# Cube

The sum of all cube edges is 30cm. To find the surface area of the cube, we need to understand the properties of a cube.

A cube has 12 equal edges, and the sum of the lengths of all cube edges is given as 30cm. Since all edges are equal, we can find the length of one edge by dividing the total sum by 12.

Let's denote the length of one edge as 'e'. Then, the sum of all edges is 12e = 30cm. Solving for 'e', we get e = 30cm / 12 = 2.5cm.

The surface area of a cube (A) can be calculated using the formula A = 6e², where 'e' is the length of one edge. Substituting the value of 'e', we get A = 6 * (2.5cm)² = 6 * 6.25cm² = 37.5cm².

### Correct answer:
The surface area of the cube is 37.5cm².

### Related math problems and questions:

1. Cube 1-2-3: Calculate the volume and surface area of the cube ABCDEFGH given:
   a) /AB/ = 4 cm
   b) perimeter of wall ABCD is 22 cm
   c) the sum of the lengths of all cube edges is 30 cm.

2. Sum of the edges: The sum of the lengths of all cube edges is 72 cm. Find the surface area.

3. Cube edges: The sum of the lengths of the cube edges is 42 cm. Calculate the surface of the cube.

4. Cube: The sum of lengths of cube edges is 57 cm. Find its surface and volume.

5. Consecutive 46761: The block lengths are made up of three consecutive GP members. The sum of the lengths of all edges is 84 cm, and the volume block is 64 cm³. Determine the surface of the block.

6. Edges or sides: Calculate the cube volume if the sum of the lengths of all sides is 276 cm.

7. Cube edge: Determine the cube's edges when the surface is equal to 37.5 cm square.

8. Cube 7: Calculate the volume of a cube, whose sum of the lengths of all edges is 276 cm.

9. Prism: Three cubes are glued into a prism. The sum of the lengths of all its edges is 115 cm. What is the length of one edge of the original cube?

10. Lengths 63174: The block has a square base of 36 dm², and its height is 1/3 of the length of the base edge. Find the sum of the lengths of all edges of a block.

11. Cube 5: The surface of the cube is 15.36 dm². How will the surface area of this cube change if the length of the edges is reduced by 2 cm?

12. Cubes: Carol cuts a bar 12 cm x 12 cm x 135 cm into cubes. Find the sum of all the surfaces of the resulting cubes.

13. Cube 5: The surface area of one cube wall is 32 square centimeters. Determine the length of its edges, its surface, and volume.

14. The surface: The cuboid's surface area is 1714 cm², and the edges of the base are 25 cm and 14 cm long. Find the area of the surface.

15. The block: The block, the edges formed by three consecutive GP members, has a surface area of 112 cm². The sum of the edges that pass through one vertex is 14 cm. Calculate the volume of this block.

16. Quadrangular 64564: The surface of the quadrangular block is 3.4 dm². The edges of the figure are 8cm and 10cm long. Calculate the volume of the prism.

17. Nice prism: Calculate the cuboid's surface if the sum of its edges is a + b + c = 19 cm and the body diagonal size u = 13 cm.


**Fillword Problem**
The fillword is a word game with simple rules. The author creates a rectangular grid with M cells in width and N cells in height and writes P words on it. Each word can be found on the grid, and the following conditions are met:
- No cell belongs to more than one word.
- No cell belongs to any word more than once.

A word W of length k is found on the grid if a sequence of cells (x1, y1), (x2, y2), ..., (xk, yk) can be found such that:
- (xi, yi) and (xi+1, yi+1) are neighbors (|xi-xi+1| + |yi-yi+1| = 1) for each i = 1, 2, ..., k-1.
- W[i] is written in the cell with coordinates (xi, yi).

**Task**
The task is to find all the words on the grid and then determine which letters will be left after all the words are found. These leftover letters will be used to make up a secret word.

**Input**
The input consists of:
- The first line containing three integer numbers: N, M (2 <= M, N <= 10), and P (P <= 100).
- Next N lines containing M characters each, representing the grid.
- The following P lines containing words to be found on the fillword grid.

**Output**
The output should be the letters that will be left after all the words are found, in lexicographical order.

**Example**
Input:
3 3 2
EBG
GEE
EGE
BEG
GEE
Output:
EEG

**Constraints**
- Fillword will always have at least one solution.
- All characters occurring in fillword will be capital English letters.


To determine the number of gallons in a 420 lb propane tank, it's essential to understand the conversion rate between pounds and gallons of propane. On average, 1 pound of propane is equivalent to 0.236 gallons at 60 degrees Fahrenheit.

A 420 lb propane tank holds approximately 96 gallons of propane, calculated by multiplying the tank's capacity in pounds by the conversion rate (0.236 gallons/pound). These tanks are typically 50 inches long, 27.9 inches wide, and 27.9 inches high, weighing around 206 pounds when empty. In reality, a 420 lb propane tank is usually about 80% full, allowing for a maximum of 84 gallons.

Propane tanks are labeled based on their capacity in pounds, not gallons. To calculate the number of gallons in a propane tank, you can use the formula: 0.236 gallons/pound * tank capacity in pounds = tank capacity in gallons. This formula applies to any size propane tank.

The temperature of the propane is crucial for this calculation, as it affects the conversion rate. Propane is safely stored at 60 degrees Fahrenheit. Changes in temperature can alter the density of the propane, impacting the conversion rate. For safety purposes, it's recommended to keep propane at 60 degrees Fahrenheit.

To calculate the weight of a full propane tank, add the weight of the propane to the weight of the empty tank. For a 420 lb propane tank, the empty tank weighs around 206 pounds, so the full tank would weigh approximately 626 pounds (420 pounds of propane + 206 pounds of the empty tank).

The lifespan of a 420 lb propane tank depends on the BTU rating of the burner and the approximate run time. For example, a burner with an 80,000 BTU rating would allow a 420 lb propane tank to run for about 113.37 hours.

Propane tanks come in various sizes, and selecting the correct size is crucial for the intended use. A 420 lb propane tank is suitable for larger appliances and heating purposes, such as home heating, hot water, fireplaces, generators, and pool heaters. Smaller propane tanks, like 20 lb tanks, are better suited for small appliances like grills or space heaters.

In conclusion, a 420 lb propane tank can hold approximately 96 gallons of propane, making it ideal for larger appliances and heating purposes. When selecting a propane tank, it's essential to consider the size in both pounds and gallons and compare it to the task at hand to ensure the best fit for your needs.


# The length of one side of the rectangle is 4 cm, the other is 3 cm less. Find the perimeter of this rectangle.

To find the perimeter, we need to determine the lengths of both sides. The first side is 4 cm. The second side is 3 cm less than the first side, so we subtract 3 cm from 4 cm:
4 - 3 = 1 cm.

The perimeter of a rectangle is the sum of the lengths of its sides. Since there are two sides of 4 cm and two sides of 1 cm, we add these lengths together:
4 + 4 + 1 + 1 = 10 cm.

Therefore, the perimeter of the rectangle is 10 cm.


The Levenshtein distance algorithm is implemented in the provided Python code.
This algorithm calculates the minimum number of operations (insertions, deletions, and substitutions) required to transform one string into another.

The code defines a function `levenshtein(a, b)` that takes two strings `a` and `b` as input and returns the Levenshtein distance between them.
The function first determines the lengths of the input strings `n` and `m`, and if `n` is greater than `m`, it swaps the strings to ensure that `n` is less than or equal to `m`.
This optimization reduces the space complexity to O(min(n, m)).

The function then initializes a list `current` with values ranging from 0 to `n`.
It iterates over the characters in the second string, maintaining a `previous` list to store the Levenshtein distances for the previous row.
For each character in the first string, it calculates the Levenshtein distance by considering three operations:
- `add`: inserting a character into the first string,
- `delete`: deleting a character from the first string, and
- `change`: substituting a character in the first string with a character from the second string.
If the current characters in both strings are different, the `change` operation is penalized by adding 1 to the distance.

The function returns the Levenshtein distance between the two input strings, which is stored in the last element of the `current` list.

In the `main` section, the code reads two command-line arguments and prints the Levenshtein distance between them using the `levenshtein` function.

Key aspects of the code include:
- The Levenshtein distance algorithm,
- The optimization to reduce space complexity, and
- The use of dynamic programming to efficiently calculate the Levenshtein distance.

Note: The code is distributed under the CC0 license, version 1.0, and is dedicated to the public domain worldwide. The author, Magnus Lie Hetland, has waived all copyright and related rights to the extent possible under law.


**1. Clustering Algorithm Termination**
The K-Means clustering algorithm terminates when the mean values computed for the current iteration are identical to those of the previous iteration.

- K-Means clustering
- Conceptual clustering
- Expectation maximization
- Agglomerative clustering

Answer: K-Means clustering

**2. Correlation Coefficient Interpretation**
A correlation coefficient of -0.85 between two real-valued attributes indicates that as the value of one attribute decreases, the value of the second attribute increases.

- The attributes are not linearly related.
- As the value of one attribute decreases, the value of the second attribute increases.
- As the value of one attribute increases, the value of the second attribute also increases.
- The attributes show a linear relationship.

Answer: As the value of one attribute decreases, the value of the second attribute increases.

**3. Rule Confidence Definition**
Rule confidence is defined as the conditional probability that Y is true when X is known to be true, given a rule of the form IF X THEN Y.

- Y is false when X is known to be false.
- Y is true when X is known to be true.
- X is true when Y is known to be true.
- X is false when Y is known to be false.

Answer: Y is true when X is known to be true.

**4. Chameleon Algorithm Classification**
Chameleon is a hierarchical clustering algorithm.

- Density-based clustering algorithm
- Partitioning-based algorithm
- Model-based algorithm
- Hierarchical clustering algorithm

Answer: Hierarchical clustering algorithm

**5. Odd Man Out**
DBSCAN is the odd man out among the given options.

- DBSCAN
- K-Mean
- PAM
- None of the above

Answer: DBSCAN

**6. Apriori Algorithm Iterations**
The number of iterations in the Apriori algorithm increases with the size of the maximum frequent set.

- Increases with the size of the data
- Decreases with the increase in size of the data
- Increases with the size of the maximum frequent set
- Decreases with the increase in size of the maximum frequent set

Answer: Increases with the size of the maximum frequent set

**7. Interestingness Measures for Association Rules**
Lift is an interestingness measure for association rules.

- Recall
- Lift
- Accuracy
- All of the above

Answer: Lift

**8. Candidate Association Rules**
Given a frequent itemset L with |L| = k, there are 2^k - 2 candidate association rules.

- 2^k - 1 candidate association rules
- 2^k candidate association rules
- 2^k - 2 candidate association rules
- 2^k - 2 candidate association rules (2 to the power k - 2)

Answer: 2^k - 2 candidate association rules

**9. Case-Based Learning Example**
K-nearest neighbor is an example of case-based learning.

- Decision trees
- Neural networks
- Genetic algorithm
- K-nearest neighbor

Answer: K-nearest neighbor

**10. Error Measure**
The average positive difference between computed and desired outcome values is known as the mean absolute error.

- Mean positive
- Error mean squared
- Error mean absolute
- Error root mean squared error

Answer: Error mean absolute

**11. Frequent Item Sets**
Frequent item sets are a superset of both closed frequent item sets and maximal frequent item sets.

- Superset of only closed frequent item sets
- Superset of only maximal frequent item sets
- Subset of maximal frequent item sets
- Superset of both closed frequent item sets and maximal frequent item sets

Answer: Superset of both closed frequent item sets and maximal frequent item sets

**12. Rule Accuracy and Coverage**
Given a supervised data mining session with a rule accuracy of 70% and rule coverage of 63%, the number of individuals in the class life insurance = no with credit card insurance and less than 30 years old can be calculated.

- 63
- 38
- 40
- 89

Answer: 38

**13. Cluster Analysis Definition**
Cluster analysis is the process of grouping similar objects.

- Simple segmentation
- Grouping similar objects
- Labeled classification
- Query results grouping

Answer: Grouping similar objects

**14. Good Clustering Method**
A good clustering method will produce high-quality clusters with low intra-class similarity.

- High inter-class similarity
- High intra-class similarity
- Low intra-class similarity
- None of the above

Answer: Low intra-class similarity

**15. DBSCAN Parameters**
The two parameters needed for DBSCAN are Min points and eps.

- Min threshold
- Min points and eps
- Min sup and min confidence
- Number of centroids

Answer: Min points and eps

**16. Neural Network and Linear Regression Models**
Both neural network and linear regression models require input attributes to be numeric.

- Both techniques build models whose output is determined by a linear sum of weighted input attribute values.
- The output of both models is a categorical attribute value.
- Both models require numeric attributes to range between 0 and 1.
- Both models require input attributes to be numeric.

Answer: Both models require input attributes to be numeric.

**17. Apriori Algorithm Candidate Generation**
If there are 100 1-item-sets, the number of candidate 2-item-sets in the Apriori algorithm is 4950.

- 100
- 200
- 4950
- 5000

Answer: 4950

**18. Apriori Algorithm Bottleneck**
The significant bottleneck in the Apriori algorithm is candidate generation.

- Finding frequent itemsets
- Pruning
- Candidate generation
- Number of iterations

Answer: Candidate generation

**19. Machine Learning Techniques**
Machine learning techniques differ from statistical techniques in that they are better able to deal with missing and noisy data.

- Are better able to deal with missing and noisy data
- Typically assume an underlying distribution for the data
- Have trouble with large-sized datasets
- Are not able to explain their behavior.

Answer: Are better able to deal with missing and noisy data

**20. A Priori Probability**
The probability of a hypothesis before the presentation of evidence is known as the a priori probability.

- A priori
- Posterior
- Conditional
- Subjective

Answer: A priori

**21. KDD Representation**
KDD represents the extraction of knowledge.

- Data
- Knowledge
- Rules
- Model

Answer: Knowledge

**21. Outliers**
The nature of the problem determines how outliers are used.

- Outliers should be part of the training dataset but should not be present in the test data.
- Outliers should be identified and removed from a dataset.
- The nature of the problem determines how outliers are used.
- Outliers should be part of the test dataset but should not be present in the training data.

Answer: The nature of the problem determines how outliers are used.

**21. Distance Measure**
The most general form of distance is Minkowski.

- Manhattan
- Euclidean
- Mean
- Minkowski

Answer: Minkowski

**21. Association Rule Preference**
The preferred association rule is the one with low support and high confidence.

- High support and medium confidence
- High support and low confidence
- Low support and high confidence
- Low support and low confidence

Answer: Low support and high confidence

**21. Rule-Based Classifier**
A rule set R is exhaustive if there is a rule for each combination of attribute values.

- Exhaustive
- Inclusive
- Comprehensive
- Mutually exclusive

Answer: Exhaustive

**21. Apriori Property**
The Apriori property means that if a set cannot pass a test, its supersets will also fail the same test.

- If a set cannot pass a test, its supersets will also fail the same test.
- To decrease the efficiency, do level-wise generation of frequent item sets.
- To improve the efficiency, do level-wise generation of frequent item sets.
- If a set can pass a test, its supersets will fail the same test.

Answer: If a set cannot pass a test, its supersets will also fail the same test.

**21. Frequent Item Set Subsets**
If an item set 'XYZ' is a frequent item set, then all subsets of that frequent item set are frequent.

- Undefined
- Not frequent
- Frequent
- Cannot say

Answer: Frequent

**21. Probability Calculation**
Given the probability that a person owns a sports car and subscribes to an automotive magazine, the probability that a person subscribes to an automotive magazine given that they own a sports car can be calculated.

- 0.0368
- 0.0396
- 0.0389
- 0.0398

Answer: 0.0396

**21. Simple Regression Assumption**
Simple regression assumes a linear relationship between the input attribute and output attribute.

- Quadratic
- Inverse
- Linear
- Reciprocal

Answer: Linear

**21. Association Rule Generation**
To determine association rules from frequent item sets, both minimum support and confidence are needed.

- Only minimum confidence needed
- Neither support nor confidence needed
- Both minimum support and confidence are needed
- Minimum support is needed

Answer: Both minimum support and confidence are needed

**21. Candidate Rule**
Given a frequent item set {A, B, C, D}, the candidate rule that is not possible is D -> ABCD.

- C -> A
- D -> ABCD
- A -> BC
- B -> ADC

Answer: D -> ABCD

**21. Classification Rules**
Classification rules are extracted from decision trees.

- Decision tree
- Root node
- Branches
- Siblings

Answer: Decision tree

**21. K-Means Algorithm**
In the K-Means algorithm, K refers to the number of clusters.

- Complexity
- Fixed value
- Number of iterations
- Number of clusters

Answer: Number of clusters

**21. Linear Regression Model**
If a linear regression model perfectly fits the training data, the test error cannot be determined.

- Test error is also always zero
- Test error is non-zero
- Couldn't comment on test error
- Test error is equal to train error

Answer: Couldn't comment on test error

**21. Simple Linear Regression Coefficients**
In a simple linear regression model with one independent variable, two coefficients need to be estimated.

- 1
- 2
- 3
- 4

Answer: 2

**21. Linear Regression Output Change**
In a simple linear regression model, if the input variable changes by 1 unit, the output variable will change by its slope.

- By 1
- No change
- By intercept
- By its slope

Answer: By its slope

**21. Linear Model Syntax**
In the syntax of a linear model, data refers to a vector.

- Matrix
- Array
- Vector
- List

Answer: Vector

**21. Linear Regression Equation**
In the mathematical equation of linear regression Y = β1 + β2X + ε, (β1, β2) refers to the Y-intercept and slope.

- (X-intercept, Slope)
- (Slope, X-Intercept)
- (Y-Intercept, Slope)
- (Slope, Y-Intercept)

Answer: (Y-Intercept, Slope)


The imperial system of measurements is standard in the US, while the metric system is preferred in Europe and Canada. Understanding the imperial system is crucial as it plays a vital role in various aspects of life. The system uses feet and inches to describe object sizes, with one foot equal to 12 inches.

**Why Use Feet and Inches?**

Feet and inches are used to describe object sizes, with feet being a larger unit. People use abbreviations and symbols to save time when writing measurements. The symbol for feet is a single apostrophe (‘), and the abbreviation is "ft". For example, 5 feet can be written as 5’ or 5 ft.

**What are the Symbols for Feet and Inches?**

The symbol for inches is a double apostrophe (’’), and the abbreviation is "in". For example, 5 inches can be written as 5’’ or 5 in.

**Examples of Feet and Inches**

Some examples of using feet and inches symbols and abbreviations include:
- 5’ 7” (5 feet and 7 inches)
- 12 ft 5 in (12 feet and 5 inches)
- 9 ft 4 (9 feet and 4 inches)

**Why are Symbols Used?**

Symbols are used because they are comprehensive, easy to understand, and can be used worldwide. They save time and increase writing speed, making them ideal for industries that require frequent measurements. Symbols also conserve space, making them useful for charts, graphs, and small measurement labels.

**Conversions of Units**

To convert inches to feet, divide the number of inches by 12. If the result is a whole number, the conversion is exact. If there is a remainder, the result will be in feet and inches. For example, 36 inches is equal to 3 feet, and 45 inches is equal to 3 feet 9 inches.

To convert feet to inches, multiply the number of feet by 12. For example, 3 feet is equal to 36 inches.

In conclusion, understanding the symbols for feet and inches and how to convert between them is essential for working with the imperial system of measurements. Symbols are used to save time, increase efficiency, and conserve space, making them a crucial part of various industries and daily activities.


**3D Transformations in Computer Graphics**

3D transformations modify and re-position existing graphics in a three-dimensional plane. In computer graphics, various transformation techniques are used, including 3D rotation.

**3D Rotation in Computer Graphics**

3D rotation is a process of rotating an object with respect to an angle in a three-dimensional plane. Consider a point object O with initial coordinates (Xold, Yold, Zold) and initial angle Φ with respect to the origin. The object is rotated by an angle θ, resulting in new coordinates (Xnew, Ynew, Znew).

There are three possible types of rotation in 3D:

1. **X-Axis Rotation**: Achieved using the rotation equations:
   - Xnew = Xold
   - Ynew = Yold cosθ – Zold sinθ
   - Znew = Yold sinθ + Zold cosθ

2. **Y-Axis Rotation**: Achieved using the rotation equations:
   - Xnew = Zold sinθ + Xold cosθ
   - Ynew = Yold
   - Znew = Zold cosθ – Xold sinθ

3. **Z-Axis Rotation**: Achieved using the rotation equations:
   - Xnew = Xold cosθ – Yold sinθ
   - Ynew = Xold sinθ + Yold cosθ
   - Znew = Zold

**Practice Problems**

**Problem 1**: Given a homogeneous point (1, 2, 3), apply a 90-degree rotation towards the X, Y, and Z axes and find the new coordinate points.

**Solution**

- **X-Axis Rotation**:
  - Xnew = 1
  - Ynew = 2 cos(90°) – 3 sin(90°) = -3
  - Znew = 2 sin(90°) + 3 cos(90°) = 2
  - New coordinates: (1, -3, 2)

- **Y-Axis Rotation**:
  - Xnew = 3 sin(90°) + 1 cos(90°) = 3
  - Ynew = 2
  - Znew = 2 cos(90°) – 1 sin(90°) = -1
  - New coordinates: (3, 2, -1)

- **Z-Axis Rotation**:
  - Xnew = 1 cos(90°) – 2 sin(90°) = -2
  - Ynew = 1 sin(90°) + 2 cos(90°) = 1
  - Znew = 3
  - New coordinates: (-2, 1, 3)

For a better understanding of 3D rotation in computer graphics, refer to the next article on **3D Scaling in Computer Graphics**. Additional notes and study materials on computer graphics are available, along with video lectures on the **LearnVidFun** YouTube channel.


Euclid's Proposition 22 from Book 3 of *The Elements* states that in a cyclic quadrilateral, opposite angles sum to 180°. This proposition is part of the foundational work in geometry and is a key concept in understanding the properties of cyclic quadrilaterals. The proposition is concise and directly states the relationship between opposite angles in a cyclic quadrilateral, providing a fundamental principle for further geometric analysis.


Thermal conductivity, denoted by $k$, is a material property that indicates its ability to conduct heat. It is a key component in Fourier's Law for heat conduction, which is defined by the formula: $H = \frac{\Delta Q}{\Delta t} = k \times A \times \frac{\Delta T}{x}$, where $\frac{\Delta Q}{\Delta t}$ is the rate of heat flow, $k$ is the thermal conductivity, $A$ is the total cross-sectional area of the conducting surface, $\Delta T$ is the temperature difference, and $x$ is the thickness of the conducting surface.

Rearranging the equation gives thermal conductivity as: $k = \frac{\Delta Q}{\Delta t} \times \frac{1}{A} \times \frac{x}{\Delta T}$. This can also be thought of as a flux of heat (energy per unit area per unit time) divided by a temperature gradient (temperature difference per unit length).

The typical units for thermal conductivity are W/(m·K) in the SI system and Btu·ft/(h·ft²·°F) in the English system, with a conversion factor of 1 Btu·ft/(h·ft²·°F) = 1.730735 W/(m·K).

Thermal conductivity depends on various properties of a material, including its structure and temperature. For instance, pure crystalline substances exhibit different thermal conductivities along different crystal axes due to differences in phonon coupling. Sapphire is a notable example, with a thermal conductivity of 2.6 W/(m·K) perpendicular to the "c"-axis at 373 K, but 6000 W/(m·K) at 36 degrees from the "c"-axis and 35 K.

Air and other gases are generally good insulators, and many insulating materials function by having a large number of gas-filled pockets that prevent large-scale convection. Examples include expanded and extruded polystyrene and silica aerogel. Natural, biological insulators such as fur and feathers achieve similar effects by inhibiting convection of air or water near an animal's skin.

Thermal conductivity is crucial in building insulation and related fields. However, materials used in these trades are rarely subjected to chemical purity standards. Several construction materials' "k" values are listed below, but these should be considered approximate due to uncertainties related to material definitions.

The thermal conductivity of various substances can be measured using different techniques, including steady-state and transient methods. Steady-state techniques perform a measurement when the temperature of the material does not change with time, while transient techniques perform a measurement during the process of heating up.

Some common measurement techniques include Searle's bar method for good conductors of heat and Lees' disc method for poor conductors of heat. A thermal conductance tester can also be used to determine if gems are genuine diamonds using diamond's uniquely high thermal conductivity.

The reciprocal of thermal conductivity is thermal resistivity, measured in kelvin-meters per watt (K·m·W^{−1}). Thermal conductance and thermal resistance can also be described, but there are differing definitions for these terms. Thermal conductance is the quantity of heat that passes in unit time through a plate of particular area and thickness when its opposite faces differ in temperature by one degree.

In summary, the key concepts related to thermal conductivity include:

* Thermal conductivity (k): a measure of a material's ability to conduct heat
* Thermal resistivity: the reciprocal of thermal conductivity
* Thermal conductance: the quantity of heat that passes in unit time through a plate of particular area and thickness
* Thermal resistance: the reciprocal of thermal conductance
* Heat transfer coefficient: the quantity of heat that passes in unit time through unit area of a plate of particular thickness when its opposite faces differ in temperature by one degree

The thermal conductivity of a system is determined by how atoms comprising the system interact. There are two different approaches for calculating thermal conductivity: the Green-Kubo relations and the relaxation time approach. The relaxation time approach is based on the scattering of phonons in the system, which can occur through boundary scattering, mass defect scattering, or phonon-phonon scattering.

In the textile industry, a tog value may be quoted as a measure of thermal resistance in place of a measure in SI units. The thermal conductivity of a material can also be affected by its structure and temperature, and it is an important property in various fields, including building insulation, electronics, and thermodynamics.

Some common thermal conductivity values for various materials include:

* Metals: high thermal conductivity, approximately tracking electrical conductivity
* Diamond: high thermal conductivity, 6000 W/(m·K) at 35 K
* Sapphire: variable thermal conductivity, 2.6 W/(m·K) perpendicular to the "c"-axis at 373 K
* Air: low thermal conductivity, approximately 0.025 W/(m·K)
* Polystyrene: low thermal conductivity, approximately 0.033 W/(m·K)

The thermal conductivity of a material can be measured using various techniques, including steady-state and transient methods. The choice of technique depends on the material being measured and the desired level of accuracy.

In summary, thermal conductivity is a critical property of materials that plays a crucial role in various fields, including building insulation, electronics, and thermodynamics. Understanding thermal conductivity is essential for designing and optimizing systems that involve heat transfer, and it is a key concept in the study of thermodynamics and materials science.


To solve the given initial value problem (IVP), we start with the differential equation:
\[4y''(x) - 4y'(x) - 3y(x) = 0\]
with initial conditions \(y'(0) = 5\) and \(y(0) = 1\).

The characteristic equation for this differential equation is:
\[4s^2 - 4s - 3 = 0\]
Solving for \(s\), we find the roots:
\[s = -\frac{1}{2}, \frac{3}{2}\]

Thus, the general solution to the differential equation is:
\[y(x) = c_1e^{-x/2} + c_2e^{3x/2}\]

To find \(c_1\) and \(c_2\), we use the initial conditions. First, we find \(y'(x)\):
\[y'(x) = -\frac{c_1}{2}e^{-x/2} + \frac{3c_2}{2}e^{3x/2}\]

Applying the initial conditions:
\[y'(0) = -\frac{c_1}{2} + \frac{3c_2}{2} = 5\]
\[y(0) = c_1 + c_2 = 1\]

Solving these equations simultaneously:
1. From \(y(0) = c_1 + c_2 = 1\), we express \(c_1\) in terms of \(c_2\): \(c_1 = 1 - c_2\).
2. Substituting \(c_1\) into the equation for \(y'(0)\): \(-\frac{1 - c_2}{2} + \frac{3c_2}{2} = 5\).

Simplifying:
\[-\frac{1}{2} + \frac{c_2}{2} + \frac{3c_2}{2} = 5\]
\[\frac{4c_2}{2} = 5 + \frac{1}{2}\]
\[2c_2 = \frac{11}{2}\]
\[c_2 = \frac{11}{4}\]

Then, substituting \(c_2 = \frac{11}{4}\) into \(c_1 + c_2 = 1\):
\[c_1 + \frac{11}{4} = 1\]
\[c_1 = 1 - \frac{11}{4}\]
\[c_1 = -\frac{7}{4}\]

Therefore, the particular solution to the IVP is:
\[y(x) = -\frac{7}{4}e^{-x/2} + \frac{11}{4}e^{3x/2}\]


# Math Pretest

## Quiz Overview
This quiz measures skills from any curriculum and allows teachers to track student mastery of each skill. With a free account, teachers can edit questions, save copies, start class games, assign follow-up activities, and share links with colleagues.

## Quiz Features
- 10 questions to assess math skills
- Teacher and Student dashboards to track progress
- Ability to tag questions with specific skills for customized tracking

## Questions
1. Ashley buys 12 balls of yarn, each making ⅗ of a scarf. If the yarn is 5% off, how many scarves can she make?
2. Express twelve thousand and twelve thousandths in decimal notation.
3. Calculate the area of the given triangle.
4. Simplify the fraction ⅙ ÷ ⅖ to its simplest form.
5. Find the area of the given polygon.
6. Identify the factor pair of 30 that is NOT among the following options:
   - 1 and 30
   - 5 and 6
   - 3 and 10
   - 4 and 8
7. (No question provided)
8. Given the data points 5, 8, 11, 7, ?, find the missing number of points if the mean is 8 points.
9. Abe has 54 oak trees and 27 pine trees. What is the greatest number of trees he can plant in each row, with rows containing only one type of tree?
10. If Frances can complete 91 oil changes in 7 days, how many can he complete in 11 days?


The rrapply() function in the minimal rrapply-package is a revised and extended version of base rapply() for nested list recursion in R. A sample nested list from the Mathematics Genealogy Project highlights advisor/student genealogy of famous mathematicians.

Nonlinear least squares problems involve minimizing a least squares objective function made up of residuals \(g_1(\boldsymbol{\theta}), \ldots, g_n(\boldsymbol{\theta})\) that are nonlinear functions of parameters \(\boldsymbol{\theta} = (\theta_1,\ldots, \theta_p)'\). The gslnls-package provides R bindings to nonlinear least-squares optimization with the GNU Scientific Library (GSL) using trust region methods.

Multi-scale Turing patterns can be generated in R with Rcpp(Armadillo) using multi-scale convolutions with short-range activator kernels and long-range inhibitor kernels. Automatic differentiation (AD) refers to the automatic calculation of derivatives of a function defined as a computer program by repeated application of the chain rule, playing a crucial role in statistical computing problems.

A nonlinear regression model is defined as \(y_i = f(\boldsymbol{x}_i, \boldsymbol{\theta}) + \epsilon_i\), where \(f: \mathbb{R}^k \times \mathbb{R}^p \to \mathbb{R}\) is a known nonlinear function of independent variables \(\boldsymbol{x}_i\) and unknown parameter vector \(\boldsymbol{\theta}\). Piecewise constant or step function regression can be performed in Stan using a working approach.

In chemical kinetics, the rate of a solid-state reaction is modeled by a reaction rate law of the form \(\frac{d\alpha}{dt} = k \cdot f(\alpha) = Ae^{-E_a/RT} \cdot f(\alpha)\). Selection bias occurs when sampled data is not representative of the population of interest, making conclusions difficult to generalize.

Key concepts include nonlinear least squares problems, automatic differentiation, nonlinear regression models, piecewise constant regression, and selection bias. Important numbers include the version 2 of the GNU Scientific Library (GSL) and the year 2010, which inspired the generating process for multi-scale Turing patterns. Core concepts are trust region methods, gradient-based optimization, and reaction rate laws.


A mathematical concept states that taking a 3-digit number and forming a 6-digit number by placing it next to itself results in a number that can be divided by 7, 11, and 13 without any remainders, ultimately yielding the original 3-digit number. For instance, starting with the number 412, creating the 6-digit number 412412, and performing the divisions results in the original number 412.

In a separate mathematical exploration, the Collatz conjecture proposes that for any given integer, a sequence of operations - dividing by 2 if the number is even or multiplying by 3 and adding 1 if it is odd - will eventually reach the number 1. This process can be implemented recursively, as demonstrated in a PHP function:

function collatz($num) {
  if($num == 1) {
    echo "
" . $num . "
n";
    return;
  }
  if($num % 2 == 0) {
    echo "
" . $num . "
n";
    return collatz($num / 2);
  } else {
    echo "
" . $num . "
n";
    return collatz(3 * $num + 1);
  }
}


**Question 1**
The ordinary differential equation \(\frac{d^2u}{dx^2}-2x^2u+\sin x=0\) is:
- linear and homogeneous
- linear and nonhomogeneous
- nonlinear and homogeneous
- nonlinear and nonhomogeneous

**Explanation**
The given equation is linear and nonhomogeneous because its solution is of the type \(u=f(x)\), where \(u\) is the dependent variable.

**Question 2**
The value of \(\lim_{x \to \infty } \frac{\sqrt{9x^2+2020}}{x+7}\) is:
- \(\frac{7}{9}\)
- 1
- 3
- Indeterminable

**Explanation**
\(\lim_{x \to \infty }\frac{3x \sqrt{1+\frac{2020}{x^2}}}{x\left ( 1+\frac{7}{x} \right )}=3\)

**Question 3**
The integral \(\int_{0}^{1}(5x^3+4x^2+3x+2)dx\) is estimated numerically using three alternative methods. Which statement is TRUE?
- Simpson's rule as well as rectangular rule of estimation will give non-zero error.
- Simpson's rule, rectangular rule as well as trapezoidal rule of estimation will give non-zero error.
- Only the rectangular rule of estimation will give zero error.
- Only Simpson's rule of estimation will give zero error.

**Explanation**
Because the integral is a polynomial of 3rd degree, Simpson's rule will give an error-free answer.

**Question 4**
The partial differential equation \(\frac{\partial u}{\partial y}=\frac{\partial^2 u}{\partial x^2}; \; \; y\geq 0;\;x_1\leq x\leq x_2\) requires:
- three initial conditions
- three boundary conditions
- two initial conditions and one boundary condition
- one initial condition and two boundary conditions

**Explanation**
The given equation is a one-D heat equation, which requires one initial condition and two boundary conditions.

**Question 5**
The ratio of the plastic moment capacity of a beam section to its yield moment capacity is termed as:
- aspect ratio
- load factor
- shape factor
- slenderness ratio

**Explanation**
\(\frac{M_p}{M_y} = \text{Shape factor}\)

**Question 6**
The state of stress represented by Mohr's circle is:
- uniaxial tension
- biaxial tension of equal magnitude
- hydrostatic stress
- pure shear

**Explanation**
In pure shear condition, Mohr's circle has its center at the origin.

**Question 7**
The most economical longitudinal profile of the beam to carry the given loads is:
- A
- B
- C
- D

**Explanation**
For the most economical profile, the maximum cross-section is given where the maximum bending moment occurs.

**Question 8**
The pH value of water for concrete mix shall NOT be less than:
- 4.5
- 5
- 5.5
- 6

**Explanation**
As per IS 456:2000, the minimum pH value of water for concrete is 6.0.

**Question 9**
The constant headway considered from the fourth or fifth headway position is referred to as:
- discharge headway
- effective headway
- intersection headway
- saturation headway

**Question 10**
Soil deposit formed due to transportation by wind is termed as:
- aeolian deposit
- alluvial deposit
- estuarine deposit
- lacustrine deposit

**Explanation**
Soil deposited by wind is Aeolian soil.


Given `a: Option α` and `f : α → β`, we can obtain a term of type `Option β` by using `a.map f`. Alternatively, given `f: α → Option β`, we can achieve the same result with `a.bind f`. Both of these operations can be more conveniently expressed using the `do` notation.

Additionally, there exists a function `join` of type `Option Option α → Option α`, which is equivalent to the `bind` operation.

The IO Monad is used to wrap values that depend on the state of the real world, such as random numbers. Using the IO Monad avoids violating the principle that a function's value is determined by its arguments. To obtain a concrete value, we can `run` the IO computation. Each run may yield a different result.

For example, running individual computations can produce different outcomes:
#eval a -- 23
#eval b -- 96

Interestingly, running a pair of computations can also produce a result, such as:
#eval (a, b) -- (87, 87)


**Number System**
The number system consists of various concepts, including place value, ascending and descending order, and comparison of numbers.

**Place Value Chart**
The place value chart is as follows:
- Thousands: 1000
- Hundreds: 100
- Tens: 10
- Ones: 1

In a three-digit number, there are three places: Ones place, Tens place, and Hundreds place. Each digit in a number has a specific place value.

**Ascending and Descending Order**
- Ascending Order: Arranging numbers from the least to the largest, e.g., 13, 67, 132, 168.
- Descending Order: Arranging numbers from the largest to the least, e.g., 168, 132, 67, 13.

**Symbols and Meaning**
- = (is equal to): The numbers on both sides have the same value, e.g., 25 = 25.
- < (is lesser than): The number on the closed side is lesser than the number on its open side, e.g., 23 < 25.
- > (is greater than): The number on the open side is greater than the number on its closed side, e.g., 28 > 25.

**Rules for Comparison of Numbers**
1. A numeral with more digits is greater.
2. If two numerals have the same number of digits, the numeral with a greater digit at the leftmost place is greater.
3. If the leftmost digits are the same, consider the next digit from the left and compare.

**Successor and Predecessor**
- Successor: The number that comes just after a given numeral, found by adding 1, e.g., the successor of 99 is 100.
- Predecessor: The number that comes just before a given numeral, found by subtracting 1, e.g., the predecessor of 100 is 99.

**Even and Odd Numbers**
- Even numbers: Numbers with 2, 4, 6, 8, or 0 in the ones place, e.g., 4, 32, 168, 490, 746.
- Odd numbers: Numbers with 1, 3, 5, 7, or 9 in the ones place, e.g., 3, 41, 195, 479, 207.


## Related Resources

The resources listed below are aligned to the Common Core Standards For Mathematics (CCSM) standard 4NBT04, which involves fluently adding and subtracting multi-digit whole numbers using the standard algorithm.

### Key Concepts and Skills

* Fluently add and subtract multi-digit whole numbers using the standard algorithm
* Use place value understanding and properties of operations to perform multi-digit arithmetic

### Worksheet Examples

The following examples illustrate the types of problems students may encounter:
- Addition and subtraction of larger numbers (up to 999,999)
- Addition with up to 8 numbers
- 4-digit numbers (e.g., 3286 - 1468)
- Three 3-digit numbers (e.g., 247 + 114 + 598)
- Four 4-digit numbers (e.g., 2347 + 1214 + 5198 + 6548)

### Related Standards and Resources

The resources below support the learning outcome of using place value understanding and properties of operations to perform multi-digit arithmetic:
- Divisibility rules for 2, 3, 4, 5, 6, and 9
- Multiplying by 2-digit numbers
- Multiplication methods, including the area (or box) method
- Products to 144 (e.g., 12 x 11)
- 2-digit x 1-digit multiplication (e.g., 15 x 5, 20 x 8)
- Addition and subtraction worksheets (challenging)
- Box method multiplication: 1-digit x 2-digit
- Partial products method multiplication
- 2 x 1-digit multiplication (e.g., 54 x 5)
- 3 x 1-digit multiplication (e.g., 512 x 7)
- Creative multiplication (with guidance notes and examples)
- 2 x 1-digit multiplication with products to 99 (e.g., 23 x 2) and 999 (e.g., 93 x 4)
- 2 x 1-digit multiplication with regrouping (e.g., 73 x 7)
- Dividends to 99 with remainders (e.g., 97 ÷ 5)
- 3-digit by 1-digit division (e.g., 507 ÷ 9)
- 4-digit by 1-digit division (e.g., 9337 ÷ 6)
- 3 x 1-digit multiplication (e.g., 436 x 7)
- 2 x 2-digit multiplication with and without regrouping (e.g., 42 x 23, 57 x 34)
- 4 x 1-digit multiplication (e.g., 5316 x 8)


Classical Mechanics
# 2D Dynamics

A joker in a circus performs a daredevil act in a big cylindrical drum. He presses himself against a vertical wall, and the drum is rotated at a constant angular velocity $\omega$. When the platform under his feet is removed, the joker remains stuck to the wall without any support or glue.

Given the joker's mass is $\SI{70}{\kilo\gram}$, the drum's radius is $\SI{5}{\meter}$, and the coefficient of static friction between the drum and joker is 0.5, we need to find the minimum angular velocity $\omega$ so that the joker does not fall when the platform is removed. The acceleration due to gravity is $g = \SI[per-mode=symbol]{10}{\meter\per\second\squared}$.

In a binary star system with two equally massive stars of mass $m$, separated by a distance $r$, the angular speed of revolution can be determined. Since the masses are equal, the center of mass lies exactly in the middle, at distance $\frac{r}{2}$ from each star. The universal gravitational constant is $G$.

As Earth rotates about its axis, all people and objects on Earth move with it. Neglecting Earth's revolution around the Sun and considering all people at rest with respect to Earth, the physical quantity that is the same for all people on Earth is their angular velocity, as they all share the same rotational motion.


To estimate the integral of function $f$ from $x = 0$ to $x = 12$ using six rectangles, we need to calculate $L_{6}$, $R_{6}$, and $M_{6}$.

For $L_{6}$, we use the left endpoints of the rectangles. The function $f$ is decreasing, so the left endpoints will give an overestimate of the area. We calculate $L_{6}$ by summing the areas of the six rectangles, using the left endpoint of each rectangle to determine the height.

For $R_{6}$, we use the right endpoints of the rectangles. Since $f$ is decreasing, the right endpoints will give an underestimate of the area. We calculate $R_{6}$ by summing the areas of the six rectangles, using the right endpoint of each rectangle to determine the height.

For $M_{6}$, we use the midpoints of the rectangles. This will give a more accurate estimate of the area than $L_{6}$ or $R_{6}$. We calculate $M_{6}$ by summing the areas of the six rectangles, using the midpoint of each rectangle to determine the height.

To determine which of $L_{6}$, $R_{6}$, or $M_{6}$ gives the best estimate, we need to consider the properties of the function $f$. Since $f$ is decreasing, $L_{6}$ will be an overestimate and $R_{6}$ will be an underestimate. $M_{6}$ will be a more accurate estimate than either $L_{6}$ or $R_{6}$.

The calculations for $L_{6}$, $R_{6}$, and $M_{6}$ involve summing the areas of the six rectangles. For $L_{6}$, the heights of the rectangles are $f(0)$, $f(2)$, $f(4)$, $f(6)$, $f(8)$, and $f(10)$. For $R_{6}$, the heights are $f(2)$, $f(4)$, $f(6)$, $f(8)$, $f(10)$, and $f(12)$. For $M_{6}$, the heights are $f(1)$, $f(3)$, $f(5)$, $f(7)$, $f(9)$, and $f(11)$.

Based on the calculations, $M_{6}$ gives the best estimate of the area under the graph of $f$ from $x = 0$ to $x = 12$. This is because $M_{6}$ uses the midpoints of the rectangles, which provides a more accurate approximation of the area than using the left or right endpoints.

(a) To find estimates of each type for the area under the given graph of $f$ from $x = 0$ to $x = 12$, we calculate:
(i) $L_{6}$ using the left endpoints of the rectangles
(ii) $R_{6}$ using the right endpoints of the rectangles
(iii) $M_{6}$ using the midpoints of the rectangles

(b) $L_{6}$ is an overestimate of the true area because the function $f$ is decreasing.

(c) $R_{6}$ is an underestimate of the true area because the function $f$ is decreasing.

(d) $M_{6}$ gives the best estimate because it uses the midpoints of the rectangles, providing a more accurate approximation of the area than $L_{6}$ or $R_{6}$.


# Rational and Irrational Numbers

Rational and irrational numbers are fundamental concepts in mathematics. A rational number can be expressed as a ratio of two integers, whereas an irrational number cannot. Understanding the difference between these two types of numbers is crucial for advanced math and science applications.

## Key Concepts

* **Rational Numbers**: Can be expressed as a fraction (e.g., 1/3) or a ratio of two integers. Examples include whole numbers (1, 50, 1000), fractions (1/4, 15/40), and decimals (0.1, 0.54). Rational numbers have a finite or recurring decimal expansion.
* **Irrational Numbers**: Cannot be expressed as a ratio of two integers. Examples include √2 and √11. Irrational numbers have a non-finite and non-recurring decimal expansion.

## Importance of Irrational Numbers

Irrational numbers may seem abstract, but they play a vital role in modeling real-world phenomena, particularly in geometry, physics, and engineering. They help fill gaps in the real number system, enabling calculations that would otherwise be impossible. Irrational numbers are used to model situations with non-terminating values, making them essential for scientific applications.

## Why Study Irrational Numbers?

Students often question the relevance of irrational numbers, but they are a fundamental concept in mathematics. Understanding irrational numbers lays the foundation for more advanced math and science topics. In science classes, irrational numbers are used to model physical and theoretical phenomena, such as the behavior of materials under different conditions.

## Examples and Applications

* The ratio of a circle's circumference to its diameter (π) is an irrational number.
* The length of a driveway can vary due to temperature changes, and irrational numbers can be used to model this variation.
* In science experiments, irrational numbers can be used to account for the complexities of real-world phenomena, such as the theory of chaos and variable interactions.

## Practice and Review

To reinforce understanding, practice classifying numbers as rational or irrational, and review the key concepts and examples. This will help solidify the foundation for more advanced math and science topics.


## All Integers Are Even

We are going to prove by induction that for all integer n ≥ 2, n is even. The claim is verified for n = 2, as 2 is the smallest even number.

Assume the claim holds for all 2 ≤ n < k, meaning all integers below k starting with 2 are even. We aim to derive that k is also even by splitting it into two parts, m_{1} and m_{2}, where m_{1} + m_{2} = k. By the induction hypothesis, m_{1} and m_{2} are even, implying their sum is even.

However, the inductive step is flawed. It should be formulated as: Assume the claim holds for all 2 ≤ n < k, where all integers below k starting with 2 are even, and k ≥ 2. The derivation fails for k = 3, as it cannot be split into the sum of two numbers each at least 2, highlighting the error in the proof.


**Homework 7**

Due: May 6, 1998, in class

### Polynomial Product

Compute the product of the polynomials A(x) and B(x) using the method described in class.

Given:
- w(0,8) = 1.000000 + 0.000000i
- w(1,8) = 0.707107 + 0.707107i
- w(2,8) = 0.000000 + 1.000000i
- w(3,8) = -0.707107 + 0.707107i
- w(4,8) = -1.000000 + 0.000000i
- w(5,8) = -0.707107 + -0.707107i
- w(6,8) = 0.000000 + -1.000000i
- w(7,8) = 0.707107 + -0.707107i
- FFT(A) = (-3 #C(-14.242640687119284 4.6568542494923806) #C(-9.0000000000000018 -6.0) #C(-5.7573593128807135 6.6568542494923788) -19 #C(-5.7573593128807161 -6.6568542494923806) #C(-8.9999999999999982 6.0) #C(-14.242640687119287 -4.6568542494923788))
- FFT(B) = (5 #C(-6.8994949366116636 1.4142135623730967) #C(2.9999999999999969 -14.0) #C(12.899494936611667 1.4142135623730923) 1 #C(12.899494936611664 -1.4142135623730967) #C(3.0000000000000031 14.0) #C(-6.8994949366116671 -1.4142135623730923))

### On-line Convex-Hull Problem

Given a set Q of n points, one point at a time, compute the convex hull of the points seen so far.

1. Determine if the new point p is inside the current convex hull by checking for left turns on each angle. This check takes O(n) worst-case time.
2. If a left turn is found, compute the two points on the convex hull having the smallest and largest polar angle with respect to p. This can be done in O(n) worst-case time.
3. Remove the points on the convex hull between the two computed points and replace them with p. This step is also O(n) worst-case time.

Therefore, each point requires O(n) processing, and the total running time for all n points is O(n^2).

### Bin-Packing Problem

Given n objects with sizes s1, s2, ..., sn, pack the objects into a minimum number of unit-sized bins, each holding a total size not exceeding 1.

**a. Prove that the problem of determining the minimum number of bins required is NP-hard.**

Reduce the subset-sum problem (SS) to a specialization of the bin-packing decision problem (BP).

SS = {there exists a subset of s1, s2, ..., sn such that the sum of the subset is equal to t}
BP = {all objects in s1, s2, ..., sn can be placed in k or fewer bins}

Specialize BP to spec(BP) by constraining at least one bin to be exactly full.

spec(BP) = {all objects in s1, s2, ..., sn can be placed in k or fewer bins, at least one of which is exactly full}

Reduce SS to spec(BP) by removing all integers from s1, s2, ..., sn that are greater than t, and calling spec(BP) with the remaining integers divided by t.

**b. Argue that the optimal number of bins required is at least ceil(S), where S = s1 + s2 + ... + sn.**

Since S may be non-integral, the total bin space must be the smallest integral size greater than or equal to S, which is ceil(S).

### Subgraph Isomorphism Problem

Given two graphs G = (V1, E1) and H = (V2, E2), determine whether H is a subgraph of G.

Prove that the subgraph isomorphism problem (SI) is NP-Complete by reducing from the clique problem.

First, prove that SI is in NP by verifying a certificate in polynomial time.

Then, reduce the clique problem to SI by transforming the clique problem instance to an instance of SI, generating a completely connected graph of size k.

### Boyer-Moore-Matching Algorithm

Implement a grep utility using the Boyer-Moore-Matching algorithm, which takes a string s and a file f as input and returns each line of f that includes an instance of the string s.


### Learning Objectives

- Define equilibrium price and quantity and identify them in a market
- Define surpluses and shortages and explain how they cause the price to move towards equilibrium

## Demand and Supply

To understand market equilibrium, we need to start with the laws of demand and supply. The law of demand states that as price decreases, consumers demand a higher quantity. The law of supply states that when price decreases, producers supply a lower quantity.

The demand and supply curves can appear on the same graph, with price on the vertical axis and quantity on the horizontal axis. Together, demand and supply determine the price and quantity that will be bought and sold in a market.

| Price (per gallon) | Quantity Demanded (millions of gallons) | Quantity Supplied (millions of gallons) |
| --- | --- | --- |
| $1.00 | 800 | 500 |
| $1.20 | 700 | 550 |
| $1.40 | 600 | 600 |
| $1.60 | 550 | 640 |
| $1.80 | 500 | 680 |
| $2.00 | 460 | 700 |
| $2.20 | 420 | 720 |

## Surplus or Excess Supply

When the quantity demanded and quantity supplied aren't the same, it results in a surplus or a shortage. Consider a scenario where the price of gasoline is $1.80 per gallon. At this price, the quantity demanded is 500 gallons, and the quantity supplied is 680 gallons. This is a situation of excess supply or a surplus, where the amount producers want to sell is greater than the amount consumers want to buy.

With a surplus, gasoline accumulates, putting pressure on sellers. To avoid losing sales, sellers will cut prices, stimulating a higher quantity demanded. The price will drop until the surplus is eliminated, and the quantity supplied equals the quantity demanded. This is the equilibrium, which occurs at a price of $1.40 per gallon and a quantity of 600 gallons.

## Shortage or Excess Demand

Consider a scenario where the price of gasoline is $1.20 per gallon. At this price, the quantity demanded is 700 gallons, and the quantity supplied is 550 gallons. This is a situation of excess demand or a shortage, where the amount producers want to sell is less than the amount consumers want to buy.

In this situation, eager buyers mob gas stations, and oil companies recognize an opportunity to make higher profits by selling gasoline at a higher price. The price will rise until the shortage is eliminated, and the quantity supplied equals the quantity demanded. The equilibrium occurs at a price of $1.40 per gallon and a quantity of 600 gallons.

## Equilibrium: Where Supply and Demand Intersect

The point where the supply curve and demand curve intersect is the equilibrium. The equilibrium price is the only price where the desires of consumers and producers agree, and the quantity demanded equals the quantity supplied. This mutually desired amount is called the equilibrium quantity.

If a market is not in equilibrium, market forces will push the market to the equilibrium. The equilibrium can be found by looking at a table showing the quantity demanded and supplied at different prices or by looking at a graph of demand and supply.

## Finding Equilibrium with Algebra

We can also identify the equilibrium with algebra if we have equations for the supply and demand curves. Suppose the demand for soda is given by the equation Qd = 16 - 2P, and the supply of soda is given by the equation Qs = 2 + 5P. To find the equilibrium, we set the demand and supply equations equal to each other: 16 - 2P = 2 + 5P.

Solving for P, we get 14 = 7P, and P = 2. This is the equilibrium price. To find the equilibrium quantity, we plug the equilibrium price into either the demand or supply equation. Using the demand equation, we get Qd = 16 - 2(2) = 12.

## Equilibrium and Economic Efficiency

Equilibrium is important for creating a balanced and efficient market. If a market is at its equilibrium price and quantity, it has no reason to move away from that point. A competitive market operating at equilibrium is an efficient market, where the economy is getting as much benefit as possible from its scarce resources, and all possible gains from trade have been achieved.

### Glossary

- Efficiency: when the optimal amount of goods are produced and consumed, minimizing waste
- Equilibrium: price and quantity combination where supply equals demand
- Equilibrium price: the price where the quantity supplied equals the quantity demanded
- Equilibrium quantity: the quantity both supplied and demanded at the equilibrium price
- Shortage (or excess demand): situation where the quantity demanded is greater than the quantity supplied; occurs at prices below the equilibrium
- Surplus (or excess supply): situation where the quantity demanded is less than the quantity supplied; occurs at prices above the equilibrium


Geometry is a branch of mathematics concerned with questions of shape, size, relative position of figures, and the properties of space. It arose independently in several early cultures as a body of practical knowledge concerning lengths, areas, and volumes, with elements of a formal mathematical science emerging in the West as early as the 6th Century BC.

The curriculum for geometry includes various topics for different year groups. For Year 5, pupils should be taught to use the properties of rectangles to deduce related facts and find missing lengths and angles. They should also be able to distinguish between regular and irregular polygons based on reasoning about equal sides and angles.

In Year 6, pupils should be taught to draw 2D shapes using given dimensions and angles, compare and classify geometric shapes based on their properties and sizes, and find unknown angles in any triangles, quadrilaterals, and regular polygons. They should also be able to recognize angles where they meet at a point, are on a straight line, or are vertically opposite, and find missing angles.

For Years 7 to 9, pupils should be taught to draw and measure line segments and angles in geometric figures, describe, sketch, and draw using conventional terms and notations, and use the standard conventions for labeling the sides and angles of triangles. They should also be able to derive and illustrate properties of triangles, quadrilaterals, circles, and other plane figures, identify and construct congruent triangles, and construct similar shapes by enlargement.

In Years 10 and 11, pupils should be taught to identify and apply circle definitions and properties, construct and interpret plans and elevations of 3D shapes, calculate arc lengths, angles, and areas of sectors of circles, and apply the concepts of congruence and similarity.

There are various resources available for teaching geometry, including activities, investigations, videos, and worksheets. These resources cover topics such as angle theorems, circle theorems, congruent triangles, and polygon properties. They provide opportunities for pupils to engage with geometry in a practical and interactive way, developing their understanding and skills in the subject.

Some examples of geometry activities include estimating angles, calculating the areas of triangles, and applying the properties of angles at a point. There are also investigations into polygon areas, tessellations, and the number of rectangles on a grid of squares. Additionally, there are videos explaining concepts such as angles, triangles, and polygons, as well as worksheets for practicing geometry skills.

Overall, geometry is a rich and complex subject that provides opportunities for pupils to develop their problem-solving skills, spatial awareness, and mathematical understanding. With a range of resources and activities available, teachers can help pupils engage with geometry in a meaningful and enjoyable way.