Question

solve this...do not type silly stuffs your answers will be reported!!​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:12 + 1 \dfrac{5}{6}x \leqslant 5 + 3x$

$\rm :\longmapsto\:12 + \dfrac{11}{6}x \leqslant 5 + 3x$

$\rm :\longmapsto\:\dfrac{11}{6}x - 3x \leqslant 5 - 12$

$\rm :\longmapsto\:\dfrac{11x - 18x}{6} \leqslant - 7$

$\rm :\longmapsto\:\dfrac{ - 7x}{6} \leqslant - 7$

$\bf\implies \:x \geqslant 6$

$\bf\implies \:x \: \in \: [6, \: \infty )$

Additional Information :-

$\green{\boxed{ \tt \: If \: a \: > \: b \: , then \: - a \: < \: - b}}$

$\green{\boxed{ \tt \: If \: a \: \geqslant \: b \: , then \: - a \: \leqslant \: - b}}$

$\green{\boxed{ \tt \: If \: a \: \leqslant \: b \: , then \: - a \: \geqslant \: - b}}$

$\green{\boxed{ \tt \: If \: a \: < \: b \: , then \: - a \: > \: - b}}$

Answer 2

Answer:

Sabida in your followings there is one girl who's dp is in some grey colour xerox type

Step-by-step explanation:

she is my enemy

she had ban me for 48 hours

&

she delete my all answer