भारतीय अर्थव्यवस्था पर महामंदी का क्या प्रभाव पड़ा​

Explanation:

a ii ,b iv,c iii,d v it is the answer

Answer:

a ii,b vi,c iii ,dv is the answer

Explanation:

Self-reliance is the ability to do things and make decisions by yourself, without needing other people to help you. Self-reliance means developing the right skills and capabilities to do one’s activities without depending on outside support.

Self-Reliance is having confidence in our abilities and being able to do things for ourselves. Covid-19 pandemic has taught a lesson to the world that being self-reliant is important.

n terms of teaching and learning, I want to share my ideas on making me and my teacher’s Self-Reliance (Aatmanirbhar) when the schools are not fully functional due to covid-19.

I should avoid blaming the situation like the covid-19 pandemic and take corrective actions on my own. I should develop the ability to think independently. I should follow strategies and tips that can help in learning.

I should understand the purpose of learning by setting argets. I should plan efficiently in the learning process. I should build confidence in teachers and their method of teaching.

On the other hand… Teachers must create a safe and positive learning environment. Teachers should be proactive in forming an action plan and implementing it. Teachers should develop their skills of teaching especially at times of online teaching.

Teachers should customize the templates, content-related videos, whiteboards, etc.., Teachers should know about their strengths and limitations. Teachers should have a positive attitude and they should learn from their mistak

In this pandemic time, teachers must be self-relian and they should encourage students to be self-reliant. This encourages emotional strength and helps to face challenges like the covid-19 pandemic which ultimately results in Self Reliant School for Self Reliant India.es.

The answer to the given question is written below.

Self Reliance

Self Reliance refers to the act of relying on no one but self. When one relies upon self other than anybody else, he/she becomes independent and can tackle any situation which life throws at him/her. We need self reliant teachers to tackle with the digital era so, that they do not have to depend on others to conduct operations. This will also provide them competitive advantage over others.

Answer:

Mrs Fitzgerald is a neighbour of Mrs Pearson. She is a fortune teller. She leared this art from the East where her husband had served in the army. She also knows the art of magic.

Explanation:

[tex]\large\underline{\sf{Solution-}}[/tex]

Given that, an arch is in the form of a semi ellipse, which is 50 metres wide at the base and has a height of 20 metres.

Let assume that the equation of ellipse be

[tex]\rm \: \dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} } = 1 \\ [/tex]

As it is given that, it is 50 m wide from the base and 20 m high from base.

[tex]\rm\implies \:2a = 50 \: \: \: and \: \: \: b = 20 \\ [/tex]

[tex]\rm\implies \:a = 25 \: \: \: and \: \: \: b = 20 \\ [/tex]

So, equation of ellipse can be rewritten as

[tex]\rm \: \dfrac{ {x}^{2} }{ {25}^{2} } + \dfrac{ {y}^{2} }{ {20}^{2} } = 1 \\ [/tex]

[tex]\rm \: \dfrac{ {x}^{2} }{625} + \dfrac{ {y}^{2} }{400} = 1 - - - (1) \\ [/tex]

Let assume that semi - ellipse is 2c metres wide at the height of 10 m from the base. So, it means (c, 10) lies on equation (1).

So,

[tex]\rm \: \dfrac{ {(c)}^{2} }{625} + \dfrac{ {10}^{2} }{400} = 1 \\ [/tex]

[tex]\rm \: \dfrac{ {c}^{2} }{625} + \dfrac{100}{400} = 1 \\ [/tex]

[tex]\rm \: \dfrac{ {c}^{2} }{625} + \dfrac{1}{4} = 1 \\ [/tex]

[tex]\rm \: \dfrac{ {c}^{2} }{625} = 1 - \dfrac{1}{4} \\ [/tex]

[tex]\rm \: \dfrac{ {c}^{2} }{625} = \dfrac{4 - 1}{4} \\ [/tex]

[tex]\rm \: \dfrac{ {c}^{2} }{625} = \dfrac{3}{4} \\ [/tex]

[tex]\rm \: {c}^{2} = \dfrac{3 \times 625}{4} \\ [/tex]

[tex]\rm\implies \:c \: = \: \dfrac{25 \sqrt{3} }{2} \: m \\ [/tex]

Hence,

[tex]\rm\implies \:Width = 2c \: = \: 25 \sqrt{3} \: m \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information :-

[tex]\begin{gathered}\boxed{\begin{array}{c|c} \bf & \bf \dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} } = 1 , \: (a > b)\\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf Vertex & \sf ( \pm \: a, \: 0) \\ \\ \sf Focus & \sf ( \pm \: ae, \: 0)\\ \\ \sf eccentricity & \sf e = \sqrt{1 - \dfrac{ {b}^{2} }{ {a}^{2} } } \\ \\ \sf Length \: of \: major \: axis & \sf 2a\\ \\ \sf Length \: of \: minor \: axis & \sf 2b\\ \\ \sf Length \: of \: latus \: rectum & \sf \dfrac{ {2b}^{2} }{a} \\ \\ \sf Distance \: between \: focus & \sf 2ae \end{array}} \\ \end{gathered} \\ [/tex]