A perfect square is a number that is obtained when a whole number is multiplied by itself. In other words, it is the result of squaring an integer. For example, $\mathrm{1=1\times 1}$ \mathrm{1=1\times 1}$$1=1×1, $\mathrm{4=2\times 2}$ \mathrm{4=2\times 2}$$4=2×2, $\mathrm{9=3\times 3}$ \mathrm{9=3\times 3}$$9=3×3, and $\mathrm{16=4\times 4}$ \mathrm{16=4\times 4}$$16=4×4. These numbers are called perfect squares because they can be arranged to form a perfect square shape.

Perfect squares are important in mathematics because they help us understand square roots, area, and patterns in numbers. The square root of a perfect square is always a whole number. For instance, the square root of 25 is 5, since $\mathrm{5\times 5=25}$ \mathrm{5\times 5=25}$$5×5=25.

in quadratic expressions, a perfect square occurs when the expression can be written as the square of a binomial. This means the quadratic can be expressed in the form:$$(a+b)2\text{or}(a-b)2$$

When expanded, these forms follow special patterns:$$(a + b)^2 = a^2 + 2ab + b^2$$$$(a – b)^2 = a^2 – 2ab + b^2$$

A quadratic expression is called a perfect square trinomial if:

1. The first term is a perfect square.
2. The last term is a perfect square.
3. The middle term is twice the product of the square roots of the first and last terms.

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Examples

x²+6x+9 is a perfect square

x² = (x) x (x) =(x)²

9 = 3 x 3 = (3)²

middle term : 6x = 2(x)(3)

Therefore:

x²+6x+9 = (x+3)²

Example 2

assume x²−10x+25 is a perfect square

x² = (x) (x) = (x)²

25 = (5)²

middle term:-10x = -2(x)(5)

therefore x²−10x+25 = (x-5)²

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Importance in Solving Quadratics

Perfect square expressions are useful because:

• They make factorization easier.
• They help in completing the square, a method used to solve quadratic equations.
• They simplify graphing, since expressions like $(x – a)^2$ clearly show the vertex of a parabola.

Example problem

factorize x² – 8x +16

solution

The constant term is a perfect square

middle term is twice the constant term

hence 16 = (4)(4)

-8 = -4-4

hence x² – 8x +16 = (x-4)(x-4) = (x-4)²

example

factorize:

solution

Example

factorize 4x²+20x+25

solution

4x²= (2x)(2x)

25=(5)(5)

hence 4x²+20x+25 = (2x+5)(2x+5) = (2x+5)²

completing the square

Completing the square is a method used to solve quadratic equations and rewrite quadratic expressions in a special form. It helps us change a quadratic expression into a perfect square trinomial, which can then be factored easily.

consider the expression x² +bx

we can add a constant term such that the expression becomes a perfect square.

To complete the square:

Take half of ,square it and then add it to the expression.

that is: add (b/2)² to the expression

Example problem on completing the square

what must be added to x² -18x to make it a perfect square?

so x² -18x is transformed to x² -18x+81 to be a perfect square.

Related Topics

• Introduction to matrices
• Gradients of line
• Quadratic equation

Upthrust in gases is the upward force exerted on an object immersed in a gas, which opposes the object’s weight. Upthrust in gases acts as an upward force on an object immersed in a gas and opposes the object’s weight. It arises from the pressure difference between the top and bottom of the object due to the gas’s density. This concept is similar to upthrust in liquids but involves the behavior of gases, which are much less dense than liquids.

When an object is placed in a gas, the pressure at the bottom of the object is greater than the pressure at the top. This difference in pressure results in an upward force (upthrust). This is what that results to upthrust in gases. The size of this force depends on the volume of the displaced gas and the density of the gas.

Just like liquids, gases exerts upthrust on objects that are in them. Air is the most common gas of interest and so it will be used extensively to illustrate upthrust in gases. We float objects on air. Sometimes, as human beings, we use parachutes to float in air.

The upthrust in gases is small because air has lower density compared to most of substances.

The density of air is about 1.3kgm^-3 or 0.0013gcm^-3.

A balloon can float in air if it contains a gas with a lower density than air. For example hydrogen has a density of 0.09 kgm^-3 whereas helium has a density of 0.18kg^m-3. Therefore a balloon filled with helium or hydrogen will rise on air provided density of the balloon fabric and air will be less than density of air.

Problems involving upthrust in gases

Consider the figure below that illustrates a balloon filled with air .

If we consider the balloon filled with air to a certain volume, the weight of air in the balloon plus its fabric is greater than the weight of air displaced. This is because the volume of air in the balloon is nearly equal to the volume of air displaced.

The upthrust force on the balloon due to the air is thus less than the weight. The balloon thus stay grounded because the it’s weight is less than the upthrust force that could set it up to float on air.

That is W-U > resultant downward forces.

If the balloon is filled with a gas that has a lower density than air, the gas and balloon fabric weigh less than the displaced air. The upthrust force U exerted by the air on the balloon is greater than the weight W of the inflated balloon. This results in the upthrust force being larger. The resultant upward force is greater than W-U and hence the balloon sets to accelerate upward.see the illustrations below.

Example Question

A meteorological balloon has a volume of of 55m³ and is filled with a helium gas of density 0.18kgm^-3. If the weight of the balloon fabric is 170N, calculate the maximum load the balloon can lift given that density of air is 1.3kgm^-3

solution

Volume of air displaced by the balloon is 55m³ which is volume occupied by the balloon.

mass of air displaced by the balloon = 55m³ x 1.3kgm^-3 = 71.5kg

weight of air displaced =71.5kg x 10 Nkg^-1 = 715N

mass of helium in the balloon = 55m³ x 0.18kgm^-3 = 9.9Kg

weight of helium in the ballon = 9.9kg x 10 Nkg^-1 = 99N

Total weight of the inflated balloon = 170N + 99N = 269 N

From the law of floatation, upthrust is the weight of air displaced which should also be weight of the balloon plus the weight of load it will carry. hence,

upthrust = weight of the balloon + load in the ballon = weight of air displaced

269N + load in the ballon = 715N

load in the balloon = (715-269)N=446N

So the total mass of the goods to be included in the balloon should never exceeded 44.6kg.

Related topics

• The Archimedes’ principle
• The law of floatation
• Causes of upthrust
• Density of some substances
• Explaining Upthrust
• Density: Concise quick Introduction
•  Volume of Regularly-shaped Solids

Alphabetical letters are the basic symbols used to form words and communicate ideas in written language. They help us read, write, and understand information in an organized and meaningful way.

capita Alphabetical letters

small alphabetical letters

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• foods
• Alphabets

Kenya has a rich tapestry of history and heritage — from ancient Swahili cities and prehistoric fossil sites to colonial forts and caves that tell the story of the fight for independence. Here are some notable historical sites in Kenya you might find fascinating:

Related topics

• Introduction to pressure
• Bomas Of Kenya

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1. Introduction

Gases, just like liquids, are fluids and can exert forces on objects placed in them. Although air cannot be seen, it has mass, occupies space, and exerts pressure. When an object is placed in air, it experiences an upward force known as upthrust. This force explains why balloons rise, smoke moves upward, and parachutes slow down falling objects.

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2. What Is Upthrust?

Upthrust (buoyant force) is the upward force exerted by a fluid (liquid or gas) on an object immersed in it.

In gases, upthrust occurs because air pressure increases with depth. The pressure acting on the bottom of an object is greater than the pressure acting on the top, resulting in a net upward force.

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3. Archimedes’ Principle in Gases

Archimedes’ principle also applies to gases and states that:

An object wholly or partially immersed in a gas experiences an upthrust equal to the weight of the gas displaced.

This means the greater the volume of air displaced by an object, the greater the upthrust acting on it.

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4. Mathematical Expression for Upthrust in Gases

The upthrust in gases can be calculated using the formula:$$U = \rho g V$$U=ρgV

Where:

• $U$U = upthrust (N)
• $\rho$ρ = density of the gas (kg/m³)
• $g$g = acceleration due to gravity (m/s²)
• $V$V = volume of gas displaced (m³)

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5. Conditions for Floating, Rising, and Falling in Air

Comparison of Forces	Result
Upthrust > Weight of object	Object rises
Upthrust = Weight of object	Object floats
Upthrust < Weight of object	Object falls

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6. Illustrations of Upthrust in Gases

(a) Balloons

A helium or hot-air balloon rises because it displaces air that is denser than itself. The upthrust acting on the balloon is greater than its weight, causing it to rise upward.

(b) Parachutes

When a parachute opens, it increases the surface area and displaces more air. The upthrust (and air resistance) acting upward increases, reducing the downward motion of the parachutist and allowing a safe landing.

(c) Smoke and Hot Air

Hot air and smoke rise because they are less dense than the surrounding cooler air. The upthrust acting on them is greater than their weight.

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7. Worked Examples

Example 1:

An object displaces 0.1 m³ of air. Given that its’ density of air = 1.2 kg/m³ and the acceleration due to gravity = 10 m/s². calculate upthrust in air.

Calculate the upthrust acting on the object.

Solution:$$U = \rho g V$$U=ρgV $$U = 1.2 \times 10 \times 0.1$$U=1.2×10×0.1 $$U = 1.2 \, \text{N}$$U=1.2N

Upthrust = 1.2 N

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Example 2: Determining Motion of an Object in Air

A balloon has a weight of 0.9 N.
Upthrust acting on it is 1.2 N.

Since:$$\text{Upthrust} > \text{Weight}$$Upthrust>Weight

The balloon will rise.

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Example 3: Volume of Air Required for Floating

An object has a weight of 2.4 N.
Density of air = 1.2 kg/m³, $g = 10 \, \text{m/s}^2$g=10m/s2

Calculate the minimum volume of air the object must displace to float.$$2.4 = 1.2 \times 10 \times V$$2.4=1.2×10×V $$V = \frac{2.4}{12}$$V=122.4​ $$V = 0.2 \, \text{m}^3$$V=0.2m3

Required volume = 0.2 m³

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8. Problems Involving Upthrust in Gases

1. A balloon displaces 0.25 m³ of air.
   Density of air = 1.2 kg/m³, $g = 10 \, \text{m/s}^2$g=10m/s2.
   Calculate the upthrust acting on the balloon.
2. An object weighs 3 N and experiences an upthrust of 2 N in air.
   State whether the object will rise, float, or fall, giving a reason.
3. Explain why a parachutist falls slowly after opening a parachute.
4. A hot-air balloon displaces 1.5 m³ of air.
   Calculate the upthrust acting on the balloon.
5. State two applications of upthrust in gases in everyday life.

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9. Conclusion

Upthrust in gases is an important concept that explains the behavior of objects in air. It depends on the density of the gas, the volume displaced, and gravity. Understanding upthrust helps explain real-life phenomena such as balloons rising, parachutes slowing descent, and hot air moving upward.

Related topics

• floating and sinking
• The law of floatation
• Upthrust in gases

Key matrices concepts include identity, determinant, inverse, and singular matrices, which help describe matrix properties and transformations.

Matrices are important tools in mathematics and science. They are used to solve systems of equations, describe transformations, and model real-life problems such as economics, physics, and computer graphics.
In this lesson, we will learn four important concepts related to square matrices:

• Identity Matrix
• Determinant of a Matrix
• Inverse of a Matrix
• Singular Matrix

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2. Identity Matrices

Definition

An identity matrix is a square matrix that has:

• 1’s on the main diagonal
• 0’s everywhere else

It is similar to the number 1 in multiplication, because multiplying any matrix by the identity matrix gives the same matrix.

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Notation of matrices

The identity matrix is denoted by I.

2 × 2 Identity Matrix

$$I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

3 × 3 Identity Matrix

$$I_3$$I3​=​100​010​001​​

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Illustration

Main diagonal →   1   1   1
Other entries →   0   0   0

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Property

$$A \times I = I \times A = A$$

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3. Determinant of a Matrix

Definition

The determinant of a square matrix is a single number that tells us important information about the matrix, such as:

• Whether the matrix has an inverse
• Whether the system of equations has a unique solution
• The scaling factor of a geometric transformation

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Determinant of a 2 × 2 Matrix

For a matrix:$$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$

The determinant is:$$|A| = ad – bc$$

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Example

$$A = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}$$A=[31​24​] $$|A| = (3 \times 4) – (2 \times 1) = 12 – 2 = 10$$∣A∣=(3×4)−(2×1)=12−2=10

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The determinant represents area scaling:

• If |A| = 2 → area doubles
• If |A| = 0 → area collapses to a line

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4. Inverse of a Matrices

Definition

The inverse of a matrix A is another matrix $A^{-1}$A−1 such that:$$A \times A^{-1} = I$$

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Inverse of a 2 × 2 Matrix

For:$$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$$$A^{-1} = \frac{1}{ad – bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$

Note: The inverse exists only if the determinant is not zero.

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Example

$$A = \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix}$$

Step 1: Find determinant:$$|A| = (2 \times 3) – (1 \times 5) = 6 – 5 = 1$$

Step 2: Find inverse:$$A^{-1} = \begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix}$$

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Illustration

Matrix A → changes a vector  
Inverse A⁻¹ → returns it back to original

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5. Singular Matrix

Definition

A singular matrix is a matrix that has no inverse.

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Condition

A matrix is singular if: ∣A∣=0

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Example

$$A = \begin{bmatrix} 2 & 4 \\ 1 & 2 \end{bmatrix}$$A=[21​42​] $$|A| = (2 \times 2) – (4 \times 1) = 4 – 4 = 0$$∣A∣=(2×2)−(4×1)=4−4=0

Therefore, A is singular and cannot be inverted.

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Illustration

Transformation squashes shape into a line → no way to reverse it

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6. Summary Table

Concept	Meaning	Key Condition
Identity Matrix	Matrix with 1’s on diagonal	Acts like number 1
Determinant	Single number describing matrix properties	The matrix is not a zero matrix
Inverse Matrix	Matrix that reverses A	Exists if A exists
Singular Matrix	Matrix without inverse

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7. Practice Questions

Question 1

Find the determinant of:$$A = \begin{bmatrix} 4 & 7 \\ 2 & 6 \end{bmatrix}$$

Answer:
$|A| = (4×6) − (7×2) = 24 − 14 = 10$

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Question 2

Write the 3 × 3 identity matrix.

Answer:$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

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Question 3

Determine whether the matrix is singular:$$A = \begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix}$$

Answer:
$|A| = (1×6) − (3×2) = 6 − 6 = 0$

Matrix is singular.

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Question 4

Find the inverse of:$$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$$

Answer:$$|A| = (1×4) − (2×3) = 4 − 6 = -2$$$$A^{-1} = \frac{1}{-2} \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$$

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8. Conclusion

Understanding identity, determinant, inverse, and singular matrices is important in solving systems of equations, physics, engineering, and computer science. These concepts help us understand when a matrix can be reversed and how it transforms.

Related Topics

• Operation on matrices
• Introduction to matrices
•  MATRICES AND TRANSFORMATIONS

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