A Second Look at Slew Rate Limiters

I recently had to pick a slew rate for a current waveform and realized I had written an article on the topic in 2014, "Slew Rate Limiters: Nonlinear and Proud of It!" The article explored the effects of two types of slew rate limiters, one feedforward and one feedback, given a particular slew rate R.

The problem statement is to choose a slew rate R for a fixed step size, where the signal input x(t) needs to go from an initial value x0 to a final value x1 = x0 + Δx with some slew rate R. The signal goes into a second-order system H(s) and out comes y(t). The goal is to reduce the overshoot while not adding too much delay for the output reaching its final value.

To analyze this, we use the normalized second-order system H(s̄) = 1 / (s̄^2 + 2ζs̄ + 1) with s̄ = τs = s/ωn. We drop the overbar to simplify the analysis.

Using numerical analysis with Python, we can find an empirical solution. We use the scipy, numpy, and matplotlib libraries to simulate the system response. The code defines functions to calculate the zeta value from overshoot, create a rate-limited step, and interpolate the maximum of a waveform.

The conclusion is that a good choice for slew rate R = Δx/T, where T = 2πατ is the time to slew from initial to final value, and α is some empirical choice, is with α = 1. This choice reduces the overshoot by a factor of about 3-3.5 and delays the peak by about 5τ for higher values of zeta (smaller overshoot).

For lower values of zeta (larger overshoot), the choice α = 1 is more optimal, limiting the overshoot to the 5.7%-7.8% range and delaying the peak by about 4.7τ-5.0τ.

As an example, suppose we want to ramp voltage from 3.0V to 5.0V with a second-order system having τ = 100 μs and ζ = 0.4559 (20% overshoot step response). With a step in voltage command, the output would reach a maximum of 5.4V at about t = πτ = 314 μs. Choosing R = ΔV/2πατ with α = 1, we get R = 3.183 V/ms, expecting an overshoot of about 5.4% → 5.11 V, delayed about 500 μs from the no-slew-rate-limit case.

Simulating this with Python, we get results that match our predictions.

In summary, if you're adding a slew rate limiter to help reduce overshoot in a second-order system with time constant τ and damping factor ζ, a good choice of slew rate limit is R = Δx / (2πατ) with α ≈ 1 for a step input of Δx.


**PART A**

1. In an ideal gas process with $dW > 0$ and $dQ = 0$, the temperature of the gas will decrease because $dU = dQ - dW$ and $dQ = 0$, so $dU = -dW < 0$, implying a decrease in temperature.

a) Temperature will decrease
b) Temperature will increase
c) Volume will decrease
d) None of the above

Answer: a) Temperature will decrease

2. A cylinder of diameter $D$ and thermal conductivity $k$ is surrounded by a cylindrical shell of inner diameter $D$ and outer diameter $2D$ made of material with thermal conductivity $2k$. The effective thermal conductivity of the system is $\frac{7k}{4}$.

a) $\frac{5k}{3}$
b) $4k$
c) $\frac{5k}{4}$
d) $\frac{7k}{4}$

Answer: d) $\frac{7k}{4}$

3. For an adiabatic process, the following are true: $dQ = 0$, $dW = -nC_vdT$, and $TV^{\gamma-1} = constant$.

a) $dQ = 0$
b) $dW = -nC_vdT$
c) $TV^{\gamma-1} = constant$
d) All of the above

Answer: d) All of the above

4. The ratio of the rms speed of $H_2$ to $O_2$ is $4:1$ because $V_{rms} = \sqrt{\frac{3RT}{M}}$.

a) $4:1$
b) $1:4$
c) $2:1$
d) $1:2$

Answer: a) $4:1$

5. The effective coefficient of linear expansion of an inhomogeneous rod with coefficients changing linearly from $\alpha$ to $\beta$ is $\frac{1}{2}(\alpha + \beta)$.

a) $\alpha + \beta$
b) $\frac{1}{2}(\alpha + \beta)$
c) $\sqrt{\alpha\beta}$
d) $\alpha - \beta$

Answer: b) $\frac{1}{2}(\alpha + \beta)$

**PART B**

1. The work done by an ideal gas taken through the cycle ABCDA is the area enclosed by the cycle, which is $2PV$.

a) $3PV$
b) $2PV$
c) $PV$
d) None of the above

Answer: b) $2PV$

2. For an ideal gas taken from state A to state B in three different paths (ABC, ADC, and AC), the work done by the gas in each path is different, with $W_1 > W_2 > W_3$.

a) $W_1 > W_2 > W_3$
b) $W_1 > W_3 > W_2$
c) $W_3 > W_2 > W_1$
d) None of the above

Answer: b) $W_1 > W_3 > W_2$

3. The work done by an ideal gas taken through the cycle ABA is the area enclosed by the cycle.

a) $2PV - \frac{1}{2}\pi V^2$
b) $2PV + \frac{1}{2}\pi V^2$
c) $2PV$
d) None of the above

Answer: d) $2PV$

4. The ratio of the thermal conductivity of a rod made of $N$ rods of equal length $L$ and equal cross-sectional area with thermal conductivities from $k$ to $nk$ to the thermal conductivity of the $N^{th}$ rod is $\frac{1}{N}$, but since the question asks for the ratio of the whole rod to the $N^{th}$ rod and given the nature of thermal conductivity in series, the correct approach leads to the answer being related to the harmonic mean, which simplifies to $\frac{1}{2}(n+1)$ for the effective thermal conductivity of the whole rod compared to the $N^{th}$ rod's conductivity $nk$, thus the ratio simplifies to $\frac{1}{2}(n+1)$ divided by $n$, but the provided options and common sense suggest focusing on the effective thermal conductivity ratio directly.

a) $\frac{1}{2}(n+1)$
b) $n+1$
c) $n$
d) $1$

Answer: a) $\frac{1}{2}(n+1)$

5. The rms speed of a gas with molecules having speeds $v$, $2v$, $3v$, and $4v$ is $v\sqrt{7.5}$ because $V_{rms} = \sqrt{\frac{v_1^2 + v_2^2 + ... + v_n^2}{N}}$.

a) $v\sqrt{7}$
b) $7v$
c) $v\sqrt{2}$
d) $v\sqrt{7.5}$

Answer: d) $v\sqrt{7.5}$

6. The amount of work done by a gas expanding from volume $V_1$ to $V_2$ is greatest when the expansion is isobaric because, for the same volume change, the area under the pressure-volume curve (which represents work done) is maximized when pressure is constant.

a) Isothermal
b) Adiabatic
c) Isobaric
d) Equal in all cases

Answer: c) Isobaric

7. The density of an ideal gas with volume $2V$, pressure $2P$, and temperature $T$ is $\frac{2Pm}{kT}$.

a) $\frac{mkT}{2V}$
b) $\frac{2P}{kT}$
c) $\frac{P}{kVT}$
d) $\frac{2Pm}{kT}$

Answer: d) $\frac{2Pm}{kT}$

8. The average kinetic energy of the molecules of Nitrogen ($EN$) and Hydrogen ($EH$) at the same temperature $T$ is the same because $E = \frac{3}{2}kT$.

a) $EN > EH$
b) $EN = EH$
c) $EN < EH$
d) Nothing can be said

Answer: b) $EN = EH$

9. For a diatomic gas with a heat capacity ratio $\gamma$, if the process is $PT^n = C$ and the heat capacity is negative, then $n$ may be $-3$ or less, but given the context, $n$ should relate to the adiabatic index $\gamma$, and for a diatomic gas, $\gamma = \frac{7}{5}$, which does not directly answer the question without considering the formula for heat capacity in terms of $n$ and $\gamma$.

a) $4$
b) $3$
c) $-4$
d) $-3$

Answer: d) $-3$

10. For an ideal gas obeying $PV^{3/2} = C$ during an adiabatic process, if the gas is adiabatically compressed to half the initial volume, its final temperature will be $2T$ because $TV^{\gamma-1} = constant$, and for this gas, $\gamma = \frac{5}{3}$.

a) $T\sqrt{2}$
b) $2T$
c) $2T\sqrt{2}$
d) $4T$

Answer: b) $2T$


# 5.4: Seismic Reflection (Single Layer)

Reflection surveys use reflected waves to determine the height (h) and velocity (v1) of a single layer subsurface. The layer geometry is used to derive a travel-time equation. Reflected waves can occur when v2 > v1 or v2 < v1.

The distance equation is given by d = v * t, and the total distance a reflected wave travels is 2d. The travel-time equation for a reflected wave is t = 2d / v1, which can be rewritten as t = √(x^2 + 4h^2) / v1.

The intercept time (t0) is given by t0 = 2h / v1. For a vertically reflected wave, if v1 is known, h can be calculated as h = v1 * t0 / 2.

The shape of the travel-time graph for a reflected wave is a hyperbola, given by the equation t^2 = x^2 / v1^2 + 4h^2 / v1^2. This equation can be rewritten as t^2 = x^2 / v1^2 + t0^2.

The equation for t0 is t0 = 2h / v1. The hyperbola's intercept is at t0, and its slope at large x is 1 / v1.

When two layers have the same thickness but different velocities, the layer with the higher velocity will have a lower intercept time (t0) and a broader hyperbola shape.

In the case of multiple layers, the travel-time equation is also a hyperbola. The Normal Move-out (NMO) method is used to analyze the data, where TNMO = t(x) - t0. Correcting for NMO allows reflections to appear as if they come from a vertical reflection.

At small x, the travel-time equation can be approximated as t(x) ≈ t0 + x / (2v1^2 * t0). Using binomial expansion, TNMO can be approximated as TNMO ≈ x^2 / (2v1^2 * t0).

The velocity (v1) can be calculated as v1 ≈ x / √(2t0 * TNMO). To analyze a travel-time graph, the following steps can be taken:

1. Measure t0, t(x), and x.
2. Calculate TNMO = t(x) - t0 if x is small.
3. Calculate v1 ≈ x / √(2t0 * TNMO).
4. Calculate h = v1 * t0 / 2.

In real processing of reflection data, the following steps are taken:

1. Guess v1 and h.
2. Calculate predicted NMO.
3. Correct arrivals.
4. Check if the corrected arrivals are flat. If not, try new values and find the peak sum.


The Scientific Research Institute for Thaumaturgy and Spellcraft (SCITS) is a research department located in Solovetz city. Despite being declassified in the 1960s, SCITS continues to conduct research work. Currently, a 3rd-level wizard, Vitka Korneev, is testing a new battle-spell of fireball in his lab.

The new fireball is an ingenious invention with a zero radius due to the incongruence of transgression inside the incub-transformation's psy-field. Its greatest characteristic is the ability to remain stable during a predefined number of collisions with obstacles, known as *N*-stability. The fireball is *N*-stable if it stays stable after *N* collisions but explodes after the (*N* + 1)th collision. After each collision with a wall, the fireball loses one level of stability and becomes (*N* − 1)-stable.

The *N*-stable fireball has unusual behavior, rebounding only from concrete walls after collisions. This allows it to fly through other obstacles, making it a valuable military invention. The fireball's trajectory is determined by the law "angle of incidence equals angle of reflection," and after collision with a room's corner, it rebounds exactly in the opposite direction.

To calculate the direction of the fireball's throw, a special program is needed. The program must determine the direction that allows the fireball to rebound from the walls exactly *N* times and hit the target with the shortest trajectory. The scientists of SCITS provide the following input:

* The first line contains two numbers: the width and length of the room.
* The second line contains the number *N*, the *N*-stability of the fireball.
* The third line contains four numbers: the coordinates of points *A* and *B*.

The output should be the angle in degrees (with 2 digits after decimal point) that gives the desired direction of the fireball. If there are multiple such angles, the program should output the minimal one.

**Input**

* Width and length of the room (integers, separated by one or more spaces)
* *N*-stability of the fireball (integer, between 0 and 10 inclusive)
* Coordinates of points *A* and *B* (four integers, separated by one or more spaces)

**Output**

* Angle in degrees (with 2 digits after decimal point) that gives the desired direction of the fireball

**Sample**

Input | Output |
---|
1000 10
3
101 5 128 8 | 45.00 |

**Notes**

The problem is based on the books "Monday Begins on Saturday" by Arkady and Boris Strugatsky and the tetralogy "Night Watch", "Day Watch", "Twilight Watch", and "Final Watch" by Sergey Lukyanenko. The problem was authored by Pavel Egorov and was part of the 10th Collegiate Programming Contest of the High School Pupils of the Sverdlovsk Region in 2004.


## RD Sharma Solutions for Class 7 Maths Chapter 15 - Properties of Triangles

## FAQs on RD Sharma Class 7 Maths Solutions Chapter 15 - Properties of Triangles

**1. What are the important Properties of a Triangle discussed in this Chapter?**

The important Properties of a Triangle discussed in this Chapter are:
- Angle sum property
- Exterior angle property
- The third side is always lesser than the sum of the lengths of two sides of a Triangle
- Pythagoras theorem

**2. What are the three main types of Triangles generally?**

The three main types of Triangles based on the length of the sides are:
- Equilateral Triangle
- Isosceles Triangle
- Scalene Triangle
Note that right-angled Triangles can be isosceles or scalene.

**3. What are the parts of a Triangle?**

The three main parts of a Triangle are:
- Vertices
- Sides
- Angles
Understanding these parts is crucial for solving Triangle-related questions.

**4. What is the Pythagoras theorem?**

The Pythagoras theorem states that the square on the hypotenuse equals the sum of the squares of the legs, allowing us to find the length of one side of a Triangle if the other two sides are known.


### Multiply Fractions Practice

#### Question 1

Multiply the fractions ^{1}/_{2} and ^{3}/_{9} and simplify the answer.


# Tour: Subgroup Containment Relation Equals Divisibility Relation on Generators

This article adapts material from the main article: subgroup containment relation in the group of integers equals divisibility relation on generators.

## Statement

Let denote the group of integers under addition. Suppose H and K are subgroups of . Suppose  and . Then, if and only if , i.e.,  is a multiple of .

Note that every nontrivial subgroup of the group of integers is cyclic on its smallest element, so subgroups H and K can be written in the form  and  respectively.

## Related Facts

### Corollaries

- The intersection of two subgroups H and K is the subgroup generated by an element with the property that , and if , then . This integer is termed a **least common multiple** of  and  (unique if only nonnegative integers are allowed).
- The join of two subgroups H and K is the subgroup generated by an element with the property that , and if , then . This integer is termed a **greatest common divisor** of  and  (unique if only nonnegative integers are allowed).
- The greatest common divisor of  and  can be written as  for some integers  and , since it is in the subgroup generated by  and .

To understand this concept, read and comprehend the main statement and its corollaries, and convince yourself of their validity.


I recently attended Lean for the Curious Mathematician 2020, an online meeting organized by Johan Commelin and Patrick Massot. The event was a unique conference where mathematicians learned to use Lean, a proof assistant, to do mathematics ranging from basic logic to category theory and differential geometry.

The conference consisted of three sessions per day, each lasting two hours. A session typically began with a 15-30 minute Zoom talk introducing a part of Lean's math library, mathlib, followed by breakout rooms where participants worked on exercises with expert guidance. This format allowed for interactive learning and problem-solving.

One of the key takeaways from the conference was that technology like Zoom can be highly effective for workshops of this nature. Additionally, providing a large number of exercises involving familiar mathematical concepts can be an excellent way to teach mathematicians how to use Lean. The conference organizers have made all the talks and exercises available online, including a playlist on the Leanprover community YouTube site and a GitHub repository.

The event highlighted the importance of accessible exercises and hints for learning Lean. Scott Morrison's efforts to teach category theory have resulted in a host of accessible exercises, which will be valuable for beginners. The conference also demonstrated that smart people can quickly pick up Lean, as seen in the case of Sophie Morel, who learned to use the system rapidly.

The conference organizers are working on a new book, Mathematics In Lean, which will be a valuable resource for mathematicians. The book will be accompanied by videos and will focus on Lean's mathematics library. The LftCM GitHub repository provides a useful asset for mathematicians, with exercises that can be solved in tactic mode.

The conference also identified areas where Lean can be improved, such as addressing issues that trip up beginners. Experienced users can help make Lean 3 a better system by addressing these issues, rather than waiting for Lean 4. Johannes Hoelzl, an experienced formaliser, advised against porting mathlib from Lean 3 to Lean 4 while simultaneously refactoring it.

The event showed that mathematicians enjoy Lean games, and there is potential for creating new games, such as a filter game. The max minigame, a game that can be played in a browser, is another example of an engaging Lean game.

Overall, the conference was a valuable learning experience, and the organizers have made the materials available online for those who missed it. The event demonstrated the effectiveness of interactive learning and the importance of accessible exercises and hints for teaching mathematicians how to use Lean.


The set difference operation is used in sets to find elements in one set that are not present in the other set. For example, for two sets `a` and `b`, `a-b` results in a set with elements of set `a` that are not present in set `b`.

## Numpy `setdiff1d()` Function

The numpy `setdiff1d()` function is used to find the set difference of two arrays. The syntax is:
```python
import numpy as np
diff = np.setdiff1d(ar1, ar2, assume_unique=False)
```
It returns a numpy array with the unique values in the first array that are not present in the second array. For numpy arrays of higher dimensions (2 or more), the arrays are first flattened, and then the set difference operation is applied.

If the input arrays contain unique values, you can pass `True` to the `assume_unique` parameter to speed up the calculation. Note that it is `False` by default.

### Set Difference between Two 1D Numpy Arrays

To find the set difference between two 1D numpy arrays, pass the two arrays as arguments to the numpy `setdiff1d()` function.
```python
import numpy as np
ar1 = np.array([3, 1, 1, 2, 0, 2])
ar2 = np.array([2, 3, 3, 4])
ar1_2_diff = np.setdiff1d(ar1, ar2)
print(ar1_2_diff)  # Output: [0 1]
```
The returned array contains elements of `ar1` that are not present in `ar2`. Note that the set difference operation `a-b` is not the same as `b-a`, so the order in which you pass the two arrays matters.

### Set Difference between Two 2D Numpy Arrays

If the input arrays are not 1D, they will be flattened before calculating the set difference.
```python
ar1 = np.array([[1, 2, 3], [4, 4, 1]])
ar2 = np.array([[3, 4, 3], [5, 6, 7]])
ar1_2_diff = np.setdiff1d(ar1, ar2)
print(ar1_2_diff)  # Output: [1 2]
```
We get a 1D numpy array with elements of `ar1` that are not present in `ar2`.

For more on the numpy `setdiff1d()` function, refer to its documentation.


## DAV Class 3 Maths Chapter 1 Worksheet 3 Solutions

The DAV Class 3 Maths Book Solutions and DAV Class 3 Maths Chapter 1 Worksheet 3 Solutions of Numbers up to 9999 provide comprehensive answers to textbook questions.

### Question 1: Reading Numbers

Read the following numbers aloud:

(a) 2000 = Two thousand
(b) 4321 = Four thousand three hundred twenty-one
(c) 7701 = Seven thousand seven hundred one
(d) 2508 = Two thousand five hundred eight
(e) 8006 = Eight thousand six
(f) 9998 = Nine thousand nine hundred ninety-eight
(g) 7256 = Seven thousand two hundred fifty-six
(h) 6066 = Six thousand sixty-six

### Question 2: Writing Number Names

Write the number names for the following:

(a) 7000 = Seven thousand
(b) 7312 = Seven thousand three hundred twelve
(c) 6806 = Six thousand eight hundred six
(d) 4509 = Four thousand five hundred nine
(e) 9009 = Nine thousand nine
(f) 9993 = Nine thousand nine hundred ninety-three
(g) 4356 = Four thousand three hundred fifty-six
(h) 2020 = Two thousand twenty

### Question 3: Writing Numerals

Write the numerals for the following:

(a) Four thousand five hundred nine = 4509
(b) Three thousand six = 3006
(c) Five thousand nine hundred fifty = 5950
(d) Four thousand eighty-nine = 4089
(e) Four thousand eight hundred five = 4805
(f) Two thousand twenty = 2020
(g) Six thousand six hundred sixty-six = 6666
(h) Nine thousand nine = 9009

### DAV Class 3 Maths Chapter 1 Worksheet 3 Notes

Key number names and their numerals:

- 4000 = Four thousand
- 6500 = Six thousand five hundred
- 7341 = Seven thousand three hundred forty-one


Primes such that when they are concatenated with their 10's complement (which also must be prime), the result is a brilliant number. The list of such primes includes:

11, 257, 509, 929, 2243, 2897, 3911, 4409, 7121, 9413, 10739, 11411, 13217, 17783, 19319, 20849, 21377, 32507, 32957, 41729, 47279, 48761, 87041, 93083, 93263, 93911, 95027, 95603, 96221.

For example, 257 is a term in this sequence because it is prime, its 10's complement (743) is also prime, and the concatenation of the two (257743) is a brilliant number, as it can be factored into 373 * 691. This concept is discussed in the reference "Brilliant Numbers and a Few Sequences" by J. Earls, published in 2009.


The area of a rectangular garden is 24 square yards. The whole-number dimensions of this garden are the factors of 24. The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, and 24, since each of these numbers divides exactly into 24. The pairs of factors of 24 are: 1 and 24, 2 and 12, 3 and 8, and 4 and 6.

To find the factors of a whole number, follow this procedure: keep dividing until a variable repeats, then list all factors separated by commas.

Example 1: Find the factors of 12.
1 | 12 ÷ 1 = 12 | 1 x 12 |
2 | 12 ÷ 2 = 6 | 2 x 6 |
3 | 12 ÷ 3 = 4 | 3 x 4 |
4 | 12 ÷ 4 = 3 | 4 x 3 |
The factors of 12 are 1, 2, 3, 4, 6, and 12.

Example 2: Find the factors of 20.
1 | 20 ÷ 1 = 20 | 1 x 20 |
2 | 20 ÷ 2 = 10 | 2 x 10 |
3 | 20 ÷ 4 = 5 | 4 x 5 |
4 | 20 ÷ 5 = 4 | 5 x 4 |
The factors of 20 are 1, 2, 4, 5, 10, and 20.

Example 3: Find the factors of 49.
1 | 49 ÷ 1 = 49 | 1 x 49 |
2 | 49 ÷ 7 = 7 | 7 x 7 |
The factors of 49 are 1, 7, and 49.

The Greatest Common Factor (GCF) of two or more whole numbers is the largest factor common to all whole numbers in the set. To find the GCF, list the factors of each number, then identify the common factors and choose the largest one.

Example 4: Find the GCF of 12 and 20.
The common factors of 12 and 20 are 1, 2, and 4. The GCF is 4.

Example 5: Find the GCF of 24 and 36.
The common factors of 24 and 36 are 1, 2, 3, 4, 6, and 12. The GCF is 12.

Example 6: Find the GCF of 56 and 63.
The common factors of 56 and 63 are 1 and 7. The GCF is 7.

Example 7: Find the GCF of 12, 30, and 54.
The common factors of 12, 30, and 54 are 1, 2, 3, and 6. The GCF is 6.

### Exercises

1. Find the GCF of 18 and 36.
Answer: GCF = 18

2. Find the GCF of 30 and 48.
Answer: GCF = 6

3. Find the GCF of 42 and 56.
Answer: GCF = 14

4. Find the GCF of 16, 48, and 72.
Answer: GCF = 8

5. One rectangular garden has an area of 18 square meters. Another rectangular garden has an area of 81 square meters. What is the largest possible whole-number dimension common to both gardens?
Answer: GCF = 9 m


The Area converter can be used to instantly convert between various units of area, including Acres (ac), Hectares (ha), Square Feet (ft2), Square Inches (in2), Square Kilometers (km2), Square Meters (m2), Square Miles (mi2), and Square Yards (yd2).

In the International System of Units (SI), the standard unit of area is the square meter (m2), which is the area of a square with sides of one meter long. A shape with an area of three square meters would have the same area as three such squares. The unit square is defined to have an area of one, and the area of any other shape or surface is a dimensionless real number.

The acre (ac) is a unit of land area used in the imperial and US customary systems, defined as the area of 1 chain by 1 furlong (66 by 660 feet), equal to 1/640 of a square mile, 43,560 square feet, approximately 4,047 m2, or about 40% of a hectare. One international acre is defined as exactly 4,046.8564224 square meters.

The hectare (ha) is an SI accepted metric system unit of area, equal to 100 ares (10,000 m2), primarily used to measure land as a metric replacement for the imperial acre. An acre is about 0.405 hectare, and one hectare contains about 2.47 acres. The hectare remains a non-SI unit accepted for use with SI units.

The square foot (ft2) is an imperial unit and US customary unit of area, defined as the area of a square with sides of 1 foot, equivalent to 144 square inches, 1/9 square yard, or 0.09290304 square meters. One acre is equivalent to 43,560 square feet.

A square inch (in2) is a unit of area, equal to the area of a square with sides of one inch, commonly used in the United States and the United Kingdom.

The square kilometer (km2) is a multiple of the square meter, equal to 1,000,000 square meters or 100 hectares, approximately equal to 0.3861 square miles or 247.1 acres.

The square meter (m2) is the SI derived unit of area, defined as the area of a square with sides of one meter.

The square mile (mi2) is an imperial and US unit of measure for an area, equal to the area of a square with a side length of one statute mile, equivalent to 4,014,489,600 square inches, 27,878,400 square feet, or 3,097,600 square yards.

The square yard (yd2) is an imperial unit of area, defined as the area of a square with sides of one yard (three feet, thirty-six inches, 0.9144 meters) in length, still in widespread use in the US, Canada, and the UK.

Key conversion factors include:
- 1 acre = 43,560 square feet = 4,047 m2 = 0.405 hectare
- 1 hectare = 10,000 m2 = 2.47 acres
- 1 square foot = 144 square inches = 0.09290304 square meters
- 1 square kilometer = 1,000,000 square meters = 100 hectares
- 1 square mile = 4,014,489,600 square inches = 27,878,400 square feet = 3,097,600 square yards
- 1 square yard = 9 square feet = 0.83612736 square meters

Understanding these units and conversion factors is essential for accurately measuring and comparing areas in various contexts.


Cryptography is the backbone of distributed ledger technologies like blockchain and other consensus-oriented distributed networks. It is essential to understand the wallet generation and transaction signing processes, which rely heavily on underlying cryptographic protocols. In addition to distributed networks, cryptography is a critical component of cybersecurity, with cybercrime on the rise and cyberattacks costing businesses an average of $2.3 million per malware attack.

Modern cryptography is mathematically robust and plays a critical role in securing sensitive information like bank account numbers and social security numbers. The concepts are graspable without a complete understanding of the encryption algorithms. One of the key components of cryptography is ciphers, which are algorithms used to conceal (encrypt) and decrypt information. The encrypted information is called ciphertext, while decrypted information is called plaintext. There are two types of ciphers: symmetric ciphers and asymmetric ciphers.

The earliest known cipher is the Caesar Cipher or shift cipher, a symmetric cipher where the same key used to encrypt information is used to decrypt it. The cipher works by mapping the current character of the plaintext to a character of a predetermined distance (the key) adjacent in the alphabet. For example, if the key is a right rotation by three, then 'A' gets mapped to 'D.' The encryption function can be mathematically represented as En = (x + n) mod 26, and the corresponding decryption function is Dn = (x - n) mod 26.

However, enumerating over 26 possible keys isn't very hard, and the cipher is easily broken. The ciphers used today that make distributed ledgers like blockchain possible are known as asymmetric ciphers. Asymmetric ciphers use a different key to encrypt information than they do to decrypt information. These keys are known as the public and private key pair. They contrast symmetric cryptography, where the same key used to encrypt information also decrypts information.

In blockchain wallets, asymmetric cryptography is used to secure transactions. When a user authorizes a transaction, they prove that they are the owner of the private key corresponding to the funds in the wallet. This is done using a Digital Signature Algorithm. The user signs a transaction with their private key, creating a digital signature that reveals no information about the private key but can be verified with the corresponding public key.

The currently best-performing asymmetric encryption algorithm is the Elliptic Curve, providing fast, secure encryption with a smaller key than RSA. For those interested in learning more, resources on RSA, Elliptic Curve, DSA, and ECDSA are available. Additionally, there are various open-source cryptography libraries, including Crypto-js, elliptic, Cryptography, OpenSSL, Botan, and Bouncy Castle, which provide access to digital signature algorithms and commonly used ciphers.

Some recommended books on the topic of cryptography include "The Code Book" and "Code Girls: The Untold Story of the American Women Code Breakers of World War II." These resources can provide a deeper understanding of cryptography and its applications. It's worth noting that OpenSSL is commonly used by system administrators to generate TLS certificates and perform various cryptographic functions, and the correct terminology is TLS (Transport Layer Security), not TSL.


## Experience

Experience is earned linearly, with approximately every $6.72 billion gaining 1 more experience.

## Effect on Earnings

The effect on earnings is not linear. The following table shows a sampling of earnings multipliers:

| Experience | Multiplier |
| --- | --- |
| 5 | 2 |
| 10 | 2.75 |
| 20 | 3.63 |
| 25 | 4 |
| 100 | 7.27 |
| 125 | 8 |
| 625 | 16 |
| 1,000 (1K) | 19.59 |
| 3,125 | 32 |
| 15,625 | 64 |
| 75,125 | 128 |
| 1,000,000 (1M) | 383.8 |
| 1,000,000,000 (1G) | 7,518 |
| 1,000,000,000,000 (1T) | 147,271 |
| 1,000,000,000,000,000 (1P) | 2,885,016 |
| 1,000,000,000,000,000,000 (1E) | 56,516,862 |
| 1,000,000,000,000,000,000,000 (1Z) | 1,107,153,199,296 |

## Formula

The formula to determine Multiplier based on experience is 2^(log₅ Experience)


**Introduction to Fibonacci Ratios in Forex Trading**

Fibonacci ratios are a crucial concept in Forex trading, used to identify potential support and resistance levels. The Fibonacci sequence, introduced by Leonardo Fibonacci, is a series of numbers in which each number is the sum of the two preceding numbers: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, and so on.

**Fibonacci Numbers and Ratios**

The Fibonacci sequence has several key ratios, including 61.8% and 38.2%, which are derived by dividing each number by its successor. For example, 987 ÷ 1597 = 0.618, and 987 ÷ 2584 = 0.382. These ratios are significant in trading, as they represent potential support and resistance levels.

**Fibonacci Ratios in Nature**

The Fibonacci ratios are prevalent in nature, appearing in the structure of plants, animals, and even human faces. This phenomenon is known as the Fibonacci spiral, where each segment of the spiral is related to the Fibonacci sequence. The human eye is accustomed to seeing these ratios, which is why they are often perceived as harmonious.

**Implementing Fibonacci Analysis in Forex Trading**

Fibonacci analysis can be applied to Forex trading by identifying potential support and resistance levels using Fibonacci retracements and fans. When a trend reverses, the price often hesitates or stops at 38.2% or 61.8% of the previous trend. Traders can use Fibonacci tools to measure these ratios on a chart and make informed trading decisions.

**Fibonacci Retracements**

Fibonacci retracements are a popular tool used to identify potential support and resistance levels. The tool consists of a line that is used to locate the basic trend, and horizontal lines that indicate the Fibonacci levels: 0.00, 23.6, 38.2, 50.0, 61.8, and 100. Traders can use these levels to predict potential price movements and make trading decisions.

**Fibonacci Fan**

The Fibonacci fan is another useful tool that uses diagonal lines to indicate potential support and resistance levels. The fan is based on the Fibonacci ratios and can be used to identify areas where the price is likely to bounce or break through. Traders can use the Fibonacci fan in conjunction with other technical analysis methods to increase the accuracy of their trading decisions.

**Conclusion**

In conclusion, Fibonacci ratios are a powerful tool in Forex trading, used to identify potential support and resistance levels. The Fibonacci sequence and its ratios are prevalent in nature and are perceived as harmonious by the human eye. By using Fibonacci retracements and fans, traders can make informed trading decisions and increase their chances of success. It is essential to use Fibonacci tools in conjunction with other technical analysis methods and to look for confluence around these levels to increase the probability of a successful trade setup.


Building Space refers to three-dimensional geometric shapes with volume. These shapes have boundaries in the form of flat and curved planes. A spatial shape must have certain characteristics, including the presence of sides, edges, and vertices.

There are two main categories of building spaces: flat plane shapes and curved plane shapes. Flat plane shapes include cubes, blocks, prisms, and pyramids, while curved plane shapes include cylinders, cones, and spheres.

The parts of a building space include:
1. Side (Field): separates the inside from the outside and can be flat or curved.
2. Ribs: formed by the intersection of two facets and can be straight or curved.
3. Corner Point: the meeting point between three or more ribs.
4. Diagonal Side: connects two vertices on different edges on one side of the plane.
5. Diagonal Space: the area of the line that connects two corner points on the top and base sides.
6. Diagonal Field: bounded by two diagonal sides and opposite sides, typically found in cubical and rectangular shapes.

There are seven types of building spaces:
1. Cube: has 6 square sides, 12 ribs, and 8 vertices.
2. Block: has 6 rectangular sides, 12 ribs, and 8 vertices.
3. Prism: a polyhedron with two parallel and congruent base planes, and several rectangular side planes.
4. Pyramid: a geometric shape bounded by a base plane and several triangular planes that meet at the apex.
5. Cylinder: a special prism with a circular base and a curved side.
6. Cone: bounded by a circular base and a curved side that tapers to a point.
7. Sphere: a three-dimensional shape with a single curved side and no edges or vertices.

Each type of building space has its own properties and formulas for calculating volume, surface area, and other characteristics. Understanding these properties and formulas is essential for working with building spaces in mathematics and real-world applications.

Examples of building spaces can be found in everyday objects, such as:
- Cube: dice, Rubik's cube
- Block: cardboard box, Jenga toy
- Prism: pencil, book
- Pyramid: birthday hat, ice cream cone
- Cylinder: drink can, pipe
- Cone: birthday hat, ice cream cone
- Sphere: ball, globe

Formulas for calculating the volume of different building spaces include:
- Cube: V = s^3, where s is the length of a side
- Block: V = lwh, where l, w, and h are the length, width, and height
- Prism: V = Bh, where B is the area of the base and h is the height
- Pyramid: V = (1/3)Bh, where B is the area of the base and h is the height
- Cylinder: V = πr^2h, where r is the radius and h is the height
- Cone: V = (1/3)πr^2h, where r is the radius and h is the height
- Sphere: V = (4/3)πr^3, where r is the radius

Understanding building spaces and their properties is crucial for various mathematical and real-world applications, including architecture, engineering, and design.


**Measurement of Length**

The shape and size of everything are measured by a measuring instrument. Rulers are well-known devices for measuring length. Measured lengths are expressed in units, including kilometer, meter, centimeter, and millimeter. The SI unit for measured length is the meter. Small lengths are measured in meters, centimeters, and millimeters.

The length of an object can be measured in different units, depending on its size. For example, the length of a finger is typically measured in inches, centimeters, or millimeters, while the distance between two cities is measured in kilometers. The conversion between units is as follows: 1 meter = 100 centimeters, and 1000 meters = 1 kilometer.

**Units of Measurement**

The following units are used to measure length:
- Kilometers (km)
- Meters (m)
- Centimeters (cm)
- Millimeters (mm)

Kilograms, on the other hand, are the SI unit of mass and are not used to measure length.

**Conversion Table**

The conversion between units can be summarized in the following table:
- 10 millimeters (mm) = 1 centimeter (cm)
- 10 centimeters = 1 decimeter (dm)
- 10 decimeters = 1 meter (m)
- 10 meters = 1 decameter (dam)
- 10 decameters = 1 hectometer (hm)
- 10 hectometers = 1 kilometer (km)

**Conversion Between Units**

To convert between units, the following steps can be used:
- To convert meters to centimeters, multiply by 100.
- To convert centimeters to meters, divide by 100.

For example, to convert 45 meters to centimeters, multiply 45 by 100, which equals 4500 centimeters.

**Multiple Choice Questions**

1. Which of the following units is not used to measure length?
(a) Centimeters
(b) Kilometers
(c) Kilograms
(d) None of these
(e) All of these

Answer: (c) Kilograms

2. Convert 45 meters to centimeters.
(a) 5400 cm
(b) 4500 cm
(c) 240 cm
(d) None of these
(e) All of these

Answer: (b) 4500 cm

3. Convert 225 meters to centimeters and 3445 centimeters to meters.
(a) 22500 cm, 34.45 m
(b) 4500 cm, 32.34 m
(c) 2250 cm, 30.45 m
(d) None of these
(e) All of these

Answer: (a) 22500 cm, 34.45 m

4. Convert 85 meters 55 centimeters to centimeters.
(a) 2500 cm
(b) 8500 cm
(c) 8555 cm
(d) None of these
(e) All of these

Answer: (c) 8555 cm

5. If the length of a wire is 56.70 meters, find the length of the same wire in centimeters.
(a) 5670 cm
(b) 8500 cm
(c) 8555 cm
(d) None of these
(e) All of these

Answer: (a) 5670 cm

6. The side length of a table is 23 meters 4 centimeters. Express the length of the table in decimal numbers.
(a) 32.04 m
(b) 8.5 m
(c) 23.04 m
(d) None of these
(e) All of these

Answer: (c) 23.04 m


## Step 1: Remove duplicate text
The provided text contains many duplicate sections, which can be removed to refine the content.

## Step 2: Identify key concepts and remove irrelevant information
Key concepts include vector addition, unit conversion, mechanical properties of materials, mass and weight, Newton's laws, and types of motion. Irrelevant information, such as redundant phrases and sentences, can be removed.

## Step 3: Reorganize content into a logical structure
The refined text should be organized into sections that flow logically, covering topics such as vector addition, unit conversion, mechanical properties, and types of motion.

## Step 4: Refine multiple-choice questions and answers
Multiple-choice questions should be reviewed and refined to ensure they are clear and concise, with accurate answer options.

## Step 5: Maintain original length as much as possible
While removing irrelevant information, the refined text should still cover the essential concepts and ideas, maintaining a similar length to the original text.

The process for adding vectors works even when the vectors do not point along the same straight line. If solving a two-dimensional problem graphically, a protractor is needed to draw the vectors at the correct angle and measure the direction and magnitude.

When changing units, such as length from feet to meters or time from days to seconds, it is essential to treat unit symbols as multiplicative factors. For example, 'cm' (centimeter) or 'yr' (year) can be used to convert between units.

Mass is the amount of material present in an object, describing how much material makes up the object. Often, mass and weight are confused due to similar units, but weight is a derived unit and a measurement of the force exerted on an object by gravity.

One law that holds good in all natural processes is Newton's gravitation law, which states that every body in nature attracts every other body with a force of attraction directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Vectors can be associated with a point in space or not, depending on the context. For instance, the velocity of a particle is a vector directly associated with a point in space.

A system comprising a block and a spring can be considered closed if there are no changes in motion or state in the environment. When the system is defined as only the spring, the work done by the block on the spring can be positive, negative, or zero, depending on the direction of the force.

One-dimensional motion occurs when only one coordinate specifying the position of an object changes with time, resulting in motion along a straight line path. Two-dimensional motion occurs when two coordinates change, and the object moves in a plane.

Simple Harmonic Motion (SHM) is a type of oscillatory motion where the net force can be described by Hooke's law. A simple harmonic oscillator is a system that exhibits SHM, and such systems are the simplest oscillatory systems.


## Texas Go Math Grade 1 Module 16 Assessment Answer Key

**Vocabulary**

To divide a shape into halves, draw a line in the center. This will give you two equal halves of the shape.

**Drawing Halves**

* To divide a circle into halves, draw a line in the center.
* To divide a square into halves, draw a line in the center.
* To divide a rectangle into halves, draw a line in the center.

**Drawing Fourths**

* To divide a square into fourths, draw two lines to create four equal parts.
* To divide a rectangle into fourths, draw two lines to create four equal parts.
* To divide a circle into fourths, draw two lines to create four equal parts.

**Concepts and Skills**

* Color the shapes that show unequal shares.
* Color the shapes that show equal shares.

**Multiple Choice Questions**

1. Which shape shows halves?
A) A circle divided into four parts
B) A square divided into two equal parts
C) A rectangle divided into three parts
D) A triangle divided into two equal parts

Answer: B) A square divided into two equal parts

2. Which shape shows unequal parts?
A) A circle divided into two equal parts
B) A square divided into four equal parts
C) A rectangle divided into three unequal parts
D) A triangle divided into two equal parts

Answer: C) A rectangle divided into three unequal parts

3. Which shading shows half of a square?
A) A square divided into four equal parts
B) A square divided into two unequal parts
C) A square divided into two equal parts
D) A square divided into three equal parts

Answer: C) A square divided into two equal parts

4. Which shape shows fourths?
A) A circle divided into two equal parts
B) A square divided into four equal parts
C) A rectangle divided into three equal parts
D) A triangle divided into two equal parts

Answer: B) A square divided into four equal parts

5. Which shading shows a quarter of a rectangle?
A) A rectangle divided into two equal parts
B) A rectangle divided into four equal parts
C) A rectangle divided into three equal parts
D) A rectangle divided into four unequal parts

Answer: B) A rectangle divided into four equal parts


#### Answer

$$ y'(x) =\frac{y-1+3 \sin (3 x-y)}{\sin (3 x-y)-x}$$

#### Work Step by Step

Given the equation $$x+\cos (3 x-y)=x y$$, differentiate with respect to $x$:
\begin{align*}
1-3 \sin (3 x-y)+y^{\prime} \sin (3 x-y)&=y+x y^{\prime}\\
y^{\prime}[\sin (3 x-y)-x]&=y-1+3 \sin (3 x-y)
\end{align*}
This yields the derivative:
$$ y'(x) =\frac{y-1+3 \sin (3 x-y)}{\sin (3 x-y)-x}$$


## 8.18 Sorting

### 8.18.1 Constants

The sorting module defines four constants:

*   **ASCENDING**: The default sort order, where the first element is the smallest and the last element is the largest.
*   **NORMAL_ORDER**: The normal sort order used by custom comparison routines.
*   **DESCENDING**: The reverse of **ASCENDING**, where the first element is the largest and the last element is the smallest.
*   **REVERSE_ORDER**: Reverses the sense of the order returned by a custom comparison routine.

### 8.18.2 Routines

#### 8.18.2.1 sort

The **sort** function sorts a sequence into ascending order.

*   **Parameters**:
    *   `x`: The sequence to be sorted.
    *   `order`: The sort order (default is **ASCENDING**).
*   **Returns**: A new sequence with the elements sorted in ascending order.
*   **Comments**:
    *   The standard **compare** routine is used to compare elements.
    *   The function uses the "Shell" sort algorithm, which is not stable.
*   **Example**:

```euphoria
constant student_ages = {18, 21, 16, 23, 17, 16, 20, 20, 19}
sequence sorted_ages
sorted_ages = sort(student_ages)
-- result is {16, 16, 17, 18, 19, 20, 20, 21, 23}
```

#### 8.18.2.2 custom_sort

The **custom_sort** function sorts a sequence according to a user-defined order.

*   **Parameters**:
    *   `custom_compare`: The routine ID of the user-defined comparison routine.
    *   `x`: The sequence to be sorted.
    *   `data`: Optional user data (default is {}).
    *   `order`: The sort order (default is **NORMAL_ORDER**).
*   **Returns**: A new sequence with the elements sorted according to the user-defined order.
*   **Errors**: If the user-defined routine does not return a valid comparison result, an error occurs.
*   **Comments**:
    *   The user-defined routine must return an integer comparison result.
    *   The function uses the "Shell" sort algorithm, which is not stable.
*   **Examples**:

```euphoria
constant students = {{"Anne", 18}, {"Bob", 21}, {"Chris", 16}, {"Diane", 23}, {"Eddy", 17}, {"Freya", 16}, {"George", 20}, {"Heidi", 20}, {"Ian", 19}}

procedure byage(object a, object b)
    if equal(a[2], b[2]) then
        return compare(upper(a[1]), upper(b[1]))
    end if
    return compare(a[2], b[2])
end procedure

sequence sorted_byage
sorted_byage = custom_sort(routine_id("byage"), students)
-- result is {{"Chris", 16}, {"Freya", 16}, {"Eddy", 17}, {"Anne", 18}, {"Ian", 19}, {"George", 20}, {"Heidi", 20}, {"Bob", 21}, {"Diane", 23}}

sorted_byage = custom_sort(routine_id("byage"), students, {}, REVERSE_ORDER)
-- result is {{"Diane", 23}, {"Bob", 21}, {"Heidi", 20}, {"George", 20}, {"Ian", 19}, {"Anne", 18}, {"Eddy", 17}, {"Freya", 16}, {"Chris", 16}}
```

#### 8.18.2.3 sort_columns

The **sort_columns** function sorts the rows in a sequence according to a user-defined column order.

*   **Parameters**:
    *   `x`: The sequence to be sorted.
    *   `column_list`: A list of column indexes to sort by.
*   **Returns**: A new sequence with the rows sorted according to the user-defined column order.
*   **Comments**:
    *   The function uses the "Shell" sort algorithm, which is not stable.
    *   Non-existent columns are treated as coming before existing columns.
*   **Example**:

```euphoria
sequence dirlist
dirlist = dir("c:\\temp")
sequence sorted
sorted = sort_columns(dirlist, {-D_SIZE, D_NAME})
```

#### 8.18.2.4 merge

The **merge** function merges two pre-sorted sequences into a single sequence.

*   **Parameters**:
    *   `a`: The first pre-sorted sequence.
    *   `b`: The second pre-sorted sequence.
    *   `compfunc`: The routine ID of a user-defined comparison function (default is -1).
    *   `userdata`: Optional user data (default is {}).
*   **Returns**: A new sequence with the elements of `a` and `b` merged together.
*   **Comments**:
    *   If `a` or `b` is not already sorted, the resulting sequence might not be sorted either.
    *   The user-defined comparison function must return an integer comparison result.
*   **Example**:

```euphoria
sequence X, Y
X = sort({5, 3, 7, 1, 9, 0})
Y = sort({6, 8, 10, 2})
? merge(X, Y)
-- result is {0, 1, 2, 3, 5, 6, 7, 8, 9, 10}
```

#### 8.18.2.5 insertion_sort

The **insertion_sort** function sorts a sequence and optionally another object together.

*   **Parameters**:
    *   `s`: The sequence to be sorted.
    *   `e`: An object to be sorted with `s` (default is {}).
    *   `compfunc`: The routine ID of a user-defined comparison function (default is -1).
    *   `userdata`: Optional user data (default is {}).
*   **Returns**: A new sequence with the elements of `s` and `e` sorted together.
*   **Comments**:
    *   The function is usually faster than the standard sort when `s` and `e` are mostly sorted before calling the function.
    *   The user-defined comparison function must return an integer comparison result.
*   **Example**:

```euphoria
sequence X = {}
while true do
    newdata = get_data()
    if compare(-1, newdata) then
        exit
    end if
    X = insertion_sort(X, newdata)
    process(new_data)
end while
```


# Greatest Common Factor Calculator

To understand the concept of the greatest common divisor (GCD), it's essential to know what a common divisor is. A common divisor of several integers is a number that can divide each number in the series without a remainder.

## Definition and Examples

For example, the numbers 8 and 12 have common divisors of 1 and 4, as shown by the mathematical expressions: 8 = 4 ⋅ 2 and 12 = 3 ⋅ 4. Every number has at least two common divisors: 1 and itself.

## Determining the Greatest Common Divisor

The GCD of two natural numbers is the largest number that can divide both numbers without a remainder. If the GCD of two numbers is 1, they are called coprime. The GCD can be written as gcd(a, b) = c or (a, b) = c, with the first option being more commonly used.

For instance, the GCD of 4 and 16 is 4, as shown by the steps:

- List the divisors of 4: 4, 2, 1
- List the divisors of 16: 16, 8, 4, 2, 1
- Identify the common divisors: 4, 2, 1
- Choose the largest common divisor: 4

The GCD can also be found for three or more integers, which is the largest integer that can divide all numbers in the series. For example, the GCD of 6, 12, 18, and 42 is 6.

## GCD Properties

The GCD has several properties:

1. **Commutative Property**: gcd(a, b) = gcd(b, a)
2. **Divisibility Property**: If a is divisible by b, then gcd(a, b) = b
3. **Euclidean Property**: If a = bq + c, then gcd(a, b) = gcd(b, c)
4. **Multiplicative Property**: gcd(ma, mb) = m ⋅ gcd(a, b) for any natural number m
5. **Coprime Property**: If p is a common divisor of a and b, then gcd(a / p, b / p) = gcd(a, b) / p
6. **Existence of Common Divisor**: Any two numbers have at least one common divisor, which is 1

Understanding the theoretical foundations and practical applications of GCD is essential for working with ordinary fractions and is closely related to the concept of the least common multiple.


**Positioning a School to Take Advantage of the Sun's Path**

When designing a school, architects consider the amount of sunlight the building will receive throughout the day. The size and location of windows in classrooms impact the amount of electric lighting needed. As the sun rises and warms a location, the air temperature increases, affecting the amount of heating or cooling needed.

The sunrise and sunset times vary significantly between locations closer to the equator and those further north. Hillsborough, California, and Anchorage, Alaska, have distinct sunrise and sunset times throughout the year.

**Analyzing Sunrise and Sunset Data**

Using the sunrise data for March in Hillsborough, CA, and Anchorage, AK, we can answer the following questions:

1. On March 3, the sunrise time in Hillsborough, CA, is 6:15 AM, and the sunset time is 5:45 PM.
2. On March 3, the sunrise time in Anchorage, AK, is 7:15 AM, and the sunset time is 5:15 PM.

To calculate the hours of sunlight, we separate the day into hours before and after 12:00 PM. For example, if the sun rises at 6:15 AM and sets at 7:30 PM, the total sunlight hours are calculated as follows:
- Before noon: 6:15 AM - 12:00 PM = 5 hours 45 minutes
- After noon: 12:00 PM - 7:30 PM = 7 hours 30 minutes
- Total sunlight hours: 5 hours 45 minutes + 7 hours 30 minutes = 13 hours 15 minutes

3. On March 30, Hillsborough, CA, receives 12 hours 45 minutes of sunlight.
4. On March 30, Anchorage, AK, receives 13 hours 30 minutes of sunlight.
5. On March 20, Anchorage receives 13 hours and 9 minutes of sunlight.
6. This amount of time can be expressed as a decimal: 13.15 hours.
7. The rate of change of sunrise time from March 1 to March 5 in Hillsborough, CA, is -0.025 hours/day, and in Anchorage, AK, it is -0.042 hours/day. The units are hours/day, and the negative sign indicates that the sunrise time is getting earlier.
8. Between Saturday, March 7, and Sunday, March 8, the sunrise and sunset times jump due to the change from standard time to daylight saving time.

**Using Scatter Plots and Linear Models**

Using the scatter plots and linear models, we can answer the following questions:

9. When calculating the rate of change in sunrise times between March 7 and March 17, we encounter the problem of the time change on March 8. To solve this, we can adjust the data to account for the time change.
10. The estimated sunrise times for Anchorage, AK, on March 1, March 8, March 22, and March 29, are 7:15 AM, 6:45 AM, 6:15 AM, and 5:45 AM, respectively.
11. The average rate of change for Hillsborough, CA, is -0.025 hours/day, and for Anchorage, AK, it is -0.042 hours/day.
12. The difference in the rate of change of sunlight between the two cities is due to their geographic locations. Hillsborough, CA, has a latitude of 37° North, while Anchorage, AK, has a latitude of 61° North.

**Linear Models and Predictions**

Using the linear models, we can predict the sunrise times for March 27 in Hillsborough, CA, and Anchorage, AK:
- Hillsborough, CA: y = -0.025x + 6.7
- Anchorage, AK: y = -0.042x + 8

13. The predicted sunrise times for March 27 are 6:30 AM in Hillsborough, CA, and 6:00 AM in Anchorage, AK.
14. The models can be used to predict outside of the data set, but the accuracy may decrease. Predicting the sunrise time for December 1, which is 275 days after March 1, may not be accurate due to the changing sunrise times throughout the year.

**Sinusoidal Models and Predictions**

Using the sinusoidal models, we can predict the sunrise times for December 1:
- Hillsborough, CA: regEQ(x) = 1.31784*sin(27.7163*x+97.4785)+6.16655
- Anchorage, AK: regEQ(x) = 3.2396*sin(29.3408*x+91.3703)+6.84801

25. The sinusoidal regression models can be used to estimate the sunrise times for December 1.
26. The predicted sunrise times for December 1 using the sinusoidal models are 7:15 AM in Hillsborough, CA, and 8:30 AM in Anchorage, AK. Comparing these predictions to the real data chart, we can assess the accuracy of the models.

**Additional Questions**

17. The graphs for the two cities will not look the same for next year or last year due to the changing sunrise and sunset times throughout the year.
18. The sunrise time is changing most slowly around the equinoxes (March 20 and September 22).
19. The sunrise time is changing most rapidly around the solstices (June 21 and December 21).
20. In Hillsborough, CA, the 8:00 AM first-period classroom will have no light from the sun during the winter months (December to February).
21. In Anchorage, AK, the 8:00 AM first-period classroom will have no light from the sun during the winter months (November to January).
22. Sunrise and sunset times can affect a school's schedule, building operation costs, and the daily life of students in several ways, including energy consumption, classroom lighting, and outdoor activities.
23. On June 21, Hillsborough, CA, receives 15 hours 30 minutes of sunlight, and on December 21, it receives 9 hours 30 minutes of sunlight.
24. On June 21, Anchorage, AK, receives 18 hours 30 minutes of sunlight, and on December 21, it receives 6 hours 30 minutes of sunlight.


# Feedback Amplifier Design

This example demonstrates the design of a noninverting feedback amplifier circuit using Control System Toolbox. The design is built around the operational amplifier (op amp), a standard building block of electrical feedback circuits.

## Op Amp Description

The op amp is a differential voltage amplifier with extremely high dc gain, typically in the range of 1e5 to 1e7. The open-loop transfer function of the op amp can be represented as a linear time-invariant (LTI) system:

$$a(s)=\frac{{a}_{0}}{(1+s/{\omega}_{1})(1+s/{\omega}_{2})}$$

where $a_0 = 1e5$, $\omega_1 = 1e4$, and $\omega_2 = 1e6$.

## Creating a Transfer Function Model

A transfer function model of the op amp can be created using the `tf` command:
```matlab
s = tf('s');
a = a0/(1+s/w1)/(1+s/w2)
```
The frequency response of the op amp can be viewed using the `bodeplot` command.

## Feedback Amplifier

A resistive feedback network is added to the op amp to create a noninverting amplifier. The feedback network is a voltage divider with input `Vo` and output `Vn`. The transfer function of the feedback network is:

$$b = \frac{Vn}{Vo} = \frac{R1}{R1 + R2}$$

The closed-loop gain of the amplifier is:

$$A = \frac{Vo}{Vp} = \frac{a}{1 + ab}$$

If the product of `a` and `b` is sufficiently large, the closed-loop gain can be approximated as:

$$A = \frac{1}{b}$$

## Designing the Amplifier

To design an amplifier with a dc gain of 10, we can solve for `R2`:
```matlab
A0 = 10;
b = 1 / A0;
R1 = 10000;
R2 = R1 * (1/b - 1)
```
The resulting value of `R2` is 90 kOhm.

## Constructing the Closed-Loop System

The closed-loop system can be constructed using the `feedback` command:
```matlab
A = feedback(a, b);
```
The frequency responses of the open-loop and closed-loop systems can be plotted using the `bodemag` command.

## Gain vs. Bandwidth Tradeoff

The use of negative feedback reduces the low-frequency gain but increases the system bandwidth. The gain versus bandwidth tradeoff is a powerful tool in the design of feedback amplifier circuits.

## System Sensitivity

The system sensitivity represents the sensitivity of the closed-loop gain to variation in the op amp's natural gain. The system sensitivity can be calculated as:

$$S = \frac{1}{1 + ab}$$

The system sensitivity can be plotted using the `bodemag` command.

## Feedback Lead Compensation

To improve the stability margin of the amplifier, feedback lead compensation can be used. This involves adding a capacitor in parallel with the feedback resistor. The value of the capacitor can be chosen to introduce a phase lead to the feedback transfer function near the crossover frequency.

## Compensated Closed-Loop System

The compensated closed-loop system can be constructed using the `feedback` command:
```matlab
A_comp = feedback(a, b_array(:,:,6));
```
The step response of the compensated closed-loop system can be plotted using the `stepplot` command.

## Summary

The final component values for the noninverting feedback amplifier circuit are:

* R1 = 10 kOhm
* R2 = 90 kOhm
* C = 2 pF

The resistive feedback network was selected to yield a broadband amplifier gain of 10. Feedback lead compensation was used to tune the loop gain near the crossover frequency, resulting in a maximum phase margin of about 58 degrees.


## One-Way and Two-Way ANOVA

In ANOVA, one-way and two-way refer to the number of independent variables. A one-way ANOVA has one independent variable, while a two-way ANOVA has two independent variables. The concept of group levels is also crucial, where levels refer to different groups within the same independent variable. For example, if the independent variable is a brand of cereal, the levels might be Lucky Charms, Raisin Bran, and Cornflakes.

Consider a study in a psychiatric hospital, where the goal is to test the effectiveness of a combination of alcoholic support group and individual counseling in reducing alcohol consumption. Participants can be split into three levels: Counseling only, Counseling and medication, and Medication only. The dependent variable is the number of alcoholic drinks consumed in a day. If the levels have a hierarchical structure, a nested ANOVA is suitable.

Replication involves duplicating the test with multiple groups. In a 2-way ANOVA with replication, two groups are used, and individuals within those groups perform more than one task. For instance, two groups of students from different elementary schools taking two assessment tests. Replication is unnecessary if only one group takes two tests.

## Choosing the Right ANOVA Test

There are two primary types of ANOVA: one-way and two-way tests. The two-way test can be with or without replication. One-way ANOVA is used to test two groups and determine if there's a difference between them. However, it won't identify which group is different. An ad hoc test, such as the least significant difference, is necessary to determine which group has a difference in mean.

Two-way ANOVA without replication is suitable for one group being double-tested, such as a doctor testing patients before and after medication. A two-way ANOVA can be considered an extension of one-way ANOVA, used when there's one measurement variable and two nominal variables. It's ideal for experiments with a quantitative outcome and two categorical explanatory variables.

## Main Effect and Interaction Effect

The main effect and interaction effect are calculated from the results of a two-way ANOVA. The main effect, also known as a simple effect, is the impact of one independent variable on the dependent variable, similar to a one-way ANOVA. It considers the effect of each factor separately.

Interaction effects occur when an independent variable interacts with another independent variable on a response variable. These effects are easily tested if each cell has more than one observation. Understanding main and interaction effects is crucial in interpreting the results of a two-way ANOVA.


**GillMatsuno.jl**

## Introduction

GillMatsuno.jl is a Julia package that numerically solves the Shallow-Water Equations on a β-plane and allows custom heat-forcing Q. It can be installed via `] add GillMatsuno`.

## Key Features and Usage

The numerical solution to the Gill-Matsuno Model is called using `GMcalc()`, which accepts the following keyword arguments:

* `xmin` and `xmax`: left and right boundaries of the domain
* `ymin` and `ymax`: lower and upper boundaries of the domain
* `δx` and `δy`: grid-spacing in the x- and y- directions
* `nt` and `δt`: number of timesteps and time between each step
* `A` and `L`: amplitude and length of the heat-forcing Q
* `α`: damping coefficient on the winds induced by the heat-forcing (default: `α = 0.1`)
* `β`: Coriolis Factor (default: `β = 0.5`)
* `g` and `H`: gravity and height of the domain (default: `g = 1.0` and `H = 1.0`)

## Example Usage

An example of `GMcalc()` with default values is:
```
GMcalc(xmin=-25.0, xmax=50.0, δx=0.1, ymin=-10.0, ymax=10.0, δy=0.1, nt=5000, δt=0.001,
       A=1, L=2, α=0.1, β=0.5, g=1.0, H=1.0)
```


## Physics - Sound Waves

Sound waves that are audible to humans have frequencies between 20 hertz and 20,000 hertz, equivalent to 20 cycles per second to 20,000 cycles per second. Frequency is also referred to as pitch. Sounds with pitches above the audible range are called ultrasonic, while those below the audible range are called infrasonic.

### Loudness

Loudness is closely related to two quantities: intensity and loudness. Intensity is measured in watts per square meter (W/m2), while loudness is measured in decibels (dB). The relationship between loudness and intensity is given by the equation: Loudness in decibels equals ten times the log of the intensity over the intensity at the threshold of hearing, which is 10^-12 watts per m2.

Multiplying the intensity by 10 adds 10 decibels to the associated loudness, multiplying by 100 adds 20 decibels, and multiplying by 1000 adds 30 decibels. This logarithmic relationship highlights the non-linear connection between intensity and perceived loudness. Understanding these concepts is essential for grasping the fundamentals of sound waves and their characteristics.


## 5.1 Sequences

### 5.1.1 Definition

A sequence is an infinitely long list of real numbers. For example, the sequence of positive even integers is \(\<2,4,6,8,\dots\>\). An explicit formula \(a_n\) is a rule defining each term of the sequence, where \(n=0\) yields the first term, \(n=1\) gives the next term, and so on. The sequence generated by the formula \(a_n\) is written as \(\<a_n\>_{n=0}^\infty=\<a_0,a_1,a_2,\dots\>\). Occasionally, the first term of the sequence may be given by an integer different from \(0\), in which case the sequence is written like \(\<a_n\>_{n=1}^\infty\).

Examples of sequences include:
- \(\<1,3,5,7,9,\dots\>\)
- \(\<3,-6,9,-12,15,\dots\>\)
- \(\<0,1,4,9,16,\dots\>\)
- \(\<a_n\>_{n=0}^\infty\) defined by \(a_n=4n\)
- \(\<b_n\>_{n=0}^\infty\) defined by \(b_n=\frac{(-1)^n}{n^2+2}\)
- \(\<c_n\>_{n=0}^\infty\) defined by \(c_n=\cos(\frac{\pi}{2}n)\)

### 5.1.2 Recursive Formulas

A recursive formula for a sequence defines one or more initial terms of the sequence and then defines future terms of the sequence by using previous terms. Examples include:
- The Fibonacci sequence defined by \(f_0=1,f_1=1,f_{n+2}=f_n+f_{n+1}\)
- The factorial sequence defined by \(!_0=1,!_{n+1}=(n+1)!_n\), which can also be written as \(n!=1\times2\times3\times\dots\times n\)
- The sequence \(\<a_n\>_{n=0}^\infty\) defined recursively by \(a_0=3,a_{n+1}=\frac{a_n}{2}\), which has an explicit formula \(a_n=\frac{3}{2^n}\)

### 5.1.3 Limits, Convergence, and Divergence

The sequence \(\<a_n\>_{n=i}^\infty\) converges to a limit \(L\) if for each \(\epsilon>0\), there exists an integer \(N\) such that \(|a_n-L|<\epsilon\) for all \(n\geq N\). This is written as \(\lim_{n\to\infty}a_n=L\) or \(a_n\to L\). Examples include:
- The harmonic sequence \(\<a_n\>_{n=1}^\infty\) defined by \(a_n=\frac{1}{n}\)
- The sequence defined by \(g_n=\frac{2^n}{2^{n+1}}\)
- A sequence diverges when it doesn’t converge to any limit, such as the sequence defined by \(b_n=(-1)^n\frac{n+1}{n+2}\)

### Review Exercises

1. Guess the next three terms of the sequences \(\<1,5,9,13,17,\dots\>\), \(\<1,\frac{1}{4},\frac{1}{9},\frac{1}{16},\frac{1}{25},\dots\>\), and \(\<\frac{1}{3},-1,3,-9,27,\dots\>\), and create an explicit formula for each.
2. Write the first five terms of the sequences \(\<a_n\>_{n=0}^\infty\), \(\<b_n\>_{n=0}^\infty\), and \(\<c_n\>_{n=0}^\infty\) defined by \(a_n=3n+2\), \(b_n=2(-\frac{1}{3})^n\), and \(c_n=\frac{n}{1+n^2}\).
3. Write the first six terms of the sequence \(\<q_n\>_{n=0}^\infty\) defined by \(q_0=0\) and \(q_{n+1}=q_n+2n+1\), and prove that \(q_n=n^2\) is an explicit formula for this sequence.
4. Write the first six terms of the sequence \(\<b_n\>_{n=1}^\infty\) defined by \(b_1=4\) and \(b_{n+1}=\frac{b_n}{2}\), and prove that \(b_n=\frac{8}{2^n}\) is an explicit formula for this sequence.
5. Guess the limit of the sequences:
   - The alternating harmonic sequence \(\<b_n\>_{n=1}^\infty\) defined by \(b_n=\frac{(-1)^n}{n}\)
   - The geometric sequence \(\<g_n\>_{n=0}^\infty\) defined by \(g_n=2^{-n}\)
   - The sequence \(\<a_n\>_{n=3}^\infty\) defined by \(a_n=\frac{3n+2}{2n+1}\)
6. Determine whether the sequences appear to be converging or diverging:
   - The sequence defined by \(c_n=\frac{n!}{n^2+1}\)
   - The sequence defined by \(s_n=\sin(\frac{\pi n}{3})\)
7. Sketch a picture to explain why \(\lim_{n\to\infty} \sin(\pi n)=0\) as the limit of a sequence, but \(\lim_{x\to\infty}\sin(\pi x)\) does not exist as a limit of a function.
8. Find the first five terms of the sequences:
   - \(\<r_n\>_{n=1}^\infty\) defined explicitly by \(r_n=\frac{n+2}{3+n^2}\)
   - \(\<w_n\>_{n=0}^\infty\) defined recursively by \(w_0=1\), \(w_1=2\), \(w_{n+2}=2w_n+w_{n+1}\)
9. Choose the most appropriate statement for describing the sequence \(\<1,\frac{3}{4},\frac{5}{8},\frac{9}{16},\frac{17}{32},\dots\>\):
   - The sequence appears to converge to \(\frac{1}{2}\).
   - The sequence appears to diverge to \(\frac{1}{2}\).
   - The sequence appears to neither converge nor diverge.


# Additional Examples of Solving Linear Congruences 2

Recall the following properties regarding linear congruences:

**1)** If $ac \equiv bc \pmod m$ and $(c, m) = 1$, then $a \equiv b \pmod m$.

**2)** If $ac \equiv bc \pmod m$ and $(c, m) = d$, then $a \equiv b \pmod {m/d}$.

**3)** Solutions to linear congruences are regarded as least residues.

**4)** The linear congruence $ax \equiv b \pmod m$ has no solutions if $(a, m) \nmid b$.

**5)** The linear congruence $ax \equiv b \pmod m$ has 1 solution if $(a, m) = 1$.

**6)** The linear congruence $ax \equiv b \pmod m$ has multiple solutions if $(a, m) \mid b$.

## Example 1

**Solve the linear congruence $81x \equiv 33 \pmod {145}$.**

We note that $(81, 145) = 1$, hence there is 1 solution to this congruence. We can find this solution by finding a modular inverse of 81 (mod 145) using the division algorithm. The modular inverse of 81 (mod 145) is -34. Thus, we get:

$81x \equiv 33 \pmod {145}$

$x \equiv -34 \cdot 33 \pmod {145}$

$x \equiv 38 \pmod {145}$

Our solution is 38 (mod 145).

## Example 2

**Solve the linear congruence $64x \equiv 28 \pmod {122}$.**

We note that $(64, 122) = 2$, and 2 | 28. Hence, we will have precisely 2 solutions (mod 122). First, let's reduce this congruence down using property 2 to get that $32x \equiv 14 \pmod {61}$. Now, let's solve this congruence by finding an inverse of 32 (mod 61) with the division algorithm. The modular inverse of 32 (mod 61) is 21. Thus, we get:

$32x \equiv 14 \pmod {61}$

$x \equiv 21 \cdot 14 \pmod {61}$

$x \equiv 50 \pmod {61}$

Our solutions are 50 and 50 + 61 = 111 (mod 122).

## Example 3

**Solve the linear congruence $20x \equiv 36 \pmod {124}$.**

We first note that $(20, 124) = 4$, and 4 | 36, hence there are precisely 4 solutions to this congruence. We will first simplify this congruence down by reducing the congruence by 4: $5x \equiv 9 \pmod {31}$. Now, we will find a modular inverse of 5 (mod 31) with the division algorithm. The modular inverse of 5 (mod 31) is -6. Thus, we get:

$5x \equiv 9 \pmod {31}$

$x \equiv -6 \cdot 9 \pmod {31}$

$x \equiv 8 \pmod {31}$

Our solutions are 8, 8 + 31, 8 + 31(2), and 8 + 31(3) (mod 124), which are 8, 39, 70, and 101.

## Example 4

**Solve the linear congruence $410x \equiv 232 \pmod {1332}$.**

We note that $(410, 1332) = 2$, and that 2 | 232, hence our linear congruence has precisely 2 solutions. Dividing our congruence by 2, we obtain that $205x \equiv 116 \pmod {666}$. Now, we'll apply the division algorithm to find an inverse of 205 (mod 666). The modular inverse of 205 (mod 666) is 13. Thus, we get:

$205x \equiv 116 \pmod {666}$

$x \equiv 13 \cdot 116 \pmod {666}$

$x \equiv 176 \pmod {666}$

Our solutions are 176 and 176 + 666 = 842 (mod 1332).

## Example 5

**Solve the linear congruence $1232x \equiv 339 \pmod {1442}$.**

We note that $(1232, 1442) = 14$. However, 14 does not divide 339, hence there are no solutions to this linear congruence.


The rate at which borrowers take money and the interest on that money or loan is charged at this rate. In other words, it is some percentage of the principal that is paid at fixed intervals on a regular basis during a fixed time duration, which can be monthly, yearly, weekly, or daily. The rate of interest is measured in percentages.

## Formulas

Since interest is of two types - simple interest and compound interest - the formulas for measuring interest rates vary with interest types:

**Simple Interest:**

The formula for finding simple interest is:

I = (P * R * T) / 100

Where:
- I = interest
- P = principal (the amount of loan given)
- R = rate of interest at which the loan is given
- T = time for which the loan is given

The rate of interest can be found using:

R = (100 * I) / (P * T)

**Compound Interest:**

The formula for finding the final amount in compound interest is:

A = P (1 + r)^n

Where:
- A = final amount after the complete tenure
- P = principal (the starting amount)
- r = rate of interest in accordance to the tenure of compounding (in decimals)
- n = number of compounding intervals

The rate of interest can be deduced as:

(1 + r)^n = A / P

1 + r = (A / P)^(1/n)

r = (A / P)^(1/n) - 1

## Examples

**Finding the Rate of Interest:**

**Example 1:** A person earned $460 in a bank account when they deposited $400 for 3 years (not compounded). Find the rate of interest applied.

Given:
- Principal (P) = $400
- Amount (A) = $460
- Time (T) = 3 years

First, find the interest: I = A - P = $460 - $400 = $60

Using the simple interest formula: I = (P * R * T) / 100

R = (100 * I) / (P * T) = (100 * 60) / (400 * 3) = 5

The rate of interest applied is 5% per annum.

**Example 2:** A woman deposited $900 in a bank to earn $1024 after a tenure of 2 years, compounded annually. Find the rate of interest applied.

Given:
- Principal (P) = $900
- Amount (A) = $1024
- Time (T) = 2 years
- Compounding intervals (n) = 2 (since it's annual compounding)

Using the compound interest formula: A = P (1 + r)^n

1024 = 900 (1 + r)^2

(1 + r)^2 = 1024 / 900

Taking the square root of both sides: 1 + r = (1024 / 900)^(1/2) = 32/30 = 16/15

r = (16/15) - 1 = 1/15 = 0.0667

To express 'r' as a percentage, multiply by 100: r = 0.0667 * 100 = 6.67%

The rate of interest applied is 6.67%.


## Homework Statement
For what value of x does [tex] \frac{1}{x} + \frac{1}{1-x} > 0 [/tex]

## Step 1: Identify restrictions on x
The expression is undefined when x = 0 or x = 1, as these values result in division by zero.

## Step 2: Combine the fractions
Combining the fractions yields [tex] \frac{1}{x} + \frac{1}{1-x} = \frac{1-x+x}{x(1-x)} = \frac{1}{x(1-x)} [/tex]. For the expression to be positive, the denominator x(1-x) must be positive.

## Step 3: Define a function for the denominator
Let f(x) = x(1-x), which is a parabola. This function changes signs at its roots, x = 0 and x = 1.

## Step 4: Determine the intervals where f(x) is positive
Testing values in the intervals defined by the roots, we find that f(x) is positive when 0 < x < 1.

## Step 5: Evaluate the solution
The given solution states that the expression is positive for 0 < x < 1 or x < 0 or x > 1, but testing x = 2 (which is greater than 1) yields [tex] \frac{1}{2} + \frac{1}{1-2} = \frac{1}{2} + \frac{1}{-1} = -\frac{1}{2} [/tex], which is not greater than 0. This indicates an error in the given solution for x > 1.

## Step 6: Conclusion
Given the analysis, the correct solution for [tex] \frac{1}{x} + \frac{1}{1-x} > 0 [/tex] is 0 < x < 1.


Rational and irrational numbers are two types of real numbers. A rational number can be written as the ratio of two integers, whereas an irrational number cannot.

Examples of rational numbers include terminating decimals, such as 0.25 or 1.3, and non-terminating but repeating decimals, such as 0.66… or 3.242424…. These numbers can be expressed as fractions, with the denominator not equal to zero.

On the other hand, irrational numbers are non-terminating and non-repeating decimals. Examples of irrational numbers include √2, √3, √5, √11, √21, and π (Pi).

To determine if a number is rational or irrational, we can check if it can be written as a fraction. If it can, then it is rational; otherwise, it is irrational. For instance, the decimal 1.0227 is a rational number because it is a terminating decimal. Similarly, 1.3333 is a rational number because it can be written as 4/3.

In contrast, numbers like 3.141141114… are irrational because they are non-terminating and non-repeating.

Other examples of rational numbers include 0.5555, 3.1416, and 0.456 repeating. These numbers can be expressed as fractions, making them rational.

It's worth noting that all integers are rational numbers, as they can be written as fractions with a denominator of one. Additionally, any decimal number that is repeating can be written in the form a/b, where b is not equal to zero, making it a rational number.

In summary, rational numbers are those that can be expressed as fractions, including terminating and repeating decimals, while irrational numbers are non-terminating and non-repeating decimals.

Multiple choice questions:

1. Is 1.3 a rational or irrational number?
A) Rational
B) Irrational
Answer: A) Rational

2. Which of the following is an irrational number?
A) 0.25
B) 0.66…
C) √2
D) 3.1416
Answer: C) √2

3. Is 1.0227 repeating a rational number?
A) Yes
B) No
Answer: A) Yes

4. Is 3.14159 a rational number?
A) Yes
B) No
Answer: A) Yes

5. What is an example of an irrational number?
A) 0.5555
B) √5
C) 1.3
D) 3.1416
Answer: B) √5


The system of Vedic Mathematics was rediscovered from ancient Sanskrit texts and has been fully structured for use in schools. It uses sixteen word-formulae that relate to how we use our mind. One of these formulae is Vertically and Crosswise, which has a wide range of applications, including multiplication.

**Multiplication**

To multiply numbers using Vertically and Crosswise, consider the example of 33 multiplied by 44:

3 3

4 4 ×

1 4 25 12 answer

Multiplying vertically on the right gives 3×4 = 12, so we put down 2 and carry 1. Then, we multiply crosswise and add the two results: 3×4 + 3×4 = 24. Adding the carried 1 gives 25, so we put 5 and carry 2. Finally, we multiply vertically on the left, get 3×4 = 12, and add the carried 2 to get 14.

This method can also be carried out from left to right, which has many advantages. To find 33 × 44 from left to right:

3 3

4 4 ×

12 44 5 2 answer

Vertically on the left, 3×4 = 12, so we put 1 and carry 2 to the right. Crosswise, we get 3×4 + 3×4 = 24, add the carried 2 as 20, to get 44. Finally, vertically on the right, 3×4 = 12, add the carried 4 as 40, to get 52.

When calculating from left to right, we add a zero to the carried figure because the first product is really 30×40 = 1200, and the 200 is 20 tens. This method is better because we write and pronounce numbers from left to right, making it easier to get our answers the same way. It can be extended to products of numbers of any size and allows us to combine operations, such as adding two products.

The same vertical and crosswise method can be used for algebraic multiplications. For example, (2x + 5)(3x + 1):

2x + 5

3x + 1 ×

6×2 + 17x + 5 answer

Using the method from the left, we have: 2x × 3x = 6×2, 2x × 1 + 5 × 3x = 17x, and 5 × 1 = 5. This method can be applied to various mathematical operations, including the calculation of sines, cosines, tangents, and their inverses, as well as the solution of polynomial and transcendental equations.


# 15. Metal Corrosion and Soil Acids

An investigation determined the relationship between metal corrosion and exposure time to soil acids. The percentage of corrosion and exposure time were measured weekly.

**Questions:**

**a.** What percentage of the total variation is explained by the quadratic regression of y on x?

**b.** Is the regression on x and x2 significant at the 5% level of significance?

**c.** Is the linear regression coefficient significant when x2 is in the model?

**d.** Is the quadratic regression coefficient significant when x is in the model?


One to one maths interventions built for KS4 success

Weekly online one to one GCSE maths revision lessons are now available for students who are confident with negative numbers, addition and subtraction, multiplication and division, decimals, fractions, and rounding.

The topic of mean, median, and mode is relevant for KS4 students. These three averages are measures of central tendency, used to represent a set of data.

To calculate the mean, find the total of the values and divide the total by the number of values. For example, the mean of a set of numbers can be calculated as:
\text{Mean}=\frac{\text{total}}{\text{number of values}}.

The median is the middle number in a set of values. To find the median, arrange the values in numerical order and find the middle value. For example, the median of a set of numbers can be found by arranging the numbers in order and selecting the middle value.

The mode is the most common number in a set of values. To find the mode, identify the most frequently occurring item in the data set. For example, the mode of a list of numbers can be found by identifying the number that occurs most frequently.

To find the mean, median, or mode, follow these steps:
- Mode: consider how many times the values occur, and note that there may be two modes (bimodal).
- Median: ensure the list of values is in numerical order and find the middle of the set.
- Mean: find the total and divide by the number of values.

Here are some examples of finding the mode, median, and mean:
- Find the mode: consider the number of times different items occur. If one item occurs more than any other, that item is the mode.
- Find the median: ensure the list of numbers is in numerical order, then find the middle of the data set. If there is an even number of values, the median is the average of the two middle values.
- Find the mean: find the total and divide by the number of values.

Some key points to note when working with averages:
- A data set can have one mode, two modes (bimodal), or no mode at all.
- When finding the median, ensure the list of numbers is in numerical order.
- The mean can be a decimal or fraction and may need rounding.

Here are some practice questions:
1. Find the mode:
12
17
18
16
The most common item is ‘18’, which occurs twice.

2. Find the mode:
19
21
23
19 and 23
The most common items are ‘19’ and ‘23’, both occurring twice.

3. Find the median:
18
19
17
31
Put the numbers in order: 17, 18, 19, 31. The middle value is 19.

4. Find the median:
35
36
37
38
Put the numbers in order: 35, 36, 37, 38. The middle pair of values is 36 and 37, and their average is 36.5.

5. Find the mean:
17
12.5
16.4
15.5
\text{Mean}= \frac{\text{total}}{\text{number of values}} = \frac{17+12.5+16.4+15.5}{4} = \frac{61.4}{4} = 15.35

6. Find the mean:
6
8
7
5
\text{Mean}= \frac{\text{total}}{\text{number of values}} = \frac{6+8+7+5}{4} = \frac{26}{4} = 6.5

Here are some multiple-choice questions:
1. Here is a list of numbers:
(a) Write down the mode
(b) Work out the median
(c) Work out the mean

Answers:
(a) 15
(b) 16
(c) (18+15+19+16+15)÷5 = 16.6

2. Here is a list of numbers:
(a) Write down the mode
(b) Work out the median
(c) Work out the mean

Answers:
(a) 1.4
(b) 1.6
(c) (1.2+1.4+1.4+1.6+1.8+1.9+2.0)÷7 = 1.61 (to 3 sf)

3. There are five cards with some numbers on them.
The number on one of the cards is 2.
The mean of 5 numbers is 7.
The mode is 10.
The median is 8.
Find the numbers on the other four cards.

Answers:
10 and 10 (using the mode)
8 (using the median)
(7×5)-(2+8+10+10) = 5 (using the mean)

By learning about mean, median, and mode, students can develop a deeper understanding of data analysis and representation. Other relevant topics include range, representing data, and frequency tables. With practice and review, students can master these concepts and achieve success in their KS4 maths GCSEs.


## Curriculum Framework | Adopted: 2016

### AR.Math.Content.8.NS: The Number System

* Know that there are numbers that are not rational, and approximate them by rational numbers.
* Understand that every number has a decimal expansion.
* Write a fraction a/b as a repeating decimal and a repeating decimal as a fraction.
* Use rational approximations of irrational numbers to compare the size of irrational numbers, locate them approximately on a number line diagram, and estimate the value of expressions.

### AR.Math.Content.8.EE: Expressions and Equations

* Work with radicals and integer exponents.
* Know and apply the properties of integer exponents to generate equivalent numerical expressions.
* Use square root and cube root symbols to represent solutions to equations.
* Evaluate square roots of small perfect squares and cube roots of small perfect cubes.
* Use numbers expressed in the form of a single digit times an integer power of 10 to estimate very large or very small quantities.
* Perform operations with numbers expressed in scientific notation.

### AR.Math.Content.8.EE.B: Understand the connections between proportional relationships, lines, and linear equations

* Graph proportional relationships, interpreting the unit rate as the slope of the graph.
* Compare two different proportional relationships represented in different ways.
* Show why the slope m is the same between any two distinct points by creating similar triangles.
* Write the equation y = mx for a line through the origin and y = mx + b for a line intercepting the vertical axis at b.

### AR.Math.Content.8.EE.C: Analyze and solve linear equations and pairs of simultaneous linear equations

* Solve linear equations in one variable.
* Give examples of linear equations in one variable with one solution, infinitely many solutions, or no solutions.
* Solve pairs of simultaneous linear equations.
* Find solutions to a system of two linear equations in two variables.

### AR.Math.Content.8.F: Functions

* Define, evaluate, and compare functions.
* Understand that a function is a rule that assigns to each input exactly one output.
* Compare properties of two functions each represented in a different way.
* Identify the unique characteristics of functions by comparing their graphs, equations, and input/output tables.
* Construct a function to model a linear relationship between two quantities.
* Describe the functional relationship between two quantities by analyzing a graph.

### AR.Math.Content.8.G: Geometry

* Understand congruence and similarity using physical models, transparencies, or geometry software.
* Verify experimentally the properties of rotations, reflections, and translations.
* Understand that a two-dimensional figure is congruent to another if the second can be obtained from the first by a sequence of rotations, reflections, and translations.
* Understand and apply the Pythagorean Theorem.
* Model or explain an informal proof of the Pythagorean Theorem and its converse.
* Apply the Pythagorean Theorem to determine unknown side lengths in right triangles.
* Apply the Pythagorean Theorem to find the distance between two points in a coordinate system.

### AR.Math.Content.8.SP: Statistics and Probability

* Investigate patterns of association in bivariate data.
* Construct and interpret scatter plots for bivariate measurement data.
* Know that straight lines are widely used to model relationships between two quantitative variables.
* Use the equation of a linear model to solve problems in the context of bivariate measurement data.
* Understand that patterns of association can also be seen in bivariate categorical data by displaying frequencies and relative frequencies in a two-way table.


# Numerical Modelling - Constitutive Models

Constitutive models describe the stress-strain behavior of materials in response to applied loads, particularly post-failure. These models predict plastic and elastic straining and physical properties of materials. Pre-failure behavior is often linear elastic, while post-failure behavior is more complex.

Rock materials exhibit three possible behaviors: strain softening, elastic-perfectly plastic, and elastic-brittle. Strain softening occurs in average quality rock, where stored energy is released as kinetic energy after failure. Elastic-perfectly plastic behavior is seen in poor quality rock, while elastic-brittle behavior is characteristic of hard, good quality rock.

The path dependency of nonlinear materials must be considered in modeling. Numerical solutions should accommodate different loading paths to apply constitutive models correctly. Nonlinearity of the stress-strain response is another important factor, with complexity increasing in the post-failure region.

Established constitutive models are classified into three groups: algebraic, differential, and integral constitutive relations. Algebraic relations include linear and nonlinear elastic models, while differential relations include plastic and viscoelastic models. Integral relations include hereditary viscoelastic models.

The strains of a material can be divided into elastic and plastic strains. Elastic strains develop during loading and disappear upon unloading, while plastic strains remain after unloading. Plastic strains start to develop when the material reaches its yielding limit, dictated by the yield criterion.

## Constitutive Model Groups

### Null Model Group

The null material model represents removed or excavated material, with stresses set to zero and no body forces acting on it.

### Elastic Model Group

Elastic models are characterized by reversible deformations upon unloading, with linear and path-independent stress-strain laws.

#### Elastic, Isotropic Model

The elastic, isotropic model is the simplest model of material behavior, with linear stress-strain behavior and no hysteresis on unloading. The model is expressed by Robert Hooke's law: σ = E ε, where σ is stress, E is Young's modulus, and ε is strain.

#### Elastic, Transversely Isotropic Model

The elastic, transversely isotropic model has a plane of elastic symmetry, with material properties the same in all directions.

### Plastic Model Group

Plastic models are characterized by nonlinear stress-strain relations and failure criteria. Different plasticity models are classified based on shear yield and potential functions, non-associated flow rules, and stress corrections.

#### Drucker-Prager Model

The Drucker-Prager model is used to model soft clays with low friction angles, with a failure envelope based on the Drucker-Prager criterion.

#### Mohr-Coulomb Model

The Mohr-Coulomb model is a common constitutive model, describing a linear relationship between shear stress and normal stress at failure. The model has two defining parameters: friction angle (φ) and cohesion (c).

#### Ubiquitous-Joint Model

The ubiquitous-joint model represents weak planes in an anisotropic plasticity model, with failure criteria depending on the orientation of the weak plane, material properties, and stress.

#### Strain-Hardening/Softening Model

This model represents the nonlinear behavior of strain softening or hardening materials, with the yield surface changing size, shape, and position.

#### Bilinear Strain-Hardening/Softening Ubiquitous-Joint Model

This model represents the softening and hardening behavior of materials, with the failure envelope and weak plane incorporated with ubiquitous-joint model properties.

#### Double-Yield Model

The double-yield model represents materials with significant irreversible compaction and shear yielding, such as hydraulically placed backfill or lightly cemented granular material.

#### Hoek-Brown Model

The Hoek-Brown criterion is an empirical nonlinear refinement of the Mohr-Coulomb criterion, specifically designed for rock-like materials. The model is based on stress conditions in intact rock and rock masses that lead to failure.

## Summary

The most common elastic-plastic constitutive models are summarized in the table below, along with representative materials and applications.

| Model | Description | Representative Material | Application |
| --- | --- | --- | --- |
| Mohr-Coulomb | Linear relationship between shear stress and normal stress | Rock, soil | Geotechnical engineering |
| Drucker-Prager | Failure envelope based on Drucker-Prager criterion | Soft clays | Geotechnical engineering |
| Ubiquitous-Joint | Weak planes in anisotropic plasticity model | Rock, soil | Geotechnical engineering |
| Strain-Hardening/Softening | Nonlinear behavior of strain softening or hardening materials | Rock, metal | Geotechnical engineering, materials science |
| Hoek-Brown | Empirical nonlinear refinement of Mohr-Coulomb criterion | Rock-like materials | Geotechnical engineering |

## References

1. Itasca Consulting Group. (2011). UDEC, Universal Distinct Element Code.
2. Strafield, M., & Cundall, P. (1988). Towards a methodology for rock mechanics modelling. International Journal of Rock Mechanics and Mining Science.
3. Smed, E. M., & Cundall, P. (2012). Elasto-plasto Strain Hardening Mohr-Coulomb Model-Derivation and Implementation. Aalborg, Denmark.
4. Kasper, W. J. (2002). Constitutive Models for Engineering Materials. University of Colorado.
5. Kelly, P. (2014). Solid Mechanics Lecture Notes. The University of Auckland.
6. Raisa, E. (2012). A study of Geotechnical Constitutive Models using PLAXIS 2D.
7. Wikipedia. (n.d.). Mohr–Coulomb theory.
8. Salvador, I. (2010). Drucker-Prager yield criterion application to study the behavior of CFRP confined concrete under compression. Department of Construction Engineering, University of Alicante, Spain.
9. Gang, L., Hong, P., Harumi, K., & Yoshiaki, M. (2003). Application of Ubiquitous Joints Models in Numerical Modeling. Chinese Journal of Rock Mechanics and Engineering.
10. PLAXIS. (2014). Material Models Manual.
11. Itasca. (n.d.). UDEC Features - Materials.


Grade 4 - Mathematics 5.2 Divisor, Dividend, Quotient, and Remainder

Multiplication and division are inverse operations. In multiplication, two numbers are given to find their product. For example, 5 × 15 = 75. This multiplication problem can be rewritten as a division problem: what number times 15 equals 75? To find the answer, we divide 75 by 15, which gives us 5.

In division, we divide a number (the dividend) into groups of a certain size (the divisor). The result is the quotient. For instance, 75 ÷ 15 = 5, where 75 is the dividend, 15 is the divisor, and 5 is the quotient. Division can also be thought of as repeated subtraction.

Key terms in division:
- Dividend: the number being divided
- Divisor: the number that divides the dividend
- Quotient: the number of times the dividend is divisible by the divisor
- Remainder: the amount left over if the dividend is not completely divisible by the divisor

To check if a division is correct, we use the formula: Dividend = Quotient x Divisor + Remainder. If the remainder is 0, the division is exact. For example, to check if 8 divided by 2 is 4, we multiply the quotient (4) and the divisor (2), which equals 8. Since 8 = 4 x 2 + 0, the division is correct.

Directions: Answer the following questions and provide at least 5 examples of your own for division with remainder and division without remainder.


# How to Add Mathematical Notation in the Rich-Text Editor

To add mathematical notation, **MathJax** must be enabled on the site. If enabled, properly formatted LaTeX expressions entered into the Rich-Text Editor will be rendered correctly in published content. Site administrators can enable **MathJax** by following the steps outlined in the Help article "How do I enable and use LaTeX math in my site?"

Once **MathJax** is enabled, there are two ways to add mathematical notation:

## Entering LaTeX Expressions in the Editor as Text

To enter LaTeX expressions directly into the editor, surround the symbol or formula with **$** for inline placement or **$$** for centered placement on its own line. Examples of properly formatted LaTeX formulas for use in the editor include:

### Inline LaTeX

- $x=\frac{1+y}{1+2z^2}$
- $\int_0^\infty e^{-x^2} dx=\frac{\sqrt{\pi}}{2}$
- $\sqrt{2} \sin x$, $\sqrt{2}\,\sin x$

## Adding LaTeX Using the Math Button

The **Math** button allows for the insertion and modification of LaTeX symbols or formulas within the editor.

### Steps to Add LaTeX Using the Math Button

1. **Position the Cursor**: Place the cursor at the desired location in the editor.
2. **Enter the LaTeX Expression**: A "Mathematics in TeX" dialog window will appear. In this window:
   - Enter the LaTeX expression in the "Write your TeX here" box without surrounding notation.
   - Sample formulas:
     - $x=\frac{1+y}{1+2z^2}$
     - $\int_0^\infty e^{-x^2} dx=\frac{\sqrt{\pi}}{2}$
     - $\sqrt{2} \sin x$, $\sqrt{2}\,\sin x$
   - A preview of the symbol or formula will be displayed.
   - Select **OK** to add the expression to the editor.

### Editing Formulas or Symbols

The editor's tools can be used to format or change expressions as needed.


A "$k$-chain" is a multi-graph obtained from a path of length $k$ by duplicating every edge, resulting in $2^k$ paths between the two endpoints.

Given a simple graph $G$ with $n$ nodes and nodes $s$ and $t$, if there are at least $n^k$ simple paths from $s$ to $t$, can a $\Omega(k)$-chain be obtained with $s$ and $t$ as endpoints through edge deletion and contraction? Alternatively, can an $\Omega(\sqrt k)$-chain or $\Omega(k^\alpha)$-chain be obtained for any $\alpha > 0$?

This problem is related to the question of whether graphs with a large number of cycles always contain a large necklace minor. Any partial answer or intuition regarding this conjecture would be appreciated.


Math 2

The concept that 0.99999999 repeating equals 1 can be proven through a simple algebraic manipulation. Let's say 0.999999 repeating equals x. Multiplying both sides by 10 gives 10x = 9.9999999999 repeating. Subtracting 1x from both sides yields 9x = 9.000000000. Dividing both sides by 9 results in x = 1.

This idea may seem counterintuitive, but it is a fundamental property of mathematics. A similar concept is the limit of 1x as x approaches 0, which also equals 1.

Another example of a counterintuitive math concept is the attempt to prove that 1 equals 2. This is often done through a flawed algebraic manipulation, such as:
Let x = y, then
xy = x^2 - y^2 = (x + y)(x - y)
y(x - y) = (x + y)(x - y)
y = x + y
y = y + y
y = 2y
1 = 2

However, this proof is incorrect and relies on a "divide by zero" trick.

A more straightforward proof of the concept that 0.99999999 repeating equals 1 is through decimal expansion. Since 1/9 in decimal is 0.1111..., multiplying both sides by 9 gives 1 = 0.999.... This proof is more intuitive and avoids the pitfalls of flawed algebraic manipulations.

It's essential to be aware of common mistakes in mathematical proofs, such as dividing by zero or ignoring negative solutions to square roots. These mistakes can lead to incorrect conclusions and undermine the validity of mathematical concepts.