Chapterwise weightage of Social Science class 9 | 2023-24

 Chapterwise weightage of social science.

1. History

Chapter 1 - 5 marks

Chapter 2 - 6 marks 

Chapter 3 - 7 marks 

Map - 2 marks

2. Democratic Politics 

Chapter 1 - 3 marks 

Chapter 2 - 3 marks 

Chapter 3 - 3 marks 

Chapter 4 - 5 marks 

Chapter 5 - 6 marks 

3. Geography 

Chapter 1 - 3 marks 

Chapter 2 - 5 marks 

Chapter 3 - 2 marks 

Chapter 4 - 4 marks 

Chapter 6 - 3 marks  

Map - 3 marks 

4. Economics

Chapter 2 - 6 marks 

Chapter 3 - 6 marks 

Chapter 4 - 8 marks

Find all positive integers x such that there exists a positive integer y satisfying: 1/x + 1/y = 1/7

 

Find all positive integers x such that there exists a positive integer y satisfying: 1/x + 1/y = 1/7

Answer.

1/x + 1/y = 1/7

(x+y)/xy = 1/7

7x + 7y = xy

xy - 7x - 7y = 0

x(y-7) - 7(y-7) = 49

(x-7)(y-7) = 49

Thus satisfying values are:

49   *   1

     1   *   49

   7   *   7

So values of x and y can be, (42, 8), (8, 42), (14, 14).

Find 3x2y2 if x and y are integers such that y2 + 3x2y2 = 30x2 + 517.

Find 3x²y² if x and y are integers such that y² + 3x²y² = 30x² + 517.

Answer.

y² + 3x²y² - 30x² – 517 = 0

3x²(y²-10) + 1(y²-10) + 10 – 517 = 0

(3x²+1)(y²-10) = 507

(3x²+1)(y²-10) = 3*13*13

To x and y be integers,

3x²+1 = 13 and y²-10 = 39

3x² = 12 and y² = 49

x² = 4  and y² = 49

To find 3x²y² we will substitute the value of x² and y². 3*4*49 = 588.

 Fortuna Audentium: De virtute et superandis periculis

Fortuna audentes iuvat, sed timidos repellit. Non in casu, sed in virtute felicitas sita est. Obstacles cedere fortis animus debet, neque segnescere periculis. Nam ex magnis difficultatibus maximae glories nascuntur. Quamvis tenuis ignis, si alitur, magnus potest evadere; quamvis robusta arbor, si negligitur, facile exarescit. Ita et ingenium nostrum, nisi colitur, languescit; contra vero, si exercetur, ad summa quaeque contendere potest. Ergo, animi fortitudo et labor assiduus ad res pulchras et magnas sunt necessariae. Surge igitur, amice, et amplectere vitam cum ardore! Fiducia in te ipso habe et nihil impossibile tibi videatur?

Let p(x) be a cubic polynomial such that (p(x))2 + (x4 + 2x2 + 2x)2 = f(x) = (x2+1)(x2+4)(x2- 2x+2)(x2+2x+2). Find the value of |p(-3)|.

Let p(x) be a cubic polynomial such that

(p(x))² + (x⁴ + 2x² + 2x)² = f(x) = (x²+1)(x²+4)(x²-

2x+2)(x²+2x+2). Find the value of |p(-3)|.

Answer.

Put f(-3)

(p(-3))² + ((-3)⁴ + 2(-3)² + 2(-3))² = f(-3) = ((-3)²+1)

((-3)²+4)((-3)²-2(-3)+2)((-3)²+2(-3)+2)

 

(p(-3))² + (81 + 18 - 6)² = (9+1)(9+4)(9 + 6 +2)(9 – 6

 + 2)

 

(p(-3))² + 8649 = (10)(13)(17)(5)

 

(p(-3))² = 11050 – 8649

 

(p(-3))² = 2401

 

p(-3) = ± 49

 

We have to find |p(-3)| . So, |± 49| = 49 = |p(-3)|

 

Let a and b be real number that satisfy

a⁴ + a²b² + b⁴ = 900

a² + ab + b² = 45

Find the value of 2ab.

Answer.

Let a⁴ + a²b² + b⁴ = 900  ------- eq(1)

& a² + ab + b² = 45 ------- eq(2)

Simplify eq(1),

Add and Subtract a²b² for square completion,

(a²)² + 2(a²)(b²) + (b²)² - (ab)² = 900

(a² + b²)² - (ab)² = 900

(a² + b² + ab)(a² + b² - ab ) = 900 ------- eq(3)

We know that, a² + b² = 45 – ab ------- eq(4)

Put eq(4) in eq(3)

(45 - ab + ab)(45 - ab - ab ) = 900

(45 - 2ab)(45) = 900

45 - 2ab = 20

-2ab = -25

2ab = 25

Let a_{1}, a_{2}, ….., be a sequence defined by a_{1} = 1 for n>=1, an+1 = √(a_{n}^2 – 2a_{n} + 3) + 1. Find a_{513}.

 Let a₁ + a₂, ….., be a sequence defined by a₁ = 1 for n>=1, a_n+1 = √(a_n² – 2a_n + 3) + 1. Find a_513.

_Ans. 

First simplify: √(a_n² – 2a_n + 3) + 1

√(a_n² – 2a_n + 1 + 2) + 1

√((a_n – 1)² + 2) + 1

Lets first find out a₂:

Put n = 1

a₁₊₁ = √(a₁² – 2a₁ + 3) + 1

a₂  = √((1-1)² + 2) + 1

a₂ = √2 + 1

Likewise put n = 2

a₂₊₁ = √((a₂ – 1)² + 2) + 1

a₃ = √((√2 + 1 – 1)² + 2) + 1

a₃ = √(4) + 1

a₃ = 3

a₃₊₁ = √((a₃ – 1)² + 2) + 1

a₄ = √((3 – 1)² + 2) + 1

a₄ = √6 + 1

a_{4 + 1} = √((a₄ – 1)² + 2) + 1

a₅ = √((√6 + 1 – 1)² + 2) + 1

a₅ = √8 + 1

Now, we can notice that as we put the values of n

we get the result in the surd in the form of:

 √(2*n) + 1

To get a₅₁₃, we have to put n = 512

So, the result directly come out to be 

                                                           = √(2*512) + 1

                                                           = √1024 + 1

                                                           = 32 + 1

                                                           = 33

So, a₅₁₃ = 33

CBSE Class 9 Science Chapter 5 - Revision Notes - followrudra

• Cell: The basic structural and functional unit of life, discovered by Robert Hooke in 1665. Cells can be classified into two types: prokaryotic and eukaryotic.
• Prokaryotic cells: Primitive cells that lack a well-defined nucleus and membrane-bound organelles. Examples are bacteria and cyanobacteria.
• Eukaryotic cells: Advanced cells that have a well-defined nucleus and membrane-bound organelles. Examples are plant and animal cells.
• Cell membrane: The outermost covering of the cell that separates the contents of the cell from its external environment. It is selectively permeable and allows the entry and exit of some materials by diffusion and osmosis.
• Cell wall: An additional rigid layer present in plant cells, made up of cellulose. It provides shape and support to the cell.
• Nucleus: The control center of the cell that contains the genetic material (DNA) and regulates the cell’s activities.
• Cytoplasm: The jelly-like substance that fills the cell and contains various organelles such as mitochondria, chloroplasts, endoplasmic reticulum, Golgi apparatus, lysosomes, etc.
• Mitochondria: The powerhouse of the cell that produces energy (ATP) by cellular respiration.
• Chloroplasts: The green organelles present in plant cells that perform photosynthesis and produce food (glucose) and oxygen.
• Endoplasmic reticulum: A network of membranous tubes that transport materials within the cell. It can be smooth (without ribosomes) or rough (with ribosomes).
• Golgi apparatus: A stack of flattened sacs that modify, sort, and package proteins and other substances for secretion or storage.
• Lysosomes: The digestive system of the cell that contain enzymes to break down waste materials and foreign invaders.
• Vacuoles: The storage sacs of the cell that store water, food, waste, or other substances. Plant cells have a large central vacuole, while animal cells have smaller vacuoles.

"Guardians of the Virtual Gate: Unveiling the Cybersecurity Challenges and Solutions in the Metaverse"

Introduction:

As the world enthusiastically embraces the concept of the metaverse, a digital universe where augmented and virtual realities converge, it becomes imperative to address the looming concerns surrounding cybersecurity. In this article, we delve into the potential risks that users may face in the metaverse and explore innovative solutions to fortify the virtual realm against cyber threats.

1. The Metaverse's Vulnerability Landscape:

   - Unveiling potential cybersecurity threats specific to the metaverse.

   - Analysis of phishing, identity theft, and other cybercrimes within the virtual space.

   - How the interconnectedness of the metaverse amplifies security challenges.

2. Protecting Your Digital Identity:

   - Best practices for securing personal and financial information in the metaverse.

   - The role of decentralized technologies (blockchain) in enhancing identity protection.

   - Exploring biometric authentication and other advanced security measures.

3. Virtual Assets and Currency Security:

   - Risks associated with virtual currencies and digital assets in the metaverse.

   - Blockchain's role in securing transactions and preventing fraud.

   - Case studies of virtual heists and lessons learned for users and developers.

4. Building Fortresses in the Digital Space:

   - Examining cybersecurity measures implemented by leading metaverse platforms.

   - The importance of end-to-end encryption and secure communication protocols.

   - How virtual security mirrors and differs from traditional cybersecurity practices.

5. Emerging Technologies and Cybersecurity Solutions:

   - Incorporating AI and machine learning to detect and prevent virtual threats.

   - The role of ethical hacking and bug bounty programs in fortifying metaverse platforms.

   - Future developments in cybersecurity technologies for the evolving metaverse.

6. User Education and Awareness:

   - The significance of educating metaverse users about cyber risks.

   - Tips for safe navigation in the virtual realm.

   - Raising awareness about potential threats and how to report incidents.

Conclusion:

As the metaverse continues to redefine the way we interact with digital spaces, ensuring a secure and resilient environment is paramount. This article aims to equip readers with the knowledge needed to navigate the metaverse safely, emphasizing the collaborative efforts required from users, developers, and technology innovators to build a cyber-secure future in the virtual world.

Phaser Game | HTML, CSS, JS

<!DOCTYPE html>

<html lang="en">

<head>

    <meta charset="UTF-8">

    <meta name="viewport" content="width=device-width, initial-scale=1.0">

    <title>Phaser Game</title>

    <script src="https://cdn.jsdelivr.net/npm/phaser@3.55.2/dist/phaser.js"></script>

</head>

<body>

    <script>

        const config = {

            type: Phaser.AUTO,

            width: 800,

            height: 600,

            physics: {

                default: 'arcade',

                arcade: {

                    gravity: { y: 0 },

                    debug: false

                }

            },

            scene: {

                preload: preload,

                create: create,

                update: update

            }

        };

        const game = new Phaser.Game(config);

        let player;

        function preload() {

            this.load.image('spaceship', 'path/to/spaceship.png');

        }

        function create() {

            player = this.physics.add.image(400, 500, 'spaceship');

            player.setCollideWorldBounds(true);

        }

        function update() {

            const cursors = this.input.keyboard.createCursorKeys();

            if (cursors.left.isDown) {

                player.setVelocityX(-160);

            } else if (cursors.right.isDown) {

                player.setVelocityX(160);

            } else {

                player.setVelocityX(0);

            }

            if (cursors.up.isDown) {

                player.setVelocityY(-160);

            } else if (cursors.down.isDown) {

                player.setVelocityY(160);

            } else {

                player.setVelocityY(0);

            }

        }

    </script>

</body>

</html>