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Geeta M. Astrology trainer in Delhi

Geeta M.

Experienced and Certified Vedic Astrologer from YOGRISH ASTRO RESEARCH INSTITUTE, INDORE

Pusa, Delhi, India - 110012.

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Overview

I am a Vedic Astrologer and I work for Astrosage as well. I pursued my passion for astrology this past year and earned my certificate from Yogesh Astro Research Centre. I've always been fond of astrology since I was a child and now that I am an adult, I am pursuing my dreams of teaching astrology. I love the idea of educating people about astrology and hence, I am here, using urban pro as my way of connecting with likeminded people who are interested in learning about Vedic astrology.

Languages Spoken

Hindi Mother Tongue (Native)

Education

Delhi University 1977

Bachelor of Arts (B.A.)

Yogrish Astro Research Centre 2023

Advance Vedic Astrology

Address

Pusa, Delhi, India - 110012

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Teaches

Astrology Classes

Class Location

Online Classes (Video Call via UrbanPro LIVE)

Student's Home

Tutor's Home

Years of Experience in Astrology Classes

3

Reviews

No Reviews yet!

FAQs

1. Which classes do you teach?

I teach Astrology Class.

2. Do you provide a demo class?

Yes, I provide a free demo class.

3. How many years of experience do you have?

I have been teaching for 3 years.

Answers by Geeta M. (1)

Answered on 26 Feb Learn CBSE/Class 8/Maths/Factorization

To factorize the quadratic expression x2+6x+8x2+6x+8, we're looking for two binomials of the form (x+p)(x+q)(x+p)(x+q) where pp and qq are numbers such that: (x+p)(x+q)=x2+(p+q)x+pq(x+p)(x+q)=x2+(p+q)x+pq In this case, we want p+qp+q to be equal to the coefficient of xx in the given expression (which... ...more

To factorize the quadratic expression x2+6x+8x2+6x+8, we're looking for two binomials of the form (x+p)(x+q)(x+p)(x+q) where pp and qq are numbers such that:

(x+p)(x+q)=x2+(p+q)x+pq(x+p)(x+q)=x2+(p+q)x+pq

In this case, we want p+qp+q to be equal to the coefficient of xx in the given expression (which is 6) and pqpq to be equal to the constant term (which is 8).

Let's find pp and qq:

p+q=6p+q=6 pq=8pq=8

The pairs of numbers that satisfy these conditions are p=2p=2 and q=4q=4.

Therefore, the factorization is:

x2+6x+8=(x+2)(x+4)x2+6x+8=(x+2)(x+4)

 
 
Answers 2 Comments
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Teaches

Astrology Classes

Class Location

Online Classes (Video Call via UrbanPro LIVE)

Student's Home

Tutor's Home

Years of Experience in Astrology Classes

3

No Reviews yet!

Answers by Geeta M. (1)

Answered on 26 Feb Learn CBSE/Class 8/Maths/Factorization

To factorize the quadratic expression x2+6x+8x2+6x+8, we're looking for two binomials of the form (x+p)(x+q)(x+p)(x+q) where pp and qq are numbers such that: (x+p)(x+q)=x2+(p+q)x+pq(x+p)(x+q)=x2+(p+q)x+pq In this case, we want p+qp+q to be equal to the coefficient of xx in the given expression (which... ...more

To factorize the quadratic expression x2+6x+8x2+6x+8, we're looking for two binomials of the form (x+p)(x+q)(x+p)(x+q) where pp and qq are numbers such that:

(x+p)(x+q)=x2+(p+q)x+pq(x+p)(x+q)=x2+(p+q)x+pq

In this case, we want p+qp+q to be equal to the coefficient of xx in the given expression (which is 6) and pqpq to be equal to the constant term (which is 8).

Let's find pp and qq:

p+q=6p+q=6 pq=8pq=8

The pairs of numbers that satisfy these conditions are p=2p=2 and q=4q=4.

Therefore, the factorization is:

x2+6x+8=(x+2)(x+4)x2+6x+8=(x+2)(x+4)

 
 
Answers 2 Comments
Dislike Bookmark

Geeta M. describes herself as Experienced and Certified Vedic Astrologer from YOGRISH ASTRO RESEARCH INSTITUTE, INDORE. She conducts classes in Astrology. Geeta is located in Pusa, Delhi. Geeta takes Online Classes- via online medium. She has 3 years of teaching experience . Geeta has completed Bachelor of Arts (B.A.) from Delhi University in 1977. She is well versed in Hindi.

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