily to serve250 patients.