Class 4 ex1 Q2

 Chapter no. 1

Exercise # 1.1

Q.NO.2 Darjzail kasoor ko fesad ki shakal mein likhiye 

vii. 23/60

=23x100/60

=2300/60%

=38.3%

viii. 8/3

=8x100/3

=800/3%

=266.7%

ix. 8/5

=8x100/5

=800/5%

=160%

x. 7/8

=7x100/8

=700/8%

=87.5%

xi. 5/8

=5x100/8

=500/8%

=62.5%

xii. 3/8

=3x100/8

=300/8%

=37.5%

Class 3 ex1 Q2

 Chapter no. 1

Exercise # 1.1

Q.NO.2 Darjzail kasoor ko fesad ki shakal mein likhiye 

i. 3/4

=3x100/4

=300/4%

=75%

ii. 3/5

=3x100/5

=300/5%

=60%

iii. 4/25

=4x100/25

=400/25%

=16%

iv. 13/20

=13x100/20

=1300/20%

=65%

v. 31/25

=31x100/25

=3100/25%

=124%

vi. 21/40

=21x100/40

=2100/40%

=52.5%

Class 2 ex1 Q1

Chapter no. 1

Exercise # 1.1

Q.NO.1   Darjzail fesad ko kasroon ki asan shakal mein waziah kijiye

viii.  33   1/2%
=33.5
Point  khatm krny sy
=335/10
=335/10x100
=335/1000
5 pr divide krny sy
=67/200

ix. 37   1/2%
=37.5
Point khatm krny sy
=375/10
=375/10x100
=375/1000
5 pr divide krny sy
=75/200
5 pr divide krny sy
=15/40
5 pr divide krny sy
=3/8

x. 87  1/2%
=87.5
Point khatm krny sy
=875/10
=875/10x100
=875/1000
5 pr divide krny sy
=175/200
5 pr divide krny sy
=35/40
5 pr divide krny sy
=7/8
xi. 5  1/4%
=5.25
Point khatm krny sy
=525/100
=525/100x100
=525/10000
5 pr divide krny sy
=105/2000
5 pr divide krny sy
=21/400

xii. 42  1/2%

=42.5
Point khatm krny sy
=425/10
=425/10x100
=425/1000
5 pr divide krny sy
=85/200
5 pr divide krny sy
=17/40

Class 1 ex1 Q1

 Chapter no. 1

Exercise # 1.1

Q.NO.1   Darjzail fesad ko kasroon ki asan shakal mein waziah kijiye

---

i.     95%

= 95/100

5 pr divide krny  sy

=19/20

---

ii.   65%   

=65/100

5 pr divide krny sy

=13/20

---

iii. 75%   

=75/100

5 pr divide krny sy

=3/4

---

iv. 25%   

=25/100

5 pr divide krny sy

=1/4

---

v. 56%r 

=56/100

2 pr divide krny sy

=28/50

2 pr divide krny sy

=14/25

---

vi. 48% 

=48/100

2 pr divide krny sy

=24/50

2 pr divide krny sy

=12/25

---

vii. 8%   

=8/100

2 pr divide krny sy

=4/50

2 pr divide krny sy

=2/25

Slide 1

Slide 2

Slide 3

Slide 1

Slide 2

Slide 3

.slider-container { width: 400px; height: 300px; perspective: 1000px; margin: 0 auto; } .slider { width: 100%; height: 100%; position: relative; transform-style: preserve-3d; animation: slide-rotate 15s infinite; box-shadow: 0 0 20px rgba(0, 0, 0, 0.3); } .slide { position: absolute; width: 90%; height: 90%; top: 5%; left: 5%; backface-visibility: hidden; transform-style: preserve-3d; display: flex; justify-content: center; align-items: center; color: white; font-size: 24px; font-weight: bold; box-shadow: 0 2px 6px rgba(0, 0, 0, 0.3); border-radius: 10px; transform: translateZ(-200px); } .slide1 { background-color: #ff6384; } .slide2 { background-color: #36a2eb; transform: translateZ(0); } .slide3 { background-color: #cc65fe; transform: translateZ(200px); } @keyframes slide-rotate { 0% { transform: rotateY(0); } 25% { transform: rotateY(-90deg); } 50% { transform: rotateY(-180deg); } 75% { transform: rotateY(-270deg); } 100% { transform: rotateY(-360deg); } }

Certainly! The code provided is an HTML and CSS implementation of a 3D slider. Here's a description of the code and its functionality:

HTML:

- The slider is contained within a `<div>` element with the class "slider-container".

- Inside the container, there is another `<div>` element with the class "slider" that holds the slides.

- The slides are represented by `<div>` elements with the classes "slide slide1", "slide slide2", and "slide slide3". Each slide contains an `<h1>` element with the slide's content.

CSS:

- The ".slider-container" class sets the dimensions of the slider container and applies a perspective effect to create the 3D view.

- The ".slider" class represents the container for the slides and specifies its positioning and appearance.

- The ".slide" class represents each individual slide. It sets the position, dimensions, and styling of the slides, such as their background color, font size, font weight, box shadow, and border radius.

- The ".slide1", ".slide2", and ".slide3" classes define the specific styles for each slide.

- The "@keyframes slide-rotate" rule specifies the animation of the slider. It rotates the slider around the Y-axis at different percentages to create a spinning effect.

The code positions the slides in a 3D space using the `translateZ` property. Each slide is positioned at a different distance from the viewer, allowing them to overlap and create a 3D effect. The slides rotate around the Y-axis to create a spinning motion.

Please note that this code provides a basic foundation for a 3D slider, and you can further customize it according to your preferences by adjusting the styles, dimensions, animation timings, or adding additional functionality.

Let me know if you need any further clarification or assistance!

find the interval satisfying the inequality 4 - (1/2)n < -7 + (1/4)n,

To find the interval satisfying the inequality 4 - (1/2)n < -7 + (1/4)n, we can solve it step by step. Let's begin:

4 - (1/2)n < -7 + (1/4)n

First, let's eliminate the fractions by multiplying the entire inequality by the least common multiple (LCM) of the denominators, which is 4. This gives:

4 * (4 - (1/2)n) < 4 * (-7 + (1/4)n)

Simplifying:

16 - 2n < -28 + n

Now, let's isolate the variable n on one side of the inequality. We can do this by adding 2n to both sides and adding 28 to both sides:

16 - 2n + 2n < -28 + n + 2n + 28

16 < 3n

Finally, divide both sides of the inequality by 3:

16/3 < n

So, the solution to the inequality is n > 16/3.

Therefore, the interval satisfying the inequality is (16/3, ∞), where n is greater than 16/3.

by using methmatical induction , prove that the eqvation 2+4+6+....2n =n(n+1) is true for all positive integers

 To prove the equation 2 + 4 + 6 + ... + 2n = n(n + 1) for all positive integers using mathematical induction, we need to show that it holds for the base case (n = 1) and then demonstrate that if it holds for some arbitrary positive integer k, it also holds for k + 1.

**Step 1: Base Case**

Let's check the equation for the base case when n = 1:

2 = 1(1 + 1)

2 = 1(2)

2 = 2

The equation holds true for the base case.

**Step 2: Inductive Hypothesis**

Assume that the equation holds true for some arbitrary positive integer k, that is, assume that:

2 + 4 + 6 + ... + 2k = k(k + 1)

**Step 3: Inductive Step**

We need to show that if the equation holds for k, it also holds for k + 1.

We start with the left-hand side (LHS) of the equation:

2 + 4 + 6 + ... + 2k + 2(k + 1)

Using the assumption from the inductive hypothesis:

= k(k + 1) + 2(k + 1)

Factoring out (k + 1):

= (k + 1)(k + 2)

This matches the right-hand side (RHS) of the equation, which is n(n + 1) when n = k + 1:

= (k + 1)((k + 1) + 1)

= (k + 1)(k + 2)

Therefore, if the equation holds for k, it also holds for k + 1.

**Step 4: Conclusion**

By proving the base case and showing that if it holds for k, it also holds for k + 1, we have established that the equation 2 + 4 + 6 + ... + 2n = n(n + 1) is true for all positive integers by mathematical induction.