Question

You are Rohan/Roma, the Students’ Representative of the School Development Committee. Next week, the Committee is meeting to prepare a Calendar of Activities for academic year 20XX-20XY. Write a notice, to be put up on your school notice board, inviting students’ suggestions on the subject.

Answer 1

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Suncity World School, Gurgaon

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \: \\ \qquad \qquad \;\; \large{\boxed{\tt{NOTICE}}}\end{gathered} \end{gathered} \end{gathered} \end{gathered}$

21st January, 20XX

Annual Development Committee Meeting – Suggestions Invited

All the students are hereby informed that the School Development Committee is going to meet on 31st January, 20XX to prepare a Calendar of Activities for the coming session.

All those students who have creative suggestions can meet the undersigned and discuss their ideas. The undersigned will meet such students on 25th January, 20XX in the Student Activity Hall in the zero period. Constructive ideas will be welcomed.

For further details contact the undersigned.

Roma,

(Students’ Representative)

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★ More to know :

Format of Notice :-

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \boxed{\begin{minipage}{6cm} \\ \sf{ N{a}me of the Institution / Organization } \\ \sf{ \dots\dots \dots NOTICE \dots\dots \dots } \\ \\ \sf{Date : Date/Month/Year } \\ \\ \tt{ Main Body } \\ \\\sf{\dots\dots\dots\dots Heading\dots\dots\dots\dots}\\ \sf[Body/Content]\\\\\sf{Signature} \\ \sf{N{a}me} \\ \sf{ Designation}\end{minipage}} \end{gathered} \end{gathered} \end{gathered}\end{gathered}</p><p>$

Note:- Kindly view the Answer from the web to see the displayed format.

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Answer 2

Explanation:

Step-by-step explanation:

Given Equation:-

⠀⠀⠀⠀ $\sf{\bigg[ \dfrac{5 {x}^{2} - 10}{12 } \bigg] }$

To find:-

value of x

Solution:-

use factor theorem

take the value of equation =0

$\\\qquad\quad\displaystyle\sf{:}\longrightarrow \left [\dfrac {5x^2-10}{12}\right]=0$

$\\\qquad\quad\displaystyle\sf{:}\longrightarrow \dfrac {5x^2-10}{12}=0$

using cross multiplication

$\\\qquad\quad\displaystyle\sf{:}\longrightarrow 5x^2-10=0$

$\\\qquad\quad\displaystyle\sf{:}\longrightarrow 5x^2=10$

$\\\qquad\quad\displaystyle\sf{:}\longrightarrow x^2=\dfrac {10}{5}$

$\\\qquad\quad\displaystyle\sf{:}\longrightarrow x^2=2$

$\\\qquad\quad\displaystyle\sf{:}\longrightarrow x=\sqrt {2}$

$\\\\\therefore\sf x=\sqrt {2}.$