How to Invest in Stocks A Beginner’s Guide

Investing in stocks can be a great way to grow your wealth over time, but it's important to do your research and approach it with a long-term mindset.

Investing is a time- tested way of putting your plutocrat to work for you, as you work to earn further of

it. fabulous investor Warren Buffett defined investing as

“ abstaining consumption now in order to have the capability to consume more at a after date. ”

By investing your plutocrat regularly, you may be suitable to increase it numerous times over with time. That is why it's important to begin investing as early as possible and as soon as you have some plutocrat saved for that purpose. likewise, the stock request is a good place to start.

Here are some steps to get started:

• Educate yourself: Learn the basics of the stock market, how it works, and the different types of stocks and investment strategies. There are plenty of online resources and courses that can help you get started.

• Set your investment goals: Determine how much money you're willing to invest, your risk tolerance, and your long-term financial goals. This will help you choose the right stocks and investment strategies.

• Open a brokerage account: A brokerage account is an online platform where you can buy and sell stocks. Choose a reputable brokerage firm that offers low fees and a user-friendly platform.

• Choose your stocks: Do your research and select individual stocks or exchange-traded funds (ETFs) that align with your investment goals and risk tolerance. Consider factors like the company's financial health, earnings growth, and market trends.

• Monitor your portfolio: Keep an eye on your investments and make adjustments as needed. Be patient and avoid the temptation to make impulsive trades based on short-term market fluctuations.

Remember, investing in stocks always comes with some level of risk, and past performance is not a guarantee of future results. It's important to diversify your portfolio and seek the advice of a financial advisor if you're unsure about any investment decisions. 

Buddhism ssc cgl notes

         BUDDHISM (600BC)

• Buddhism was founded by  Gautama Buddha.
• Buddha was born on the Vaisakh  Purnima day in 563 B.C.
• He belonged to the Sakya clan of Kshatriyas.
• His father was Suddhodana, the ruler of Kapilvastu.
• He was born in Lumbini in Kapilvastu.
• His mother was Mahamaya of the Kosala dynasty.
• Buddha got Nirvana at the age of 35 years.
• Buddha got Nirvana at Uruvela on the bank of river Niranjana
• Buddha gave his first Sermon at Sarnath.
• Buddha's first sermon is called as "Dhama Chakraparivartana
• Buddha died in 483 B.C. at Kushinagar. Kushinagar has been identified with village Kasia in Deoria dis- trict of U.P.
• Buddha's last words were "All com- posite things, strive diligently".
• Buddha was brought up by his stepmothef Gautami.
• After seeing an old man, a sick man, a corpse and an ascetic Buddha decided to become a wanderer. Asvajit,Upali, Mogallana, Sariputra and Ananda weré five disciples of Buddha.

FIVE GREAT EVENTS OF BUDDHA'S LIVE AND THEIR SYMBOLES 

1. Birth: Lotus and Bull
2. Great Renunciation: Horse
3. Nirvana: Bodhi tree
4. First Sermon: Dharmachakra or wheel
5. Parinirvana or Death: Stupa

FOUR NOBLE TRUTH'S 

1. The world is full of sorrows.
2. Desire is root cause of sorrow.
3. If desire is conquered, all sorrows can be removed.
4. Desire can be removed by following the eight-fold path.

EIGHT FOLD PATH

(1)Right understandingg

(2) Right speech

3) Right livelihood

(4)Right mindfulness

(5) Right thought

(6)Right action

(7)Right effort and

(8)Right concentration

THE RATNAS 

1) bhuddhas

2) Dharmha

3) Sangha

Code of Conduct:
(1) Do not covet the property of otherss
(2) Do not commit violence
(3) Do not speak a lie
(4) Do not indulge in corrupt practices

BUDDHIST COUNCLs

• The First Council was held in 483 BC at Saptaparni cave near Rajagriha to compile the Dhamma Pitaka and Vinaya Pitaka. Chairman Mahakassapa, Patron: Ajatshatru

• The Second Council was held at Vaisali in 383 BC. The monks of Vaisali wanted some change in the rites. Schism into Shaviravadins and Mahasanghikas. Chairman : Sabakami, Patron Kalashoka

• The Third Council was held at Pataliputra duríng the reign of Ashoka 236 years after the death of Buddha. It was held under the Presidentship of Moggliputta Tissa to revise the scriptures.

• The Fourth Council was held during the reign of Kanishka in Kashmir under the Presidentship of Vasumitra, who was helped by Aswvaghosha and resulted in the division of Buddhists into Mahayanists and Hinayanists.

SECTS OF BUDDHuSM

Hinayana:

(a) Its followers belleved in the

original teachings of Buddha

b) They sought Individual sal

vation through self-discipline

and meditation.

(c) They did not believe in idol

worship

(d) Hinayana, like Jainism. Is a

religlon without God, Kama

taking the place of God

Mahayana:

(a) Its followers believed in the

heavenliness of Buddha and

sought the salvation of all

through the grace and help

of Buddha and Bodhisatvas

b) Believes in idolworship.

c)Believes that Nirvana is not a

negattve cessation of misery

but a positve state of bliss.

C) Mahayana had two chicf philo

sophical schools: the

Madhyamika and the

Yogachara.

BUDDHIST LITERATURE

The Sutta Pitaka is divided into

five Nikayas.

The five Nikayas are Digh

Nikaya, Majjhima Nikaya.

Samyutta Nikaya, Anguttara

Nikaya and Khuddaka Nikaya.

The Khuddaka Nikaya consists of

large number of miscellanceous  worker.

BUDDHIST ARCHITECTUREE

Buddhist architecture developed

essentially in three forms, viz.

(a) Stupa (relics of the Buddha or

some prominent Buddhist

monks are preserved)

b) Chaitya (prayer hall

(c)Vihara (resldence)

NCRT math class 9th polynomial exersice 2.3

                       EXERSICE 2.3

Question 1.
Find the remainder when x3 + 3x2 + 3x + 1 is divided by
(i) x + 1
(ii) x – 
(iii) x
(iv) x + π
(v) 5 + 2x
     Solution:
     Let p(x) = x3 + 3x2 + 3x +1
(i) The zero of x + 1 is -1.
∴  p(x=-1) = (-1)^3 + 3(-1)^2 + 3(-1) +1
    = -1 + 3- 3 + 1 = 0
   Thus, the  remainder = 0

(ii) The zero of  is 

     Thus, the  remainder = 

iii) The zero of x is 0.
∴ p(0) = (0)^3 + 3(0)^2 + 3(0) + 1
= 0 + 0 + 0 + 1 = 1
Thus, the  remainder = 1.

(iv) The zero of x + π is -π.
       p(-π) = (-π)^3 + 3(- π)^2  + 3(- π) +1
       = -π3 + 3π2 + (-3π) + 1
       = – π3 + 3π2 – 3π +1
       Thus, the required remainder is                        -π3 + 3π2 – 3π+1.

(v) The zero of 5 + 2x is  .


     Thus, the remainder is  .

Question 2.
Find the remainder when x3 – ax2 + 6x – a is divided by x – a.
Solution:
We have, p(x) = x3 – ax2 + 6x – a and zero       of x – a is a.
  ∴ p(a) = (a)3 – a(a)2 + 6(a) – a
  = a3 – a3 + 6a – a = 5a
 Thus, the  remainder is 5a.

Question 3.
Check whether 7 + 3x is a factor of 3x3+7x.
Solution:
We have, p(x) = 3x3+7x. and zero of 7 + 3x is .

Since,( ) ≠ 0
i.e. the remainder is not 0.
∴ 3x3 + 7x is not divisib1e by 7 + 3x.
Thus, 7 + 3x is not a factor of 3x3 + 7x.

NCRT CLASS 9TH MATH CHAPTER 2( POLYNOMIALS EXERSICE 2.2 SOLUTION )

              EXERSICE 2.2 

Question 1.
Find the value of the polynomial 5x – 4x2 + 3 at
(i) x = 0
(ii) x = – 1
(iii) x = 2
Solution:
 *    let  p(x) = 5x – 4x2 + 3
(i) so, p(x=0) = 5(0) – 4(0)2 + 3

                       = 0 – 0 + 3  =3

     Thus, p(x=o) is 3 . 

(ii) p(x=-1) = 5(-1) – 4(-1)2 + 3
                   = – 5x – 4x2 + 3 

                   = -9 + 3 = -6

     hence , the value of 5x – 4x2 + 3 at 

     x =   -1 is -6.

(iii) p(2) = 5(2) – 4(2)2 + 3 

               = 10 – 4(4) + 3
               = 10 – 16 + 3 = -3

                hence , the value of 5x – 4x2 + 3 at x = 2 is – 3.

Question 2.
Find p (0), p (1) and p (2) for each of the following polynomials.
(i) p(y) = y2 – y +1
(ii) p (t) = 2 +1 + 2t2 -t3
(iii) P (x) = x3
(iv) p (x) = (x-1) (x+1)
Solution:
(i) Given that p(y) = y2 – y + 1.

    so,
∴  P(y=0) = (0)2 – 0 + 1 = 0 – 0 + 1 = 1
    p(y=1) = (1)2 – 1 + 1 = 1 – 1 + 1 = 1
    p(Y=2) = (2)2 – 2 + 1 = 4 – 2 + 1 = 3
(ii) Given that p(t) = 2 + t + 2t2 – t3
∴ p(t=0) = 2 + 0 + 2(0)2 – (0)3

    = 2 + 0 + 0 – 0=2
   P(t=1) = 2 + 1 + 2(1)2 – (1)3
   = 2 + 1 + 2 – 1 = 4
  p(t=2) = 2 + 2 + 2(2)2 – (2)3
   = 2 + 2 + 8 – 8 = 4
(iii) Given that p(x) = x3
∴ p(x=0) = (0)3 = 0, p(1) = (1)3 = 1
    p(x=2) = (2)3 = 8
(iv) Given that p(x) = (x – 1)(x + 1)
∴ p(x=0) = (0 – 1)(0 + 1) = (-1)(1) = -1
   p(x=1) = (1 – 1)(1 +1) = (0)(2) = 0
   P(x=2) = (2 – 1)(2 + 1) = (1)(3) = 3

Question 3.

Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x) = 3x + 1,x = –
(ii) p (x) = 5x – π, x = 
(iii) p (x) = x2 – 1, x = x – 1
(iv) p (x) = (x + 1) (x – 2), x = – 1,2
(v) p (x) = x2, x = 0
(vi) p (x) = 1x + m, x = – 
(vii) P (x) = 3x2 – 1, x = – ,
(viii) p (x) = 2x + 1, x = 
Solution:
(i) We have , p(x) = 3x + 1

(ii) We have, p(x) = 5x – π
∴ 
(iii) We have, p(x) = x2 – 1
∴ p(1) = (1)2 – 1 = 1 – 1=0
Since, p(1) = 0, so x = 1 is a zero of x2 -1.
Also, p(-1) = (-1)2 -1 = 1 – 1 = 0
Since p(-1) = 0, so, x = -1, is also a zero of x2 – 1.

(iv) We have, p(x) = (x + 1)(x – 2)
∴ p(-1) = (-1 +1) (-1 – 2) = (0)(- 3) = 0
Since, p(-1) = 0, so, x = -1 is a zero of (x + 1)(x – 2).
Also, p( 2) = (2 + 1)(2 – 2) = (3)(0) = 0
Since, p(2) = 0, so, x = 2 is also a zero of (x + 1)(x – 2).

(v) We have, p(x) = x2
∴ p(o) = (0)2 = 0
Since, p(0) = 0, so, x = 0 is a zero of x2.

(vi) We have, p(x) = lx + m

(vii) We have, p(x) = 3x2 – 1

(viii) We have, p(x) = 2x + 1
∴ 
Since,  ≠ 0, so, x =  is not a zero of 2x + 1.

Question 4.
Find the zero of the polynomial in each of the following cases
(i) p(x)=x+5
(ii) p (x) = x – 5
(iii) p (x) = 2x + 5
(iv) p (x) = 3x – 2
(v) p (x) = 3x
(vi) p (x)= ax, a≠0
(vii) p (x) = cx + d, c ≠ 0 where c and d are real numbers.
Solution:
(i)  p(x) = x + 5. 

     Since, p(x) = 0
⇒ x + 5 = 0
⇒ x = -5.
    Thus, zero of x + 5 is  -5.

ii) P(x)=x - 5

    Since, p(x)=0.

     x-5=0 ⇒x=5.

     hence, zero of x - 5 is 5.

 iii)p(x) = 2x+5

      Since, p(x)=0.

      2x+5=0⇒2x=-5

       x=-5/2.

      hence zero of 2x+5 is -5/2.

iv)  p(x)= 3x-2

      Since, p(x)=0

       3x-2=0⇒x=2/3

       hence zero of 3x-2  is 2/3.

v)   p(x)= 3x

       Since, p(x)=0

       3x=0 ⇒x=0

       hence zero of 3x is 0.

vi)   p(x) =ax

        since ,p(x) =0

        ax=o ⇒x=0

         hence zero of ax is 0.

vii)  p(x)=cx+d

        since ,p(x) =0

        cx+d =0 ⇒ cx=-d

       x=-d/c

        hence zero of cx+d is -d/c