The number of miles Ling jogged more than Trent can be expressed as 2.8x - 8, where x represents the number of hours.

Trent's expression is 2.2x + 8, and Ling's expression is 5x. The difference between Ling's and Trent's expressions is 5x - (2.2x + 8) = 5x - 2.2x - 8 = 2.8x - 8.

Therefore, Ling jogged 2.8x - 8 miles more than Trent.

A parallelogram has vertices A(-2, 4), B(4, 4), C(2, -2), and D(-4, -2). To find the area, divide the parallelogram into a quadrilateral and a triangle. The area of the quadrilateral is 4 * 6 = 24, and the area of the triangle is (2 * 6) / 2 = 6. Since there are two triangles, multiply the area by 2: 2 * 6 = 12. The total area is 24 + 12 = 36 square units.

Given a standard deck of 52 playing cards, the probability of picking three black cards in a row without replacement is 2/17 or approximately 0.1176. The probability of picking a black card on the first draw is 26/52 = 1/2. On the second draw, it's 25/51, and on the third draw, it's 24/50 = 12/25. Multiply these probabilities: (1/2) * (25/51) * (12/25) = 6/51 = 2/17.

The equation for the function P(x) representing Ayden's total pay on a day on which he sells x dollars worth of computers is P(x) = 0.01x + 90. The base pay is $90, and the commission is 1% of the sales amount.

To find the number of acres in a lot that is 1/2 mile by 1/2 mile, convert the length and width from miles to feet: 1/2 mile * 5280 feet/mile = 2640 feet. The area is 2640 * 2640 = 6,969,600 square feet. Since 1 acre = 43,560 square feet, divide the area by 43,560: 6,969,600 / 43,560 = 160 acres.

If the gross income per month is $4000, the monthly disposable income after taxes is $2920. The federal tax is 10% of $4000 = $400, the state tax is 7% of $4000 = $280, the city tax is 3% of $4000 = $120, and the social security tax is 7% of $4000 = $280. The total tax is $400 + $280 + $120 + $280 = $1080. The monthly disposable income is $4000 - $1080 = $2920.

The equation of the total bill with x number of refills is y = 5 + x. If the bill is $9, then 9 = 5 + x, so x = 4. The number of refills is 4.

The half-life of Tl-201 is 3.0 days. To find the time it takes for the activity to be one-fourth of the initial activity, use the formula: time = half-life * log2(initial activity / final activity). time = 3.0 * log2(1 / 0.25) = 3.0 * log2(4) = 3.0 * 2 = 6.0 days.

The slope-intercept form of a line is y = 2/3x + 2. A line perpendicular to this line has a slope of -3/2. The equation of the line shifted down 8 units is y = -3/2x - 6.

Expand and simplify (3x + 4)(2x + 3) = 6x^2 + 17x + 12.

The equation for the total bill with x number of refills is y = 5 + x. If the bill is $9, then 9 = 5 + x, so x = 4. The number of refills is 4.

The approximate cost of a 5-hour trip if the price of fuel is R78 per liter is R3000.68. The total distance traveled is 100 km/h * 5 h = 500 km. The total fuel consumed is 500 km / 13 km/l = 38.46 liters. The cost of fuel is 38.46 liters * R78/liter = R3000.68.

The inequality for buying pencils and an eraser with a budget of $7 is 0.25x + 0.75 ≤ 7, where x is the number of pencils.

The main cause of dysmenorrhea in adolescents is prostaglandins.

The reform movement that affected South Carolina the least during the 1800s is education.

The location of the hydrogens on carbon A relative to the hydrogens on carbon B in a 1H NMR spectrum of the given compound is not specified.

The factors that could cause two regions to develop with different economies are different land resources, different climate, different water resources, different natural resources, and different populations.

The radius of a right circular cylinder with a volume of 12 in^3 and a minimum surface area is not specified.

The percentage decrease in the total number of students on the honor roll is not specified.

The line in the poem where the speaker compares her love to the pursuit of a social cause is not specified.

Pierre and Marie Curie discovered radiation in 1900.

According to Erikson, parents establish trust in their 6-month-old by being responsive to their needs.

In a regression equation, the "predicted score" for a person with a "predictor variable" score of 0 is 7. The regression constant is 7.

The human rights issues of concern addressed by the EPA, FCC, FTC, and FDA are not specified.

The type of firm for which locational excellence is particularly important is retailers.

The depreciation for the vehicle using method 2 is not specified.

The number of gifts Jason can wrap with 4 feet of ribbon is not specified.

The volume of the right circular cylinder is 12 in^3. To find the radius, use the formula V = πr^2h, where V is the volume, π is a constant, r is the radius, and h is the height. Since the surface area is minimum, the height is equal to the diameter, which is 2r. So, V = πr^2 * 2r = 2πr^3. Now, 12 = 2πr^3. Solve for r: r^3 = 12 / (2π) = 1.91, r = ∛1.91 = 1.26 inches.

The percentage decrease in the total number of students on the honor roll is calculated as follows: initial number of boys = 100, initial number of girls = 80, total initial students = 180. After the decrease, the number of boys = 100 - 2% of 100 = 98, the number of girls = 80 - 10% of 80 = 72. Total students after decrease = 98 + 72 = 170. The percentage decrease is ((180 - 170) / 180) * 100 = 5.56%.

The line in the poem where the speaker compares her love to the pursuit of a social cause is not specified.

Pierre and Marie Curie discovered radiation in 1900.

According to Erikson, parents establish trust in their 6-month-old by being responsive to their needs.

In a regression equation, the "predicted score" for a person with a "predictor variable" score of 0 is 7. The regression constant is 7.

The human rights issues of concern addressed by the EPA, FCC, FTC, and FDA are not specified.

The type of firm for which locational excellence is particularly important is retailers.

The depreciation for the vehicle using method 2 is not specified.

The number of gifts Jason can wrap with 4 feet of ribbon is 4 / (1/3) = 12.

The volume of the right circular cylinder is 12 in^3. To find the radius, use the formula V = πr^2h, where V is the volume, π is a constant, r is the radius, and h is the height. Since the surface area is minimum, the height is equal to the diameter, which is 2r. So, V = πr^2 * 2r = 2πr^3. Now, 12 = 2πr^3. Solve for r: r^3 = 12 / (2π) = 1.91, r = ∛1.91 = 1.26 inches.

The percentage decrease in the total number of students on the honor roll is calculated as follows: initial number of boys = 100, initial number of girls = 80, total initial students = 180. After the decrease, the number of boys = 100 - 2% of 100 = 98, the number of girls = 80 - 10% of 80 = 72. Total students after decrease = 98 + 72 = 170. The percentage decrease is ((180 - 170) / 180) * 100 = 5.56%.


# Josie Martin

I am a student from Maggie Walker Governor's School in Virginia, currently enrolled in a math and art seminar. As a project, I created a piece using plywood, house paints, and acrylic paints to represent the binary code for numbers 1-100.

The binary code is based on the concept that all numbers can be formed by adding powers of two. Specifically, the numbers 1, 2, 4, 8, 16, 32, and 64 are required to create the numbers 1-100. For instance, the number 53 can be formed using the numbers 1, 4, 16, and 32. In traditional binary code, the numbers used are represented by a 1 and the numbers not used are represented by a 0. The code sequence starts with the largest number and descends. Therefore, the binary code for 53 would be 01100101.

In my piece, I replaced the traditional 0's and 1's with colors to represent the binary code. Each panel corresponds to a specific number, with colors indicating which powers of two are used. For example, the panel for 53 features yellow, blue, white, and purple. All the panels are glued onto a black board, creating a visual representation of the binary code for numbers 1-100.


**Two Proportion Hypothesis Test**
The two-proportion z-test is used when the sampling method for each population is simple random sampling, the samples are independent, and each sample includes at least 10 successes and 10 failures.

**Standard Error for 2 Sample Z Test**
The standard error used for a 2 sample z test for a difference in proportions is the proportion.

**Symbol for Sample Proportion**
The symbol for sample proportion is p̂.

**Two Sample T Test**
The two-sample t-test is used to test whether the unknown population means of two groups are equal or not.

**Testing Proportions in Statistics**
To test proportions in statistics, state the null hypothesis H0 and the alternative hypothesis HA, set the level of significance, calculate the test statistic, calculate the p-value, and make a decision.

**Null Hypothesis for Two Sample T Test**
The default null hypothesis for a 2-sample t-test is that the two groups are equal.

**Normal Distribution of Proportions**
The normal distribution of proportions can be found using the formula Z = (X - m) / s, where Z is the z-score, X is the value, m is the population mean, and s is the standard deviation.

**Interpreting Difference in Proportions**
The expected value of the difference between all possible sample proportions is equal to the difference between population proportions.

**Statistical Distribution for Testing Difference between 2 Population Proportions**
The difference of two proportions follows an approximate normal distribution.

**Difference between Proportion and Percentage**
A proportion is a quantity of something that is part of the whole amount or number, while a percentage is the amount, number, or rate of something regarded as part of a total of 100.

**Finding Sample Proportion**
The sample proportion is approximately normally distributed, with mean μ̂P = p and standard deviation σ̂P = √(pqn).

**Using T Test for Proportions**
The t test is not appropriate for proportions because the mean and variance are independent in the case of means, but not for proportions.

**Calculating Hypothesis for Population Proportion**
To calculate a hypothesis for a population proportion, the sample was obtained through a simple random sample process, n * p * (1 - p) ≥ 10, and n ≤ 0.05 * N.

**Interpreting Proportions**
Use the p-value to determine whether the population proportion is statistically different from the hypothesized proportion.

**Comparing Two Proportions in Statistics**
To compare two proportions in statistics, calculate the sample proportions, find the difference between the two sample proportions, calculate the overall sample proportion, and calculate the standard error.

**Proportions in Statistics**
A proportion refers to the fraction of the total that possesses a certain attribute.

**Finding Difference between Proportions**
To calculate a CI for the difference between two population proportions, determine the confidence level, find the sample proportion, take the difference between the sample proportions, and multiply z* times the result.

**Two Sample Test of Proportions**
The two sample proportion test is used to determine whether the proportions of two groups differ.

**Proportion Comparison**
The compare proportions test is used to evaluate if the frequency of occurrence of some event, behavior, intention, etc. differs across groups.

**One Proportion Z Test**
The one proportion z test is used to compare an observed proportion to a theoretical one, when there are only two categories.

**Meaning of Proportion**
A proportion is a harmonious relation of parts to each other or to the whole, balance, or symmetry.

**Difference between Two Independent T Test and Z Test for Two Proportions**
The two independent t test is used to compare the means of two independent samples, while the z test for two proportions is used to compare the proportions of two independent samples.

**Difference between Z Test and T Test**
The z test is used when the true variance of the population is known, while the t test is used when the true variance of the population is unknown.

**Formula of Proportion**
The formula of proportion is a/b = c/d or a:b = c:d.

**Sample Proportion P̂**
The sample proportion p̂ is the mean of all sample proportions.

**Interpreting Z Test**
The z test is used to determine whether the population proportion is statistically different from the hypothesized proportion.

**Z Test for Proportions**
The z test for proportions is used to compare an observed proportion to a theoretical one, when there are only two categories.


# Prime Palette

This past π day, I reflected on mathematical occurrences, such as prime numbers in prices, ages, or dates. I recalled a Tweet about a visualization of prime factors of the first 100 whole numbers by Sondra Eklund. The visualization uses a square box for each number, with prime numbers filled in with a unique color and non-prime numbers divided into colored rectangular slices representing their prime factors.

For example, the number 2 is colored blue and appears as a prime factor in many numbers. In the number 12, two blue rectangles represent the prime factor 2, and a red rectangle represents the prime factor 3. This creates a striking mosaic where the eye scans and multiplies colors to generate numbers.

Inspired by the initial Twitter thread, people recreated the art with Python. I set out to re-imagine the work, creating a scalable design that captures relationships between numbers. I used Streamlit's new sharing platform to deploy the app.

**Identifying and Storing Prime Factors**

I created a function `prime_factors` that takes a numerical input and returns a list of prime factors. The function uses the modulo operator to find prime factors, starting with the divisor 2 and incrementing until the dividend is reduced to 1.

```python
def prime_factors(number):
    prime_factors = []
    dividend = number
    divisor = 2
    while True:
        if dividend == 1:
            return prime_factors
        elif dividend % divisor == 0:
            prime_factors.append(divisor)
            dividend = dividend / divisor
        else:
            divisor += 1
```

I also created a function `prime_df` that returns the prime factors for a range of numbers. This function uses the `prime_factors` function to find the prime factors for each number in the range and stores them in a Pandas dataframe.

```python
def prime_df(number):
    num_prime_array = np.array(([2, 2, 1]))
    if number > 2:
        for this_num in range(3, number+1):
            num_primes = prime_factors(this_num)
            unique, counts = np.unique(num_primes, return_counts=True)
            freq = np.asarray((list(unique), list(counts))).T
            num_id = np.array([[this_num] * len(freq)]).T
            num_id_unique_counts = np.hstack((num_id, freq))
            num_prime_array = np.vstack((num_prime_array, num_id_unique_counts))
        num_prime_df = pd.DataFrame(num_prime_array)
    else:
        num_prime_df = pd.DataFrame([num_prime_array], index=[0])
    num_prime_df.columns = ['Numbers', 'Primes', 'Powers']
    return num_prime_df
```

**Depicting Prime Factors**

I used Seaborn's relplot to depict the prime factors of a set of numbers. The x-axis represents the set of numbers, and the y-axis represents the prime factors. The plot shows the frequencies between factors and the factors commonly shared across numbers. I used colored bubbles of varying size to represent the powers of each prime factor.

To customize the plot, I created a color map by blending a range of colors. I then loaded the code onto Streamlit, creating a dashboard that allows users to control the range of numbers depicted on the graph and switch up the color palette.

The dashboard provides an interactive way to explore the patterns and relationships between prime factors. I look forward to building bigger and better things and sharing them with readers.


**Year 6 Programme of Study**

**Number - Number and Place Value**
Pupils should be taught to:
- read, write, order, and compare numbers up to 10,000,000, determining the value of each digit
- round whole numbers to a required degree of accuracy
- use negative numbers in context and calculate intervals across 0
- solve number and practical problems involving the above concepts

**Number - Addition, Subtraction, Multiplication, and Division**
Pupils should be taught to:
- multiply multi-digit numbers (up to 4 digits) by two-digit whole numbers using long multiplication
- divide numbers (up to 4 digits) by two-digit whole numbers using long division and short division, interpreting remainders
- perform mental calculations with mixed operations and large numbers
- identify common factors, common multiples, and prime numbers
- use the order of operations for calculations involving the 4 operations
- solve multi-step problems in contexts, deciding on operations and methods
- use estimation to check calculation answers and determine the degree of accuracy

**Number - Fractions (including Decimals and Percentages)**
Pupils should be taught to:
- simplify fractions using common factors and express fractions in the same denomination using common multiples
- compare and order fractions, including those greater than 1
- add and subtract fractions with different denominators and mixed numbers
- multiply and divide proper fractions, writing answers in simplest form
- associate fractions with division and calculate decimal equivalents
- identify the value of each digit in numbers up to 3 decimal places and perform multiplication and division by 10, 100, and 1,000
- recall and use equivalences between simple fractions, decimals, and percentages

**Ratio and Proportion**
Pupils should be taught to:
- solve problems involving relative sizes of quantities using integer multiplication and division facts
- calculate percentages and use them for comparison
- solve problems involving similar shapes with known or unknown scale factors
- solve problems involving unequal sharing and grouping using fractions and multiples

**Algebra**
Pupils should be taught to:
- use simple formulae
- generate and describe linear number sequences
- express missing number problems algebraically
- find pairs of numbers satisfying an equation with 2 unknowns
- enumerate possibilities of combinations of 2 variables

**Measurement**
Pupils should be taught to:
- solve problems involving calculation and conversion of units of measure using decimal notation up to 3 decimal places
- use, read, write, and convert between standard units, including length, mass, volume, and time
- convert between miles and kilometres
- calculate the area of parallelograms and triangles
- calculate, estimate, and compare volumes of cubes and cuboids using standard units

**Geometry - Properties of Shapes**
Pupils should be taught to:
- draw 2-D shapes using given dimensions and angles
- recognise, describe, and build simple 3-D shapes, including making nets
- compare and classify geometric shapes based on properties and sizes
- find unknown angles in triangles, quadrilaterals, and regular polygons
- illustrate and name parts of circles, including radius, diameter, and circumference

**Geometry - Position and Direction**
Pupils should be taught to:
- describe positions on the full coordinate grid (all 4 quadrants)
- draw and translate simple shapes on the coordinate plane and reflect them in the axes

**Statistics**
Pupils should be taught to:
- interpret and construct pie charts and line graphs to solve problems
- calculate and interpret the mean as an average


Waves can be categorized into two main types: electromagnetic waves and mechanical waves. Electromagnetic waves do not require a medium to travel, whereas mechanical waves do require a medium to travel. In this unit, we will focus on mechanical waves.

Mechanical waves travel through a medium by causing the particles of that medium to oscillate up and down or side to side with the disturbance as it passes. The particles of the medium return to their original position after the disturbance passes and do not move with the wave. Mechanical waves can travel through all types of media, including gases, liquids, and solids. A wave is a means of transferring energy, but not mass, from one point to another.

There are two kinds of mechanical waves: transverse and longitudinal. A transverse wave has the following characteristics:
- Amplitude (A): the height of the wave measured from equilibrium
- Crest: the highest point on a wave
- Trough: the lowest point on a wave
- Wavelength (λ): the length of one full wave, measured from trough to trough, crest to crest, or through one full wave starting from equilibrium
- Period (T): the amount of time it takes for one wave to pass a given point
- Frequency (f): the inverse of period, the number of waves that pass a given point in 1 second
- Wave speed (v): the speed that a wave moves through a medium

The equations that relate these characteristics are:
- f = 1/T
- v = d/t
- v = λ(1/T)
- v = λf

Longitudinal waves, on the other hand, are made up of areas of stretch and compression of the particles of the medium. A stretched area is also known as a rarefaction. Longitudinal waves can be graphed with distance on the x-axis and amount of compression (density, pressure, etc.) graphed on the y-axis.

When two waves pass through the same space, they do not interact. Instead, each one acts on the particles in that space independently, and then the waves move through each other without affecting the other. This is known as superposition, where the waves simply add (or subtract) their effect on the medium as they cross.

Standing waves occur when a wave is produced on a string or slinky and travels down to one end, then reflects back. If the wave is of the right frequency, the reflected waves will constructively and destructively interfere in such a way to make a standing wave. A standing wave is usually drawn with its reflection, and each bump is ½ of a wavelength. A node is an area of maximum destructive interference, where there is no displacement of the medium, while an antinode is an area of maximum constructive interference, where there is maximum displacement of the medium.


# 2. Compare Fractions with Half and One

To compare fractions to benchmark fractions (1/2 and 1), find an equivalent fraction with the same denominator. For example, to compare 2/6 with 1/2, use the equivalent fraction 3/6. Since 2/6 < 3/6 (smaller numerator), then 2/6 < 1/2.

To compare a fraction to 1, follow similar logic. For 4/7 and 1, use 7/7 as the equivalent fraction. Since 4/7 < 7/7, then 4/7 < 1. As a shortcut, if the numerator is smaller than the denominator, the fraction is smaller than 1.

### Common Core Alignment

**4.NF.2** Compare two fractions with different numerators and denominators by creating common denominators, comparing to benchmark fractions (like 1/2), or using visual models. Recognize that comparisons are valid only when referring to the same whole. Record results using >, =, or < and justify conclusions.


Accuracy and Error
Upper and lower bounds are used to express the uncertainty in a measurement. For example, a table measured to two decimal places is 1.86m tall. The smallest possible value is 1.855m, as any smaller value would round down to 1.85m. The largest possible value is less than 1.865m, as 1.865m itself would round up to 1.87m. Therefore, the true length lies between these bounds, expressed as 1.855 ≤ L < 1.865 or 1.86 ± 0.005.

Note that when a question asks for the "upper bound," the answer is 1.865, not <1.865. The upper bound can never be reached, which is why "<" is used instead of "≤".

Relative and percentage error provide a better way to represent errors, as they take into account the importance of the error compared to the size of the object being measured. For instance, a 3-meter error could be significant when building a bridge, but negligible when measuring the distance from the Earth to the Sun.

Diagrams can be used to represent upper and lower bounds, although this is rarely asked at GCSE level.

Significant figures and decimal places are crucial when recording measurements. When dealing with decimal places, 1.372000 is more accurate than 1.372. When dealing with significant figures, the examiner may not always be able to determine the number of significant figures used. For example, 1200 can be expressed as 1 significant figure (1000), 2 significant figures (1200), or 3 significant figures (1200).

A common complication with significant figures arises when determining upper and lower bounds. For a measurement of 10cm to 2 significant figures, the upper and lower bounds are not 9.5 and 10.5, but rather 9.95 and 10.5. This is because 9.5 to 2 significant figures is still 9.5, whereas 9.95 to 2 significant figures rounds to 10.

Calculations involving upper and lower bounds require using these bounds at every stage. For example, if 20.7cm is cut from a 32.2cm piece of wood, the upper and lower bounds for the remaining wood can be calculated as follows:

* Upper bound: Assume the original piece was 32.2 + 0.05 = 32.25cm, and cut as little as possible (20.7 - 0.05 = 20.65cm) off. The upper bound of the remaining wood is 32.25 - 20.65 = 11.6cm.
* Lower bound: Assume the original piece was 32.2 - 0.05 = 32.15cm, and cut as much as possible (20.7 + 0.05 = 20.75cm) off. The lower bound of the remaining wood is 32.15 - 20.75 = 11.4cm.

Therefore, the actual value can only be expressed as "11 to 1 significant figure," as this is the most accurate value that both 11.4 and 11.6 round to. Expressing the true value as "11 to 2 significant figures" would be incorrect, as 11.6 rounds to 12 to 2 significant figures.


# Spearman and Pearson Correlations

## Analyzing Poll Results
A TV ad claims that 53% of the population supports making smoking illegal, with a margin of error of +/- 3%. This means that the true percentage of supporters is likely between 50% and 56%. Given the small margin of error, the result is relatively reliable.

## Sampling Methodology
As the marketing director of Harley-Davidson, using a survey on the company website to determine customer preferences for the next model is not a good sampling frame. This method may attract a biased sample of existing customers who are already interested in the brand, rather than a representative sample of the broader target market.

## Interpreting Unemployment Rates
The local unemployment rate dropped from 10.5% to 10.4%. This 0.1% decrease may not be statistically significant, as it is a relatively small change. To determine significance, more information about the sample size and margin of error would be needed.

## Type I and Type II Errors
A Type I Error is more serious than a Type II Error in a medical context, where a false positive diagnosis could lead to unnecessary treatment. In contrast, a Type II Error is more serious in a quality control context, where a false negative result could lead to defective products being released.

## p-Value and Null Hypothesis
The p-value indicates the probability of observing the test results assuming the null hypothesis is true. If the p-value is smaller than the level of significance, it suggests that the null hypothesis can be rejected, indicating a statistically significant result.

## Dependent Samples and Statistical Tests
A study tracking the average highway speed of 30 drivers before and after consuming coffee involves dependent samples, as the same drivers are being measured twice. To determine if there is a statistical difference in driving speeds, a paired t-test would be appropriate, as it accounts for the correlation between the two measurements.

## Correlation and t-Test
Correlation can be used in real-life scenarios, such as analyzing the relationship between coffee consumption and driving speed. However, a t-test is necessary to confirm that the correlation is statistically significant and not due to chance.

## Spearman and Pearson Correlations
Spearman and Pearson correlations are both used to measure the relationship between two variables. Pearson correlation assumes a linear relationship and is sensitive to outliers, while Spearman correlation is non-parametric and can handle non-linear relationships. Pearson correlation is suitable for continuous data, while Spearman correlation is suitable for ordinal or ranked data.


**Gradian/Year to Gon/Microsecond Conversion**

One Gradian/Year is equivalent to 3.1688087814029e-14 Gon/Microsecond. To convert Gradian/Year to Gon/Microsecond, multiply the number by 3.1688087814029e-14.

**Conversion Formula:**
1 Gradian/Year = 1 × 3.1688087814029e-14 = 3.1688087814029e-14 Gon/Microsecond

**Gradian/Year and Gon/Microsecond Definitions:**

* Gradian per year is a unit of measurement for angular velocity, representing a change in orientation of an object by one gradian every year. The symbol is grad/year.
* Gon per microsecond is a unit of measurement for angular velocity, representing a change in orientation of an object by one gon every microsecond. The symbol is gon/μs.

**Conversion Table:**

| Gradian/Year [grad/year] | Gon/Microsecond [gon/μs] |
| --- | --- |
| 1 | 3.1688087814029e-14 |
| 2 | 6.3376175628058e-14 |
| 3 | 9.5064263442087e-14 |
| 4 | 1.2675235125612e-13 |
| 5 | 1.5844043907014e-13 |
| 6 | 1.9012852688417e-13 |
| 7 | 2.218166146982e-13 |
| 8 | 2.5350470251223e-13 |
| 9 | 2.8519279032626e-13 |
| 10 | 3.1688087814029e-13 |
| 100 | 3.1688087814029e-12 |
| 1000 | 3.1688087814029e-11 |

**Gradian/Year Conversions to Other Units:**

* 1 gradian/year in degree/second: 2.8519279032626e-8
* 1 gradian/year in degree/millisecond: 2.8519279032626e-11
* 1 gradian/year in degree/microsecond: 2.8519279032626e-14
* 1 gradian/year in degree/nanosecond: 2.8519279032626e-17
* 1 gradian/year in degree/minute: 0.0000017111567419576
* 1 gradian/year in degree/hour: 0.00010266940451745
* 1 gradian/year in degree/day: 0.0024640657084189
* 1 gradian/year in degree/week: 0.017248459958932
* 1 gradian/year in degree/month: 0.075
* 1 gradian/year in degree/year: 0.9
* 1 gradian/year in radian/second: 4.9775531941431e-10
* 1 gradian/year in radian/millisecond: 4.9775531941431e-13
* 1 gradian/year in radian/microsecond: 4.9775531941431e-16
* 1 gradian/year in radian/nanosecond: 4.9775531941431e-19
* 1 gradian/year in radian/minute: 2.9865319164858e-8
* 1 gradian/year in radian/hour: 0.0000017919191498915
* 1 gradian/year in radian/day: 0.000043006059597396
* 1 gradian/year in radian/week: 0.00030104241718177
* 1 gradian/year in radian/month: 0.0013089969389957
* 1 gradian/year in radian/year: 0.015707963267949
* 1 gradian/year in gradian/second: 3.1688087814029e-8
* 1 gradian/year in gradian/millisecond: 3.1688087814029e-11
* 1 gradian/year in gradian/microsecond: 3.1688087814029e-14
* 1 gradian/year in gradian/nanosecond: 3.1688087814029e-17
* 1 gradian/year in gradian/minute: 0.0000019012852688417
* 1 gradian/year in gradian/hour: 0.0001140771161305
* 1 gradian/year in gradian/day: 0.0027378507871321
* 1 gradian/year in gradian/week: 0.019164955509925
* 1 gradian/year in gradian/month: 0.083333333333333
* 1 gradian/year in gon/second: 3.1688087814029e-8
* 1 gradian/year in gon/millisecond: 3.1688087814029e-11
* 1 gradian/year in gon/microsecond: 3.1688087814029e-14
* 1 gradian/year in gon/nanosecond: 3.1688087814029e-17
* 1 gradian/year in gon/minute: 0.0000019012852688417
* 1 gradian/year in gon/hour: 0.0001140771161305
* 1 gradian/year in gon/day: 0.0027378507871321
* 1 gradian/year in gon/week: 0.019164955509925
* 1 gradian/year in gon/month: 0.083333333333333
* 1 gradian/year in gon/year: 1
* 1 gradian/year in revolution/second: 7.9220219535072e-11
* 1 gradian/year in revolution/millisecond: 7.9220219535072e-14
* 1 gradian/year in revolution/microsecond: 7.9220219535072e-17
* 1 gradian/year in revolution/nanosecond: 7.9220219535072e-20
* 1 gradian/year in revolution/minute: 4.7532131721043e-9
* 1 gradian/year in revolution/hour: 2.8519279032626e-7
* 1 gradian/year in revolution/day: 0.0000068446269678303
* 1 gradian/year in revolution/week: 0.000047912388774812
* 1 gradian/year in revolution/month: 0.00020833333333333
* 1 gradian/year in revolution/year: 0.0025
* 1 gradian/year in sign/second: 9.5064263442087e-10
* 1 gradian/year in sign/millisecond: 9.5064263442087e-13
* 1 gradian/year in sign/microsecond: 9.5064263442087e-16
* 1 gradian/year in sign/nanosecond: 9.5064263442087e-19
* 1 gradian/year in sign/minute: 5.7038558065252e-8
* 1 gradian/year in sign/hour: 0.0000034223134839151
* 1 gradian/year in sign/day: 0.000082135523613963
* 1 gradian/year in sign/week: 0.00057494866529774
* 1 gradian/year in sign/month: 0.0025
* 1 gradian/year in sign/year: 0.03
* 1 gradian/year in mil/second: 5.0700940502446e-7
* 1 gradian/year in mil/millisecond: 5.0700940502446e-10
* 1 gradian/year in mil/microsecond: 5.0700940502446e-13
* 1 gradian/year in mil/nanosecond: 5.0700940502446e-16
* 1 gradian/year in mil/minute: 0.000030420564301468
* 1 gradian/year in mil/hour: 0.0018252338580881
* 1 gradian/year in mil/day: 0.043805612594114
* 1 gradian/year in mil/month: 1.33
* 1 gradian/year in mil/year: 16

**Disclaimer:** We strive to provide accurate conversions, but we cannot guarantee their correctness. Please validate the results with an authority before using them.


Let $S$ be a scheme. A morphism of algebraic spaces $f : X \to Y$ over $S$ is said to be *integral* if for every affine scheme $Z$ and morphism $Z \to Y$, the algebraic space $X \times_Y Z$ is representable by an affine scheme integral over $Z$. Similarly, $f$ is said to be *finite* if for every affine scheme $Z$ and morphism $Z \to Y$, the algebraic space $X \times_Y Z$ is representable by an affine scheme finite over $Z$.

The following are equivalent for a morphism $f : X \to Y$ of algebraic spaces over $S$:
1. $f$ is representable and integral (resp. finite).
2. $f$ is integral (resp. finite).
3. There exists a scheme $V$ and a surjective étale morphism $V \to Y$ such that $V \times_Y X \to V$ is integral (resp. finite).
4. There exists a Zariski covering $Y = \bigcup Y_i$ such that each of the morphisms $f^{-1}(Y_i) \to Y_i$ is integral (resp. finite).

The composition of integral (resp. finite) morphisms is integral (resp. finite). The base change of an integral (resp. finite) morphism is integral (resp. finite). A finite morphism of algebraic spaces is integral, and an integral morphism of algebraic spaces which is locally of finite type is finite.

For a morphism $f : X \to Y$ of algebraic spaces over $S$, the following are equivalent:
1. $f$ is integral.
2. $f$ is affine and universally closed.

A finite morphism of algebraic spaces is quasi-finite. For a morphism $f : X \to Y$ of algebraic spaces over $S$, the following are equivalent:
1. $f$ is finite.
2. $f$ is affine and proper.

A closed immersion is finite (and a fortiori integral). If $X_i \to Y$, $i = 1, \ldots, n$ are finite morphisms of algebraic spaces over $S$, then $X_1 \amalg \ldots \amalg X_n \to Y$ is finite too.

Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of algebraic spaces over $S$. If $g \circ f$ is finite and $g$ is separated, then $f$ is finite. If $g \circ f$ is integral and $g$ is separated, then $f$ is integral.


Mrs. Solar modeled Contractor 2's fees with the formula y = 150 + 55x and compared it to Contractor 1's fee, given by y = 200 + 50x.

(a) What does the difference between 50 and 55 mean in the real-world context?

(b) What does the difference between 200 and 150 mean in the real-world context?

In the given formulas, 50 and 55 represent the variable costs incurred during the project, which are only implemented when the variable x changes. X may represent the number of hours worked or the number of people added to complete the task. On the other hand, 200 and 150 are the fixed costs of the project, which Mrs. Solar must pay regardless of the changes in x.

To illustrate the difference, consider the following scenarios:

- If 15 additional people are added to complete the task:
  - Contractor 1: y = 200 + 50(15) = 200 + 750 = 950
  - Contractor 2: y = 150 + 55(15) = 150 + 825 = 975
  In this case, Contractor 1 is the best option.

- If 10 additional people are added:
  - Contractor 1: y = 200 + 50(10) = 200 + 500 = 700
  - Contractor 2: y = 150 + 55(10) = 150 + 550 = 700
  Here, either contractor is the best option.

- If 5 additional people are added:
  - Contractor 1: y = 200 + 50(5) = 200 + 250 = 450
  - Contractor 2: y = 150 + 55(5) = 150 + 275 = 425
  In this scenario, Contractor 2 is the best option.


The NEC TL facility is a special form of network input designed to enable modelers to place transmission lines into a model. To use TL, two wires are required to place the ends of the line. When connecting a transmission line to a source or using transmission line stubs as reactances, a second wire must be created, positioned and sized to play no significant role in the far field of the antenna. This wire should be thin, short, and lossless.

The user specifies the segments of wires representing the terminal points of the transmission line, its length, and characteristic impedance. The basic length is the electrical length, assuming a velocity factor of 1.0. Some programs allow the user to specify the physical length and velocity factor, which are then used to calculate the electrical length.

TL entries appear in parallel with sources and loads on the same segment. Since TL transmission lines are mathematical, they show no losses. If losses are desired, they must be hand-calculated or calculated using other software. The TL line will be accurate only if applied under balanced conditions, where the current on each side of the line is the same.

To implement TL transmission lines, the user may need to do little calculation or a lot, depending on the level of user help in the given implementation of NEC. EZNEC, for example, allows users to specify the two ends of the line and offers preset options for shorted or open stubs. Other programs may require the user to introduce the stub-terminating wires as added wires in the model description and specify the characteristic impedance of the transmission line.

The position of the terminating wire makes no difference to the length of the transmission line in the mathematical model. The user will always specify the length of the transmission line, which will supersede the distance created by the modeled placement of the terminating wire. The line itself plays no role in the generation of the antenna's far field, which may be a modeling convenience in many cases but a limitation in others.

For example, consider a 3-element 15-meter Yagi with a quarter wavelength matching section. The feedpoint impedance is 24.8 + j 0.4 Ohms, too low for a match to the standard 50-Ohm coaxial cable. A 35-Ohm quarter wavelength matching section can be inserted to effect the match. The line would be 11.6' or 3.54 m long electrically, and 7.655' long physically, considering the velocity factor.

Another example is a 4-element 10-meter Yagi with a beta match (shorted stub). The natural feedpoint impedance of the antenna is 25.8 - j 28.8 Ohms, a good candidate for a beta match. A 5" stub of 450-Ohm line with a velocity factor of 1.0 can be used to simulate the hair pin. The feedpoint impedance with the stub is 43.4 + j25.2 Ohms, which is not a perfect match but can be optimized by experimenting with stub lengths and characteristic impedances.

A third example is a 3-element 10-meter phased array with a director and a beta match (open stub). The feedpoint impedance is 15.5 + j23.0 Ohms, requiring an open stub to compensate for the inductive reactance. The open circuit stub is about 3.2' of 50-Ohm transmission line with a 0.66 velocity factor. The final array shows a feedpoint impedance at 28.5 MHz of 46.1 + j12.9 Ohms, which can be optimized by experimenting with stub lengths and transmission line types.

These examples demonstrate the effective use of the TL facility within NEC, considering the requirements for balance and the fact that TL transmission lines are in parallel with sources and loads. By following these guidelines and experimenting with different parameters, users can achieve accurate modeling results and optimize their antenna designs.


CAREER & COLLEGE READY

In this lesson, students will learn to count teeth on gears, calculate gear ratios, and describe gear movement. The key objectives include:

- Counting teeth on gears to calculate the gear ratio between two gears
- Comparing and contrasting drive and driven gears
- Describing the direction gears move when linked together

This lesson will be completed within 1 class period, lasting 55 minutes. Upon completion, students will be able to:

- Describe the direction various gears spin when linked together
- Calculate gear ratios by counting teeth on two gears
- Determine input and output ratios
- Calculate the final drive ratio

Through individual and group work, students will achieve a comprehensive understanding of gear ratios and movement.


# Edexcel Core Mathematics 4 Parametric Equations

## Introduction to Parametric Equations

A parametric equation of a curve is one that does not give the relationship between x and y directly but rather uses a third variable, typically t, to do so. The third variable is known as the parameter. A simple example of a pair of parametric equations is:

x = 5t + 3
y = t^2 + 2t

## Converting to Cartesian

To find the Cartesian equation of the curve from parametric equations, we need to eliminate the parameter. The easiest way to do this is to rearrange one parametric equation to get the parameter as the subject and then substitute this into the other equation.

## Parametric Equations of a Circle

A circle with an origin (a, b) has the parametric equation:

x = a + rcos(t)
y = b + rsin(t)

We can use the result sin^2(t) + cos^2(t) = 1 to derive these.

## Example 1: Finding a Cartesian Equation

The curve C is described by the parametric equations:
x = 5cos(t)
y = cos(2t)

To find a Cartesian equation for the curve C, we can use the identity:
cos(2t) = 2cos^2(t) - 1

Substituting x = 5cos(t) into the equation for y, we get:
y = 2(x/5)^2 - 1
y = (2/25)x^2 - 1

## Example 2: Finding dy/dx

The parametric equations of a curve are:
x = 14sin(t)
y = 14tcos(t)

To find dy/dx, we can use the chain rule:
dy/dx = (dy/dt) / (dx/dt)

Differentiating the parametric equations, we get:
dx/dt = 14cos(t)
dy/dt = 14cos(t) - 14tsin(t)

Substituting these into the equation for dy/dx, we get:
dy/dx = (14cos(t) - 14tsin(t)) / (14cos(t))
dy/dx = 1 - tsin(t)/cos(t)
dy/dx = 1 - ttan(t)

## Example 3: Finding dy/dx

The curve is given by the parametric equations:
x = 7sin^3(t)
y = 6cos(2t)

To find dy/dx, we can use the chain rule:
dy/dx = (dy/dt) / (dx/dt)

Differentiating the parametric equations, we get:
dx/dt = 21sin^2(t)cos(t)
dy/dt = -12sin(2t)

Substituting these into the equation for dy/dx, we get:
dy/dx = (-12sin(2t)) / (21sin^2(t)cos(t))
dy/dx = (-12sin(2t)) / (21sin^2(t)cos(t))
dy/dx = (-12)(2sin(t)cos(t)) / (21sin^2(t)cos(t))
dy/dx = -24/21
dy/dx = -8/7

## Example 4: Finding the Gradient of the Curve

The curve C is described by the parametric equations:
x = tan(t)
y = sin(2t) - 2t^2

To find the gradient of the curve at the point P where t = 3, we can use the chain rule:
dy/dx = (dy/dt) / (dx/dt)

Differentiating the parametric equations, we get:
dx/dt = sec^2(t)
dy/dt = 2cos(2t) - 4t

Substituting these into the equation for dy/dx, we get:
dy/dx = (2cos(2t) - 4t) / sec^2(t)

Let t = 3:
dy/dx = (2cos(6) - 12) / sec^2(3)

## Example 5: Finding the Equation of the Normal

The curve C is described by the parametric equations:
x = tan(t)
y = sin(2t) - 2t^2

To find the equation of the normal to the curve at the point P, we need to find the gradient of the curve at P and then use the point-slope form of a line.

The gradient of the curve at P is:
dy/dx = (2cos(6) - 12) / sec^2(3)

The slope of the normal is the negative reciprocal of the gradient:
m = -1 / ((2cos(6) - 12) / sec^2(3))

The equation of the normal is:
y - y1 = m(x - x1)

where (x1, y1) is the point P.

## Differentiating ax

The function f(x) = ax describes growth and decay, and its derivative gives a measure of the rate of change of this growth/decay.

f'(x) = a(ln(a))x^(ln(a)-1)

This result needs to be learned, and is not given in the formula sheet.

## Past Paper Questions

1. A curve has parametric equations x = 2cot(t), y = 2sin^2(t), 0 < t < 2.
(a) Find an expression for dy/dx in terms of the parameter t.
(b) Find an equation of the tangent to the curve at the point where t = 4.
(c) Find a Cartesian equation of the curve in the form y = f(x). State the domain on which the curve is defined.

2. The curve shown in Figure 2 has parametric equations x = sin(t), y = sin(6t), 2 < t < 2.
(a) Find an equation of the tangent to the curve at the point where t = 6.
(b) Show that a Cartesian equation of the curve is y = 2x^3 + 2(1 - x^2), -1 < x < 1.

3. A curve has parametric equations x = 7cos(t)cos(7t), y = 7sin(t)sin(7t), 8 < t < 3.
(a) Find an expression for dy/dx in terms of t.
(b) Find an equation of the normal to the curve at the point where t = 6.

4. A curve has parametric equations x = tan^2(t), y = sin(t), 0 < t < 2.
(a) Find an expression for dy/dx in terms of t.
(b) Find an equation of the tangent to the curve at the point where t = 4.
(c) Find a Cartesian equation of the curve in the form y^2 = f(x).

5. The curve C shown in Figure 3 has parametric equations x = t^3 - 8t, y = t^2.
Given that the point A has parameter t = 1,
(a) find the coordinates of A.
The line l is the tangent to C at A.
(b) Show that an equation for l is 2x - 5y - 9 = 0.
The line l also intersects the curve at the point B.
(c) Find the coordinates of B.

6. Figure 2 shows a sketch of the curve with parametric equations x = 2cos(2t), y = 6sin(t), 0 < t < 2.
(a) Find the gradient of the curve at the point where t = 3.
(b) Find a Cartesian equation of the curve in the form y = f(x), |x| < k, stating the value of the constant k.
(c) Write down the range of f(x).

7. A curve C has parametric equations x = sin^2(t), y = 2tan(t), 0 < t < 2.
(a) Find dy/dx in terms of t.
The tangent to C at the point where t = 3 cuts the x-axis at the point P.
(b) Find the x-coordinate of P.

8. The curve C has parametric equations x = ln(t), y = t^2 - 2, t > 0.
Find
(a) an equation of the normal to C at the point where t = 3,
(b) a Cartesian equation of C.

9. Figure 3 shows part of the curve C with parametric equations x = tan(2), y = sin(2), 0 < 2.
The point P lies on C and has coordinates (1, 3).
(a) Find the value of at the point P.
The line l is a normal to C at P. The normal cuts the x-axis at the point Q.
(b) Show that Q has coordinates (k/3, 0), giving the value of the constant k.

10. Figure 2 shows a sketch of the curve C with parametric equations x = 4sin(6t), y = 3cos(2t), 0 < t < 2.
(a) Find an expression for dy/dx in terms of t.
(b) Find the coordinates of all the points on C where dy/dx = 0.

11. Figure 2 shows a sketch of the curve C with parametric equations x = 3sin(2t), y = 4cos^2(t), 0 < t < 2.
(a) Show that dy/dx = k3tan(2t), where k is a constant to be determined.
(b) Find an equation of the tangent to C at the point where t = 3.
Give your answer in the form y = ax + b, where a and b are constants.
(c) Find a Cartesian equation of C.

12. Figure 2 shows a sketch of part of the curve C with parametric equations x = 1 + 2t, y = 2t - 1.
The curve crosses the y-axis at *View more*


### Analyzing Graph Intercepts

To predict how many times the graph of an equation will touch the x-axis, it's helpful to first rewrite the equation in a more useful form. This process often involves factoring to identify x-intercepts.

#### Steps to Identify X-Intercepts:
1. **Factor the equation**: Factoring is crucial for identifying the x-intercepts of a graph. By breaking down the equation into its factors, you can set each factor equal to zero to find the x-intercepts.
2. **Set each factor equal to zero**: Once the equation is factored, set each factor equal to zero and solve for x. The solutions will give you the x-intercepts.
3. **Use a calculator for verification**: After predicting the number of x-intercepts, use a calculator to graph the equation and verify your predictions.

#### Patterns for X-Intercepts:
- **No x-intercepts**: If an equation has no real solutions when set equal to zero, its graph will not touch the x-axis.
- **One x-intercept**: A linear equation or a quadratic equation with one real solution will have one x-intercept.
- **Two x-intercepts**: A quadratic equation with two distinct real solutions will have two x-intercepts.

#### Identifying Patterns in Parabolas:
For a parabola (a quadratic equation), the number of x-intercepts can be determined by its discriminant (b^2 - 4ac) in the quadratic formula.
- If the discriminant is greater than zero, the parabola has two distinct x-intercepts.
- If the discriminant is equal to zero, the parabola has one x-intercept (touches the x-axis at one point).
- If the discriminant is less than zero, the parabola has no real x-intercepts.

By analyzing the equation's form and applying these principles, you can predict how many times the graph will touch the x-axis.