Question

express the following in the form of p/q where p ans q are integers and q =0

1. 0.6bar​

Answer 1

$\huge\star\:\:{\orange{\underline{\green{\mathcal{Solution}}}}}$

$\hookrightarrow\:x\:=\:0.6¯\:=\:0.6666.......\:(1)$

$\:\sf\underline\red{Multiply\:(1)\:with\:(10)}\:$

$⟹\:10x\:=\:10\:×\:10.66666........(2)$

$\:\sf\underline\red{Subtract\:(1)\:From\:(2)}\:$

$⟹\:10x\:-\:x\:=\:6.666.....\:-\:0.666......$

$⟹\:9x\:=\:6$

$⟹\:x\:=\:\dfrac{6}{9}$

⤵️$\:\sf\underline\red{Simplest\:Form}\:$

$⟹\:x\:=\:\dfrac{2}{3}$

$\boxed {Thus \:0.6 \: =\:\dfrac{2}{3}}$

Answer 2

Answer:

In \triangle△ ACD,

\because∵ \angle∠ ACD and \angle∠ EDC are opposite interior angles with exterior angle as \angle∠ AED,

\therefore∴ \angle∠ ACD + \angle∠ CDE = \angle∠ AED

( Exterior Angle Property Of A Triangle )

\implies⟹ \angle∠ ACD + 54° = 132°

( Substituting their values )

\implies⟹ \angle∠ ACD = 132° - 54°

\implies⟹ \boxed{\sf \angle ACD = {78}^{\circ}}

∠ACD=78

∘

\because∵ \angle∠ ACD + \angle∠ ACB = 180°

( Linear Pair )

\implies⟹ 78° + \angle∠ ACB = 180°

\implies⟹ \angle∠ ACB = 180° - 78°

\implies⟹ \boxed{\sf \angle ACB\: = {102}^{\circ}}

∠ACB=102

∘

Now in \triangle△ ABC,

\because∵ \angle∠ ABC + \angle∠ ACB + \angle∠ BAC = 180°

(Angle Sum Property Of A Triangle)

\implies⟹ 46° + 102° + x = 180°

( Substituting their values )

\implies⟹ 148° + x = 180°

\implies⟹ x = 180° - 148°

\implies⟹ \boxed{\sf x \: = {32}^{\circ}}

x=32

∘

\sf \boxed{\sf\therefore \: x \: = \: {32}^{ \circ}}

∴x=32

∘

\red\bigstar★ Concepts Used:

Exterior Angle Property of a Triangle

Substitution of values

Linear pair

Angle Sum Property Of A Triangle

\blue\bigstar★ Extra - Information:

Sum of the complementary angles is 90°

Sum of the supplementary angles is 180°

Supplementary angles may form a linear pair

Linear pair of angles are formed when two lines intersect each other at a single point.

The sum of angles of a linear pair is always equal to 180°.

Sum of all the interior angles of a Quadrilateral is 360°