# Ellipsoid Model

## Definition

The 2D scattering intensity function for oriented ellipsoids is given by:

$$ P(q,\alpha) = \frac{\text{scale}}{V} F^2(q,\alpha) + \text{background} $$

where

$$ F(q,\alpha) = \Delta \rho V \frac{3(\sin qr - qr \cos qr)}{(qr)^3} $$

for

$$ r = \left[ R_e^2 \sin^2 \alpha + R_p^2 \cos^2 \alpha \right]^{1/2} $$

Here, $\alpha$ is the angle between the axis of the ellipsoid and $\vec q$, $V = (4/3)\pi R_pR_e^2$ is the volume of the ellipsoid, $R_p$ is the polar radius, $R_e$ is the equatorial radius, and $\Delta \rho$ is the scattering length density difference between the scatterer and the solvent.

## Orientational Average

For randomly oriented particles, the orientational average is:

$$ \langle F^2(q) \rangle = \int_{0}^{\pi/2}{F^2(q,\alpha)\sin(\alpha)\,d\alpha} $$

computed via substitution as:

$$ \langle F^2(q) \rangle = \int_0^1{F^2(q, u)\,du} $$

with

$$ r = R_e \left[ 1 + u^2\left(R_p^2/R_e^2 - 1\right)\right]^{1/2} $$

## Orientation

For 2D data from oriented ellipsoids, the direction of the rotation axis is defined using two angles $\theta$ and $\phi$. The $\theta$ and $\phi$ parameters are not used for the 1D output.

## Validation

The model was validated by comparing the output of the 1D model to the output of the NIST software. The implementation of the intensity for fully oriented ellipsoids was validated by averaging the 2D output using a uniform distribution.

## Comparison

A comparison of the intensity for uniformly distributed ellipsoids calculated from the 2D model and the intensity from the NIST SANS analysis software showed a discrepancy at high $q$ values due to the numerical integration method used by the NIST software.

## References

1. L. A. Feigin and D. I. Svergun. *Structure Analysis by Small-Angle X-Ray and Neutron Scattering*, Plenum Press, New York, 1987
2. A. Isihara. *J. Chem. Phys.*, 18 (1950) 1446-1449

## Authorship and Verification

**Author:** NIST IGOR/DANSE
**Date:** pre 2010
**Converted to sasmodels by:** Helen Park
**Date:** July 9, 2014
**Last Modified by:** Paul Kienzle
**Date:** March 22, 2017
**Verified by:** SasView Team on 07 Sep 2017

## Model Details

The 2nd virial coefficient of the solid ellipsoid is calculated based on the $R_p$ and $R_e$ values and used as the effective radius for $S(q)$ when $P(q) \cdot S(q)$ is applied. The model is included in the SasView library by default.


**Subject: C. Maths, Time: 3 hours, Full Marks: 75**

**Introduction to Blood Types**

Eight blood types are represented by the eight regions of a Venn diagram, with each circle symbolizing one of three antigens: A, B, or Rh. If A and B are absent, the blood type is O. If Rh is present, the blood type is positive (+); otherwise, it is negative (-).

**Table of Blood Types**

| Antigens | Number of People |
| --- | --- |
| A | 60 |
| B | 27 |
| Rh | 123 |
| A and B | 12 |
| A and Rh | 46 |
| B and Rh | 17 |
| A and B and Rh | 9 |

**Cardinality Notation of People for All Groups**

Let U be the universal set of all antigens.
A, B, and Rh denote the sets of people having A, B, and Rh antigens, respectively.
Thus, the cardinality of all surveyed people as the Universal Set U is n(U) = 150.

**Item-wise Solution**

1. **Cardinality Notation**
   - n(A ∩ B ∩ Rh) = n(AB+) = 9
   - n(A ∩ Rh) = n(A+) = 46 - 9 = 37
   - n(B ∩ Rh) = n(B+) = 17 - 9 = 8
   - n(A ∩ B) = n(AB) = 12 - 9 = 3
   - n(A) = n(A) = 60 - 37 - 3 - 9 = 11
   - n(B) = n(B) = 27 - 8 - 3 - 9 = 7
   - n(Rh) = n(O+) = 123 - 37 - 8 - 9 = 69

2. **Venn Diagram Presentation**
   The Venn diagram presents the above information as follows:
   - n(U) = 150
   - The diagram includes the following regions: O+, O-, A+, A-, B+, B-, AB+, AB-

3. **Number of People with O Negative Blood**
   According to the Venn diagram:
   - n(U) = 69 + 11 + 7 + 37 + 3 + 8 + 9 + x = 150
   - x = 6
   Therefore, the number of people having O negative blood is n(A ∪ B ∪ Rh) = 6.

4. **Blood Donation**
   A person having O+ blood can donate blood to people having Rh antigens, which includes people with O+, A+, B+, AB+ blood groups.
   - n(Rh) = 123
   Thus, a person having O+ blood can donate blood to 123 people.


To analyze the given quadratic functions, we need to find the vertex, axis of symmetry, y-intercept, x-intercept, domain, and range.

**Function 1: f(x) = x^2 - 4x**

* Vertex: The vertex form of a quadratic function is f(x) = a(x - h)^2 + k, where (h, k) is the vertex. To find the vertex, we complete the square: f(x) = (x - 2)^2 - 4. Therefore, the vertex is (2, -4).
* Axis of symmetry: The axis of symmetry is the vertical line x = h, where h is the x-coordinate of the vertex. So, the axis of symmetry is x = 2.
* Y-intercept: To find the y-intercept, we substitute x = 0 into the function: f(0) = 0^2 - 4(0) = 0.
* X-intercept: To find the x-intercept, we substitute y = 0 into the function: 0 = x^2 - 4x. Factoring out x, we get 0 = x(x - 4). Therefore, the x-intercepts are x = 0 and x = 4.
* Domain: The domain of a quadratic function is all real numbers, unless there are restrictions. In this case, the domain is (-∞, ∞).
* Range: The range of a quadratic function is all real numbers greater than or equal to the minimum value, or less than or equal to the maximum value. Since the vertex is (2, -4), the minimum value is -4. Therefore, the range is [-4, ∞).

**Function 2: f(x) = x^2 + 6x + 9**

* Vertex: The given function is already a perfect square: f(x) = (x + 3)^2. Therefore, the vertex is (-3, 0).
* Axis of symmetry: The axis of symmetry is the vertical line x = h, where h is the x-coordinate of the vertex. So, the axis of symmetry is x = -3.
* Y-intercept: To find the y-intercept, we substitute x = 0 into the function: f(0) = 0^2 + 6(0) + 9 = 9.
* X-intercept: To find the x-intercept, we substitute y = 0 into the function: 0 = x^2 + 6x + 9. Factoring out (x + 3), we get 0 = (x + 3)^2. Therefore, the x-intercept is x = -3.
* Domain: The domain of a quadratic function is all real numbers, unless there are restrictions. In this case, the domain is (-∞, ∞).
* Range: The range of a quadratic function is all real numbers greater than or equal to the minimum value, or less than or equal to the maximum value. Since the vertex is (-3, 0), the minimum value is 0. Therefore, the range is [0, ∞).

To find intercepts without drawing a graph, we can substitute zero for x or y and solve for the other value. For quadratics in x, there will always be a y-intercept. The number of x-intercepts depends on the values of a, b, and c in the quadratic equation. The discriminant (b^2 - 4ac) shows the nature of the roots (x-intercepts).


# Interior Point of a Set

A point is an interior point of a set if there exists an open set such that . In a topological space , a point is an interior point of a subset if it is in some open set contained in .

**Interior of a Set:**

The interior of a subset , denoted by or , is the union of all open sets contained in . It is the union of all open subsets of , and a point in the interior of is called an interior point of .

**Key Properties:**

* The interior of a set is the largest open set contained in it.
* The interior of a set is always open.
* The union of open sets is an open set.

**Example:**

Let with topology . If , then . Since there is no open set containing and a subset of , is not an interior point of . Similarly, is not an interior point of . However, is an interior point of since is an open set containing and is a subset of .

**Theorems:**

* Each point of a non-empty subset of a discrete topological space is its interior point.
* The interior of a subset of a discrete topological space is the set itself.
* The interior of a subset of a topological space is the union of all open subsets of .
* A subset of a topological space is open if and only if .
* If is a subset of a topological space , then .
* Let be a topological space and and be subsets of , then:
  1.
  2.
  3.


Binary Numbers
Binary numbers are based on a counting system that uses only two symbols: 0 and 1. These numbers dominate electronics because they correspond to simple concepts such as off and on, false and true, and switch open and closed. More complex ideas can be created by aggregating multiple binary bits into assemblies like nybbles (4 bits), bytes (8 bits), words (16 bits), and long words (32 bits).

The place value system works for binary numbers just like it does for decimal numbers. In decimal, the number 234 means 2×100 + 3×10 + 4×1. Similarly, in binary, each place has a digit (0 or 1) times the base (2) raised to an integer power. The position immediately to the left of the base^0 is valued at base^1, which for binary means 2^1 = 2.

The first few binary numbers are: 00, 01, 10, 11, which correspond to the decimal numbers 0, 1, 2, and 3. To understand the place values for binary numbers, consider the following table for 16-bit numbers (words):

| Power of 2 | Base 10 (decimal) Value |
| --- | --- |
| 2^0 | 1 |
| 2^1 | 2 |
| 2^2 | 4 |
| 2^3 | 8 |
| 2^4 | 16 |
| 2^5 | 32 |
| 2^6 | 64 |
| 2^7 | 128 |
| 2^8 | 256 |
| 2^9 | 512 |
| 2^10 | 1024 |
| 2^11 | 2048 |
| 2^12 | 4096 |
| 2^13 | 8192 |
| 2^14 | 16384 |
| 2^15 | 32768 |

Converting between decimal and binary is straightforward. To convert a decimal number to binary, find the largest power of 2 that fits into the number, place a 1 in that position, and subtract the corresponding value. Repeat this process with the remaining value until you reach 0. For example, to convert 84 to binary:
- 84 is between 2^6 = 64 and 2^7 = 128, so it will need 7 bits.
- The most significant position (2^6) gets a 1, representing 64 out of 84, leaving 20 to be encoded.
- The next position (2^5) has a decimal value of 32, which is bigger than 20, so it gets a 0.
- The next position (2^4) is 16, which is less than 20, so it gets a 1, leaving 4 to be encoded.
- Since 2^2 = 4, the 2's place gets a 1, and all other places get 0's, resulting in 1010100.

To convert binary to decimal, add up the place values where there's a 1 and ignore the places with 0's. For example, 1001001 = 1×2^6 + 1×2^3 + 1×2^0 = 64 + 8 + 1 = 73.

Characters, such as those being read now, are also encoded in binary. ASCII encoding is common for simple alphabetic characters, while Unicode is used to enable all character sets of all languages.

Exercises:
1. Express 17487 as a binary number.
2. Express 1001111000 as a decimal number.


Vedic Math - Divisibility of a Number by 4

The divisibility rule for 4 states that a number is divisible by 4 if the last two digits of the number are divisible by 4. For example:
- 85736 is divisible by 4 because 36 is divisible by 4.
- 58648 is divisible by 4 because 48 is divisible by 4.
- 85728 is divisible by 4 because 28 is divisible by 4.
- 3571916 is divisible by 4 because 16 is divisible by 4.
- 4781204 is divisible by 4 because 04 is divisible by 4.

Additionally, if a number ends with two or more zeros, it is also divisible by 4. For instance:
- 98300 is divisible by 4 because it ends in 00.
- 8573600 is divisible by 4 because it ends in 00.
- 5864800 is divisible by 4 because it ends in 00.
- 8572800 is divisible by 4 because it ends in 00.
- 357191600 is divisible by 4 because it ends in 00.
- 478120400 is divisible by 4 because it ends in 00.

To check if a number is divisible by 4, first verify if the last two digits are divisible by 4 or if the number ends in two or more zeros. If either condition is met, the number is divisible by 4.


### Course: 6th grade (WNCP) > Unit 2

Lesson 1: Number patterns - Math patterns: toothpicks

The lesson explores a pattern of creating figures with toothpicks. Sal creates a pattern to determine the total number of toothpicks required for a given number of houses. The formula used is 6 + 5(x-1), where x represents the number of houses.

The conversation surrounding the lesson reveals that an alternative formula, x5+1, can also be used to represent the same pattern. This is because the first house requires 6 toothpicks, and each additional house requires 5 toothpicks.

For example, to calculate the total number of toothpicks for 500 houses, the formula can be applied as follows: 6 + 5(500-1) = 6 + 5(499) = 6 + 2495 = 2501. Alternatively, using the x5+1 formula, the calculation is 500*5 + 1 = 2501.

The lesson also highlights the importance of understanding the pattern and how it can be applied to different scenarios. By recognizing that each house shares a common wall with the previous one, except for the first house, the formula can be derived.

The video transcript explains the process of creating the pattern and applying it to different numbers of houses. Sal starts with one house, which requires 6 toothpicks, and then adds 5 toothpicks for each additional house. The pattern is then used to calculate the total number of toothpicks for 50 houses, resulting in 251 toothpicks.

Key concepts and formulas:

- 6 + 5(x-1) = total number of toothpicks for x houses
- x5+1 = alternative formula for the same pattern
- The first house requires 6 toothpicks, and each additional house requires 5 toothpicks.

Example calculations:

- 500 houses: 6 + 5(500-1) = 2501 or 500*5 + 1 = 2501
- 50 houses: 6 + 5(50-1) = 251

The lesson aims to help students understand and apply mathematical patterns to real-world scenarios, using toothpicks as a visual aid to illustrate the concept.


The set of **costs** consists of nonnegative numbers together with infinity:
$$ [0,\infty] . $$
There's an obvious way to define \(\le\) for costs, where we decree that everything is less than or equal to \(\infty\).
There's an obvious way to add costs, where we decree that \(\infty\) plus anything equals \(\infty\).
And there's an obvious unit for this addition operation, namely \(0\).

Now comes the tricky part, which tripped me up at first.
We've got our hands on a obvious way to create a symmetric monoidal poset, which we could call \( ( [0,\infty], \le, +, 0 )\).
However, this is not the one we want!
Instead, we will to play a fiendish trick!
We'll use the "opposite" symmetric monoidal poset:
$$ \mathbf{Cost} = ( [0,\infty], \ge, + , 0) .$$

A \(\mathbf{Cost}\)-category, say \(\mathcal{X}\), consists of:
a set of objects, \(\mathrm{Ob}(\mathcal{X})\), and
an element \(\mathcal{X}(x,y) \in \mathbf{Cost} \) for any two objects \(x,y\).
So, a \(\mathbf{Cost}\)-category gives a number in \([0,\infty]\) for each pair of objects \(x,y\).

Next:
a) In any enriched category we have
$$ I\leq\mathcal{X}(x,x) \textrm{ for any object } x .$$
However, in \(\mathbf{Cost}\) the unit \(I\) is the number \(0\), and we've *fiendishly reversed the ordering*.
So, in a \(\mathbf{Cost}\)-category we have
$$ 0 \geq \mathcal{X}(x,x) \textrm{ for all objects } x .$$
But \( \mathcal{X}(x,x) \) is a cost, so we also know \( \mathcal{X}(x,x) \geq 0 \).
So, we can just say
$$ \mathcal{X}(x,x) = 0 \textrm{ for all objects } x .$$
Finally:
b) In any enriched category we have
$$ \mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z) \textrm{ for all objects } x,y,z. $$
However, in \(\mathbf{Cost}\) the monoid operation \(\otimes\) is addition, and we've *fiendishly reversed the ordering*.
So, in a \(\mathbf{Cost}\)-category we have
$$ \mathcal{X}(x,y) + \mathcal{X}(y,z) \geq\mathcal{X}(x,z) \textrm{ for all objects } x,y,z. $$
Now we're done.

But hey: that last inequality is the triangle inequality!
So maybe we should think of \(\mathcal{X}(x,y)\) as a *distance* and write it as \(d(x,y)\).
That's what Bill Lawvere did in 1973... and he invented the concept that's now called a Lawvere metric space.
In other words:
**Definition.** A **Lawvere metric space** is a \(\mathbf{Cost}\)-category.

**Theorem.** A Lawvere metric space consists of a set \(X \) together with a function
$$ d : X \times X \to [0,\infty] $$
such that
a) \( d(x,x) = 0 \) for all \( x \in X \)
and
b) \( d(x,y) + d(y,z) \geq d(x,y) \) for all \( x,y,z \in X\).

**Proof.** We've shown that any \(\mathbf{Cost}\)-category consists precisely of a set \(X = \mathrm{Ob}(\mathcal{X})\) together with a cost \(d(x,y) = \mathcal{X}(x,y) \) for each \(x,y \in X\), obeying these two inequalities.

Now, you may already have heard about metric spaces.
A metric space is a very famous concept: it's a set with a way of measuring distances that obeys a few axioms.
These include the two axioms of a Lawvere metric space, but also three more:
d) if \(d(x,y) = 0 \) then \(x = y \)
e) \(d(x,y) = d(y,x) \) for all \(x,y \in X\), and
f) \(d(x,y) < \infty \) for all \(x,y \in X\).

These extra axioms hold in a lot of examples, but Lawvere noticed that they aren't really necessary for many of the main theorems involving metric spaces!

Dropping these extra axioms also allows for new applications.
For example, we could let \(d(x,y)\) be the *cost* of going from one place \(x\) to another place \(y\), using your favorite modes of transportation.
There are places you can go for free, so axiom d) does not hold.
Sometimes it costs a different price to go from \(x\) to \(y\) than to go back, so axiom e) does not hold.
And there are places you cannot go for any prices, like the Andromeda Nebula, so axiom f) does not hold.

In fact, axioms d)-f) are also bad in some purely mathematical ways: there are some nice ways to build new Lawvere metric spaces from old ones (called "coequalizers" and "coproducts") that don't work when you include these extra axioms.

You might think that extra axioms are always good, because they let you prove more theorems.
But you have to remember they impose a price: it means the axioms apply to fewer things, so constructing structures that obey these axioms becomes harder.

So, Lawvere metric spaces are a great example of how category theory can lead us to refine and perfect existing ideas.
And when you get good at enriched category theory, it means every general result you prove will apply to Lawvere metric spaces as well as preorders!

To see what Lawvere actually did, read this paper:
- Bill Lawvere, Metric spaces, generalized logic and closed categories,
*Rendiconti del Seminario Matematico e Fisico di Milano* *XLIII* (1973), 135-166.
Reprinted in *Reprints in Theory and Applications of Categories* *1* (2002), 1-37.

It's good to look at papers by Lawvere now and then, to see how close you are to being a true category theorist.
It's inevitably humbling, at least for me... but I've decided that's a good thing.
It drives me to do better, while preventing me from getting a big head.

You may wonder what would happen if we hadn't flipped the order when defining \(\mathbf{Cost}\).

**Puzzle 90.** What's a \(\mathbf{Cost}^{\text{op}}\)-category, and what if anything are they good for?

Some of you have already answered the first part.
Section 2.3.4 of the book leads you through other examples of enriched categories.
Here's another.
Let
$$ D = \{ \textrm{Monday}, \textrm{Tuesday}, \textrm{Wednesday}, \textrm{Thursday}, \textrm{Friday}, \textrm{Saturday}, \textrm{Sunday} \} $$
and let
$$ \mathcal{V} = \prod_{d \in D} \mathbf{Cost} $$
that is, the product of 7 copies of \(\mathbf{Cost}\), one for each day of the week.

**Puzzle 91.** Show how to make the product of symmetric monoidal posets into a symmetric monoidal poset.
Thus, \(\mathcal{V} \) becomes a symmetric monoidal poset.
What is a \(\mathcal{V} \)-category like, and what are they good for in everyday life?

The economic models are "fun".
Consider the couch surfer, staying in one place too long will incur a cost but travelling is less costly.
The costs increase the longer you stay.
They expect you to pay for food and stuff.
Not quite sure how to deal with the temporal issues.

I made a diagram for **Cost** that I think helps the discussion.

Also, see Exercise 22 - Chapter 2

Where the \( X \) and \( X_{neg} \) have the conventional meaning for \( \le \) and \( \ge \) while the \( X^{op} , X_{neg}^{op} \) have the opposite meaning.

The story doesn't look very good for them :(

**Theorem.** If \(\mathcal{X}\) is a \(\mathbf{Cost}^{\text{op}}\)-enriched category, then:
$$\forall a,b. \mathcal{X}(a,b) = 0 \text{ or } \mathcal{X}(a,b) = \infty$$

**Proof**.
If every element \(\mathcal{X}(a,b) = 0\), we are done.

Next assume to the contrary.
We must show that for an arbitrary \(a\) and \(b\) that \(\mathcal{X}(a,b) = \infty\).

So observe there must be some \(\hat{a}\) and \(\hat{b}\) such that \(\mathcal{X}(\hat{a},\hat{b}) > 0\).

It must be \(\mathcal{X}(\hat{a},\hat{b}) = \infty\).
To see this, we know from the laws of enriched categories (part (b)) that:
$$ \begin{align} \mathcal{X}(\hat{a},\hat{b}) + \mathcal{X}(\hat{b},\hat{a}) & \leq \mathcal{X}(\hat{a},\hat{a}) \\
\implies \mathcal{X}(\hat{a},\hat{b}) & \leq \mathcal{X}(\hat{a},\hat{a}) \end{align} $$
However, then we have
$$ \begin{align} \mathcal{X}(\hat{a},\hat{b}) + \mathcal{X}(\hat{a},\hat{a}) & \leq \mathcal{X}(\hat{a},\hat{b}) \\
\implies 2 \mathcal{X}(\hat{a},\hat{b}) & \leq \mathcal{X}(\hat{a},\hat{b}) \end{align} $$
This can only happen if \(\mathcal{X}(\hat{a},\hat{b}) = \infty\) or \(\mathcal{X}(\hat{a},\hat{b}) = 0\).
But we know \(\mathcal{X}(\hat{a},\hat{b}) > 0\) so it must be \(\mathcal{X}(\hat{a},\hat{b}) = \infty\) .

Next observe from the enriched category theory law (b) that:
$$ \mathcal{X}(a,\hat{a}) + \mathcal{X}(\hat{a},\hat{b}) \leq \mathcal{X}(a,\hat{b}) $$
So it must be that \(\mathcal{X}(a,\hat{b}) = \infty\).
But then
$$ \mathcal{X}(a,\hat{b}) + \mathcal{X}(\hat{b},b) \leq \mathcal{X}(a,b) $$
Hence \(\mathcal{X}(a,b) = \infty\).

After reading Matthew's proof in #8, I think that a \(\mathbf{Cost}^{\text{op}}\)-category is equivalent to a \( (\{ \tt{true}, \tt{false}\}, \implies, \tt{false}, \tt{or}) \)-category, that Daniel described in #10 of Lecture 29.

Both Daniel and Matthew point out that either, \(\mathcal X(x,y) = \tt{true}\) for all pairs \(x,y)\) or \(\mathcal X(x,y) = \tt{false}\) for all pairs \(x,y)\).
Thinking about the set of objects as a discrete points and drawing an arrow between objects where \(\mathcal X(x,y) = \tt{true}\), we could visualize every \(\mathbf{Cost}^{\text{op}}\)-category as either a discrete set of points or as a fully connected graph.

Somehow this has the feeling of adjoints, since the discrete set of points under-estimates the connections between objects and the fully connected graph over-estimates the connections between objects.

**Puzzle 91**
Let's start with the first part
> Show how to make the product of symmetric monoidal posets into a symmetric monoidal poset.
Let \((X, \leq_X, I_X, \otimes_X)\) and \((Y, \leq_Y, I_Y, \otimes_Y)\).
We can define a new symmetric monoidal poset \( (X\times Y, \leq_{X \times Y}, I_{X \times Y}, \otimes_{X\times Y} ) \) where
$$(x, y) \leq_{X \times Y} (x', y') \textrm{ iff } x \leq_X x' \textrm{ and } y \leq_Y y' $$
$$I_{X \times Y} = (I_X, I_Y)$$
$$(x_1, y_1) \otimes_{X \times Y} (x_2, y_2) = (x_1 \otimes_X x_2, y_1 \otimes_Y y_2)$$
Proving that this is a symmetric monoidal poset follows pretty immediately from breaking down each property into a property about \((X, \leq_X, I_X, \otimes_X)\) and a property about \((Y, \leq_Y, I_Y, \otimes_Y)\) and then applying the fact that they are both symmetric monoidal posets.

Here is an example:
We want to show that if \( (x_1, y_1) \leq_{X \times Y} (x_1', y_1') \) and \( (x_2, y_2) \leq_{X \times Y} (x_2', y_2') \), then
\[(x_1, y_1)\otimes_{X \times Y} (x_2, y_2) \leq_{X \times Y} (x_1', y_1')\otimes_{X \times Y} (x_2', y_2') .\]

By the definition of \(\leq_{X \times Y}\), the hypotheses imply that
$$ x_1 \leq_X x_1' ,\quad x_2 \leq_X x_2' ,\quad y_1 \leq_Y y_1' , \quad \textrm{ and } y_2 \leq_Y y_2' .$$
Since \((X, \leq_X, I_X, \otimes_X)\) and \((Y, \leq_Y, I_Y, \otimes_Y)\) are monoidal posets we have that
$$x_1 \otimes_X x_2 \leq_X x_1' \otimes_X x_2' \textrm{ and } y_1 \otimes_Y y_2 \leq_Y y_1' \otimes_Y y_2' $$
And so by the definition of \( \leq_{X \times Y}\),
\[ ( x_1 \otimes_X x_2, y_1 \otimes_Y y_2 ) \leq_{X \times Y} (x_1' \otimes_X x_2', y_1' \otimes_Y y_2' ).\]
Of course applying the definition of \(\otimes_{X \times Y}\) gives us our desired result:
\[(x_1, y_1)\otimes_{X \times Y} (x_2, y_2) \leq_{X \times Y} (x_1', y_1')\otimes_{X \times Y} (x_2', y_2') .\]

Now onto the next part
> Thus, \(\mathcal{V} \) becomes a symmetric monoidal poset.
What is a \(\mathcal{V} \)-category like, and what are they good for in everyday life?
A \(\mathcal{V} \)-category assigns to every pair of objects a cost for each day of the week.
Projecting to just the costs for one of the days of the week will give you a Lawvere metric space.
This might be useful when thinking about commutes.
For example, my commute to and from work is very bad on Monday - Thursday, it's a little better on Friday because some people work from home, and on Saturday and Sunday it's very good.
So as a objects of a \(\mathcal{V} \)-category, I might represent this as
$$ \mathcal{X}( \textrm{home}, \textrm{work}) = (30, 30, 30, 30, 25, 15, 15) $$

Yes, that's the best way to make \(X \times Y\) into a symmetric monoidal poset!
If we ever get around to talking about the category of symmetric monoidal posets, this will be the product in that category!

A \(\mathcal{V} \)-category assigns to every pair of objects a cost for each day of the week.
Projecting to just the costs for one of the days of the week will give you a Lawvere metric space.
This might be useful when thinking about commutes.
For example, my commute to and from work is very bad on Monday - Thursday, it's a little better on Friday because some people work from home, and on Saturday and Sunday it's very good.
So as a objects of a \(\mathcal{V} \)-category, I might represent this as
$$ \mathcal{X}( \textrm{home}, \textrm{work}) = (30, 30, 30, 30, 25, 15, 15) $$

You can use Haskell and MySQL for CRUD web apps.
I like Servant for routing.
There's also scotty for routing which is inspired by Ruby's Sinatra.
For MySQL I would use MySQL-simple.

If you want to learn Haskell, you might try what Prof. Robby Findler at Northwestern University does.
He organizes mostly student-run seminars every other semester.
He explores stuff like the K Framework, Coq and other cutting edge topics.

Python is great for algorithms, and I agree along with JavaScript it is the lingua franca of the web.

I will try my hardest to use Python to communicate ideas.

You might consider formatting your coding blocks with
`<pre>...</pre>`
rather than backticks.

Moreover, should you ever become a professional software engineer, you should know using unicode a lot can be hard on coworkers who aren't into that sort of thing.

Luckily I'm only interested in doing so as a means to make programming easier to teach to mathematicians, more specifically, myself.

It's nice because,
<pre>
{min(x y) ≝ (if {x ≤ y} x y) }
</pre>
is basically how a mathematician defines a function,
\[ \begin{align} \min(x,y) \overset{\mathrm{def}}{=} \texttt{if } [x \leq y] \\ \texttt{then } x ; \\ \texttt{else } y. \\ \end{align} \]

I must admit that I prefer Haskell written without Unicode, to Haskell with it.

That isn't Haskell.
It's a Scheme variant, Racket.

Yes Haskell would use a lot fewer parentheses. :)
But Haskell also allows Unicode operators, and i mostly prefer just using 8 bit characters for source code.
Though the space of 1 or 2 character operators is a bit overfull.

The use of operators instead of words is one of the things that confuses and scares people off from both Haskell in General and off of a number of libraries.
Like lenses, which has a well organized but very large collection of them.


The figure shows a solid formed by joining the bases of two square pyramids to opposite faces of a cube. Each edge of the solid has a length of 5 inches. The total surface area of this solid is calculated as follows: the surface of the solid is made up of 4 square faces of the cube plus the triangular (lateral) faces of the pyramids. Each face of the cube has an area of 25 square inches, so the 4 faces of the cube have a total surface area of 100 square inches.

Each lateral face of the pyramids is an equilateral triangle, which can be divided into two 30°,60°,90° triangles. Each 30°,60°,90° triangle has a base of 2.5 inches and a height of 2.5√3 inches. The area of each triangle is 1/2 * base * height = 1/2 * 2.5 * 2.5√3 = 3.125√3 square inches. Each lateral face of each of the pyramids is made up of two congruent triangles, so the area of each lateral face of the pyramids is 2 * 3.125√3 = 6.25√3 square inches.

There are a total of 8 equilateral triangular faces on the solid, so the total area of triangular faces is 8 * 6.25√3 = 50√3 square inches. Therefore, the total surface area of the solid is 100 + 50√3 square inches.

In a parallelogram, opposite sides are parallel and equal in length. If the points (x1, y1), (x2, y2), (x3, y3), and (x4, y4) lie in the xy coordinate plane and form a parallelogram, then x2 - x1 = x4 - x3 and y2 - y1 = y4 - y3. This simplifies to (x2 - x1) / (y2 - y1) = (x4 - x3) / (y4 - y3). Since opposite sides are equal in length, (x2 - x1)^2 + (y2 - y1)^2 = (x4 - x3)^2 + (y4 - y3)^2.

Given the equation x^2 + y^2 = 25, we can find examples of x and y that satisfy the equation while also satisfying other statements. For example, x = 3 and y = 4 satisfy the equation, as do x = 4 and y = 3, and x = 0 and y = 5.

If x and y are two different numbers selected from the integers from 500 to 1000, inclusive, the greatest possible value for (x + y) / (x - y) is achieved when the numerator is maximized and the denominator is minimized. This occurs when x = 1000 and y = 999, giving a value of 1999.

Two functions f and g are defined by f(x) = ax^2 + bx + c and g(x) = ax^2 + bx + c + 1. If f(0) = 2 and g(0) = 3, then c = 2 and c + 1 = 3. Subtracting the corresponding sides of these equations gives 1 = 1, which is true. Since f(1) = 3 and g(1) = 4, we have a + b + c = 3 and a + b + c + 1 = 4. Subtracting the corresponding sides of these equations gives 1 = 1, which is true. Hence, a = 3 - b - c. Since f(2) = 10 and g(2) = 11, we have 4a + 2b + c = 10 and 4a + 2b + c + 1 = 11. Subtracting the corresponding sides of these equations gives 1 = 1, which is true.

The line graph shows the number of students in the computer lab during seven periods on one school day. Each of the students in the computer lab during lunch was also in the lab during two of the other periods. The sum of the number of students who were in the computer lab periods throughout the day is 140. These students were counted 3 times each, for a total of 24 counts. In determining the maximum number of different students who could have been in the classroom that day, each student can only be counted once. Therefore, students who were in the computer lab during lunch should be counted as 8 students instead of 24, and the sum 140 is at least more than the number of different students. Since 140 - 16 = 124, it follows that 124 is the maximum number of different students who could have been in the computer lab that day.

The figure represents four offices that will be assigned randomly to four employees, one employee per office. If Karen and Tina are two of the four employees, the probability that each will be assigned an office indicated with an X is 1/6. This is because two of the four employees will have a chance to get the X office, so the probability is 2/4 or 1/2, and then this must be multiplied by the second probability, which is now 1/3. The answer is 1/6.

Multiple Choice Questions:

1. What is the total surface area of the solid formed by joining the bases of two square pyramids to opposite faces of a cube, where each edge of the solid has a length of 5 inches?
(A) 100 + 25√3
(B) 100 + 50√3
(C) 150 + 25√3
(D) 200
(E) 150 + 50√3

2. If the points (x1, y1), (x2, y2), (x3, y3), and (x4, y4) lie in the xy coordinate plane and form a parallelogram, what is one possible value of (x4 - x3) / (y4 - y3)?
Any number greater than 3 and less than 5.

3. If x^2 + y^2 = 25, which of the following could be true?
I. x = 3 and y = 4
II. x = 4 and y = 3
III. x = 0 and y = 5
(A) I only
(B) II only
(C) I and III only
(D) II and III only
(E) I, II, and III

4. If x and y are two different numbers selected from the integers from 500 to 1000, inclusive, what is the greatest possible value for (x + y) / (x - y)?
The correct answer is 1999.

5. Two functions f and g are defined by f(x) = ax^2 + bx + c and g(x) = ax^2 + bx + c + 1. If f(0) = 2 and g(0) = 3, what is the value of a?
The correct answer is 3.

6. The line graph shows the number of students in the computer lab during seven periods on one school day. What is the maximum number of different students who could have been in the computer lab that day?
(A) 156
(B) 148
(C) 132
(D) 124
(E) 116

7. The figure represents four offices that will be assigned randomly to four employees, one employee per office. If Karen and Tina are two of the four employees, what is the probability that each will be assigned an office indicated with an X?
(A) 1/16
(B) 1/12
(C) 1/6
(D) 1/4
(E) 1/2


Binary is a fundamental format in computers and electronics, represented by 0s and 1s. It acts as the foundation of computers and computational operations. The use of binary in computing and electronics is common, but there are still misconceptions and questions about the number format.

### 1. The Electrical Current

The history of computers reveals that early computers used mechanical switches, which had two positions: on and off. This led to the use of 0s and 1s, where 1 represents on and 0 represents off. Base 2 binary is used in computers today due to tradition and its ability to simplify complex computing tasks. Binary has been used as the low-level language of computers for a long time, making it difficult to switch to a new approach.

### 2. The Ternary Logic

Attempts to replace binary have been made, with Ternary logic being a notable example. Introduced in the 1950s, Ternary computers use multiple voltage levels to perform calculations. However, the added complexity made binary a simpler and more efficient choice. Boolean logic can be mapped to a binary system, making it a more practical option.

### 3. Four Rules

Binary is a dependable low-level language because it requires only four rules, resulting in four possible outcomes for any binary calculation. This simplicity makes it efficient for computing. In contrast, a ternary system requires 19,683 operators, compared to binary's 16. The scalability of binary is also due to its ability to work with minimal variables, making it a fundamental component of modern computer chips and electronic hardware.

### 4. Conversions

While computers understand binary natively, it can be converted into various other number and string formats. Tools like binary to decimal converters make this process easier. Conversions to Octet, a group of 8 bits, are also possible. Each bit is a binary digit, and a byte represents a character, such as the letter A, which is represented by 01100001 in binary.

### 5. A Coded Message

Binary can be used for creative purposes, such as sending coded messages. By converting text to binary and back to strings, messages can be hidden and revealed. Online tools, like binary to text translators, can be used to convert binary messages. This method is also used in internet Easter eggs, online threads, and hidden messages in games and movies.

### 6. It’s Part of Our Culture

Binary has become a part of our culture, appearing in video games and movies as a way to hide Easter eggs or as a plot device. For example, the movie Wanted uses binary embedded in fabric to identify targets, while Interstellar uses binary as a communication method. Binary's presence in modern civilization is often unnoticed, but it is the foundation of many technological advancements.

### 7. Braille and Other Languages

Binary is adopted into various languages, including braille, which uses a combination of raised and not-raised dots to represent characters. Other languages, like Bagua, use broken and unbroken lines to represent characters. Coding systems like ASCII and binary-coded decimals are also based on binary, with the defining factor being the grouping and encoding of values.

### 8. Older Than You Think

The binary system is older than expected, dating back to the 1560s. Thomas Harriot, an English mathematician, discovered the binary system, and Francis Bacon used it in the development of the biliteral steganography method in the 1600s. The use of binary in ancient Egyptian calculations and 19th-century cyphers demonstrates its long history. The electro-mechanical binary computer developed by Alan Turing in 1937 was also based on past use cases of the binary system.


**Graphlib: Functionality to Operate with Graph-like Structures**
===========================================================

The `graphlib` module provides functionality to operate with graph-like structures, specifically topological sorting of directed acyclic graphs (DAGs).

**TopologicalSorter Class**
-------------------------

The `TopologicalSorter` class provides functionality to topologically sort a graph of hashable nodes. A topological order is a linear ordering of the vertices in a graph such that for every directed edge `u -> v` from vertex `u` to vertex `v`, vertex `u` comes before vertex `v` in the ordering.

### Initialization

The `TopologicalSorter` class can be initialized with an optional `graph` argument, which must be a dictionary representing a DAG where the keys are nodes and the values are iterables of all predecessors of that node in the graph.

### Methods

*   `add(node, *predecessors)`: Adds a new node and its predecessors to the graph. Both the node and all elements in `predecessors` must be hashable.
*   `prepare()`: Marks the graph as finished and checks for cycles in the graph. If any cycle is detected, a `CycleError` is raised.
*   `is_active()`: Returns `True` if more progress can be made and `False` otherwise.
*   `done(*nodes)`: Marks a set of nodes returned by `get_ready()` as processed, unblocking any successor of each node in `nodes` for being returned in the future by a call to `get_ready()`.
*   `get_ready()`: Returns a tuple with all the nodes that are ready.
*   `static_order()`: Returns an iterator object which will iterate over nodes in a topological order.

### Example Usage

```python
>>> graph = {"D": {"B", "C"}, "C": {"A"}, "B": {"A"}}
>>> ts = TopologicalSorter(graph)
>>> tuple(ts.static_order())
('A', 'C', 'B', 'D')
```

### Parallel Processing

The `TopologicalSorter` class is designed to easily support parallel processing of the nodes as they become ready.

```python
topological_sorter = TopologicalSorter()
# Add nodes to 'topological_sorter'...
topological_sorter.prepare()
while topological_sorter.is_active():
    for node in topological_sorter.get_ready():
        # Worker threads or processes take nodes to work on off the 'task_queue' queue.
        task_queue.put(node)
    # When the work for a node is done, workers put the node in 'finalized_tasks_queue' so we can get more nodes to work on.
    node = finalized_tasks_queue.get()
    topological_sorter.done(node)
```

### Exceptions

The `graphlib` module defines the following exception classes:

*   `CycleError`: Subclass of `ValueError` raised by `TopologicalSorter.prepare()` if cycles exist in the working graph.

The detected cycle can be accessed via the second element in the `args` attribute of the exception instance and consists of a list of nodes, such that each node is, in the graph, an immediate predecessor of the next node in the list. In the reported list, the first and the last node will be the same, to make it clear that it is cyclic.


This is a variation of the popular puzzle Sudoku, where the word "RAMANUJAN" occupies 89 cells. The puzzle uses numbers 1 to 9, each used 9 times, and the remaining cells are filled with numbers 1 to 8, following the properties of normal Sudoku.

The puzzle is constructed as follows:
1. The word "RAMANUJAN" is written using 119 squares, with numbers 1 to 9 filled inside each letter.
2. Numbers 1 to 9 are written a total of 13 times, and the remaining cells are filled with numbers 1 and 2.
3. Each 9x9 Sudoku is filled individually, with the total number of covered squares for each word/letter filled with numbers 1 to 9.
4. The adjacent Sudoku to the right commences with the next numbers, from the last used number, filling in the alphabet in that 9x9 Sudoku, and the next word/letter starts with the filled number.

For example, the first Sudoku contains 20 squares filled with numbers 1 to 9 twice, and the balance filled with numbers 1 and 2. Consequently, the next Sudoku commences with numbers 3 to 9, followed by 1 to 9, and so on.


**Geometry and Symmetry**

1. **Perimeter of Shapes**: Each shape is made from arcs of a circle of radius r. What is the perimeter of a shape with 3, 4, 5, and n "nodes"?

2. **Watch Face Symmetry**: A watch face has identical hands and marks for hours, opposite a mirror. When is the time read directly and in the mirror exactly the same between 6 and 7?

3. **Tablecloth Designs**: Charlie likes tablecloths with symmetry and many colors. How many colors are needed for different designs?

4. **Frieze Patterns**: A gallery of cast ironwork friezes in Australia with a mathematical discussion on classification.

5. **Symmetry of Traces**: Points off a rolling wheel make traces. What makes those traces have symmetry?

6. **Overlapping Squares**: A red square overlaps a blue square, covering a quarter of it. Prove this regardless of orientation.

7. **Line Symmetry**: How many ways can three pieces fit together to make shapes with line symmetry?

8. **Rotation Symmetry**: Streamers have rotation symmetry. Can you find a design with only one center of rotation?

9. **Symmetrical Shapes**: How many different symmetrical shapes can be made by shading triangles or squares?

10. **Repeating Designs**: Create a strip that matches itself after rotation or reflection.

11. **Tetrahedron Volume**: An irregular tetrahedron has two opposite sides of length a, with a perpendicular line of length b joining their midpoints. What is the volume?

12. **Trominoes on a Chessboard**: Can all but one square of an 8x8 chessboard be covered by trominoes?

13. **Pattern Units**: Images from the Topkapi Palace show patterns. What is the basic unit of each pattern? Can you continue the pattern?

14. **Celtic Knots**: Create Celtic knots and explore the mathematics behind them.

15. **Rotating Shapes**: Create attractive images by rotating shapes through multiples of 90, 30, 72 degrees, or other angles.

16. **Frieze Patterns Proof**: Proofs of the seven frieze patterns involve group theory, but the symmetries of a cylinder provide an easier approach.

17. **Tournament Scheduling**: Scheduling games can be challenging. What tournament formats do sport schedulers use?

18. **Solids with Squares and Triangles**: Each solid has 3 squares and a triangle around each vertex, with 18 square faces and 8 equilateral triangle faces. How many faces, edges, and vertices are there?

19. **Geometric Proof**: Given a square ABCD, M is the midpoint of AD, and CP is perpendicular to MB with P on MB. Prove DP = DC.

20. **Repeating Patterns**: Patterns that repeat in a line are interesting. How many types are there, and how do you tell one type from another?

21. **Vertical Symmetry**: When a strip has vertical symmetry, there always seems to be a second mirror line. Can you find a design with only one mirror line?

22. **Symmetry and Square Dancing**: An article on symmetry and square dancing. What do the symmetries of the square have to do with a dos-e-dos or a swing?

23. **Geometry in Car Tyres**: Car tyres and wheels are a rich source of geometry. What geometric opportunities can be found in a car park?

24. **Holly Leaf Geometry**: The "holly leaf" is made of 10 arcs of circles with a radius of 1 cm. Find the perimeter and area of the holly leaf.

25. **Mathematics and Patterns**: Mathematics is the study of patterns. What opportunities are there to observe, hypothesize, experiment, discover, and create?

26. **Prime Number Grid**: Place numbers 1-9 on a 3x3 grid so that all rows and columns add up to a prime number. How many solutions are there?

27. **Sharing a Pizza**: Can you show that a square pizza can be shared equally between two people by cutting it four times using vertical, horizontal, and diagonal cuts?

28. **Edge-Up Dice**: What if dice that land edge-up are not rolled again? What are the implications?

29. **3D Geometry**: An article introducing a rich area for practical exploration and discovery in 3D geometry.

30. **Circumscribing a Shape**: An equilateral triangle sits on top of a square. What is the radius of the circle that circumscribes this shape?

31. **Dominoes Square**: Using 8 dominoes, make a square where each column and row adds up to 8.

32. **Tile Patterns**: Photos of tiles at a school in Gibraltar show patterns. What questions can be asked about them?

33. **Symmetrical Fabric Design**: Create a symmetrical fabric design based on a flower motif using Logo.

**Multiple Choice Questions**

* What is the minimum number of colors needed for a tablecloth with symmetry?
A) 2
B) 3
C) 4
D) 5

* How many frieze patterns are there?
A) 5
B) 6
C) 7
D) 8

* What is the volume of the irregular tetrahedron?
A) (1/3)ab
B) (1/2)ab
C) (2/3)ab
D) ab

**Answers**

* 2
* 7
* (1/3)ab


Completion Percentage = (200 / 500) * 100 = 40%
So, in this case, you would have completed 40% of the project.
This formula can be applied to various scenarios, such as tracking the completion of tasks in a project management tool, evaluating student progress in a course, or measuring the completion of goals in personal development plans.

Percent Work Complete = (Actual Work / Work) * 100
If you type a value in the % Work Complete field, Project automatically calculates actual work and remaining work.
Similarly, entering a value in the Actual Work or Remaining Work field automatically recalculates the other fields.

A project complete percentage is a calculation of the duration of a complete project divided by what tasks you already completed in the project.
Project complete percentages display the amount of work on a project that a team has completed and how much work they have left to accomplish.

A time management calculator is a tool that enables you to understand, plan, and organize your time.
It helps you understand how you spend your time.
It can be applied to any given period of time — meaning that you can use it on a daily, weekly, monthly, or even yearly basis.
The 2 main purposes of using time management calculators

To calculate the percentage complete for a project with a list of tasks, you can use a simple formula based on the COUNTA function.
In the example shown, the formula in F6 is:
= COUNTA(C5:C11) / COUNTA(B5:B11)
Generic formula
= COUNTA(range1) / COUNTA(range2)
Explanation

How to Calculate Percent Complete
We've discussed what percent complete is and why it's important in project management.
The next step is figuring out how to calculate for percent complete.
It's a simple formula.
Percent Complete = Actual Duration/Duration (PC = AD/D)

Formula
Time Percentage (%) = ( Time A [sec] / Time B [sec] ) × 100
Where:
Time A [sec] = Seconds + 60 × Minutes + 3600 × Hours + 86400 × Days
Time B [sec] = Seconds + 60 × Minutes + 3600 × Hours + 86400 × Days
Example:
What percentage of 5 minutes is 30 seconds?
There are 300 seconds in 5 minutes (60 sec × 5 = 300 sec.)

On the Gantt chart, click a task and click Task and the percentage complete you want to show.
For example, to show a task as 25 completed, click Task > 25% Complete.
To see the task progress, point to the dark line inside the bar.

One of the simplest ways to calculate task progress is to use the formula =completed/total, where completed is the number of tasks that are done, and total is the number of tasks that are planned ...

Percent Complete Tracking #3: Arbitrary amounts.
The final way of working out what to put in the percent complete column is by using an arbitrary, fixed figure.
For example:
when a task begins, you give it 20%.
When the task finishes, you make it up to 100%.
The task would never be 25% or 90%.
It could only ever be 0%, 20% or 100%.

Entry Type
Calculated
Description
The % Work Complete field contains the current status of all tasks assigned to a resource expressed as the total percentage of the resource's work that has been completed.

How Calculated
When a resource is first added, the percent work complete for the resource is zero percent.
As soon as tasks are assigned to the resource, and any actual work or percent work complete is entered for any of the resource's assignments, the total percent work complete for the resource is calculated.
Project calculates the resource's percent work complete as follows:

Best Uses
Add the % Work Complete field to a resource sheet when you want to display or filter that percent work complete for a resource.
This can help you review how a resource is progressing on all assigned tasks.

Example
Jamie is assigned to four tasks for a total of 40 hours of work.
Until Jamie reports any actual work on any of the tasks, the percent work complete is zero.
Jamie reports that five hours of work has been done on one task, and 15 hours of work has been done on another task.
Because 20 hours of actual work has been performed out of 40 hours of total work scheduled, Project calculates that Jamie has 50 percent work complete on all assigned tasks.

Percent (%) Work Complete (assignment field)

Description
The % Work Complete field contains the current status of an assignment expressed as the percentage of the assignment's work that has been completed.
You can enter percent work complete, or you can have Project calculate it for you based on actual work on the assignment.

How Calculated
When an assignment is first made, the percent work complete is zero.
If you enter actual work for the assignment, Project calculates percent work complete as follows:

If you type a value in the % Work Complete field, Project automatically calculates actual work and remaining work.
If the % Work Complete field is set to a value greater than zero, the Actual Start field is set to the scheduled start date if you have not yet entered an actual start date.
If the % Work Complete field is set to 100, the Actual Finish field is set to the scheduled finish date.
Similarly, entering a value in the Actual Work or Remaining Work field automatically recalculates the other fields.

Best Uses
Add the % Work Complete field to the sheet portion of the Task Usage or Resource Usage view when you want to display, filter, or edit percent work complete for assignments.

Example
Jamie is assigned to the "Write proposal" task for 40 hours of work.
So far, Jamie has reported 20 hours of actual work on this assignment.
Project calculates that the assignment is 50 percent work complete.

In addition, Jamie is assigned to 32 hours of work for the "Conduct client meetings" task.
Jamie has reported that 16 hours have just been completed.
In the % Work Complete field, you enter 50.
The Actual Work field changes to 16 hours.

When you enter % Work Complete, Project might adjust actual and remaining work around the current status date.
If you prefer, you can leave these tracking fields in the project as originally scheduled, even if completed work is shown in the future or remaining work is shown in the past.
You can do this in the Project Options dialog box.

Percent work complete indicates the status of the amount of work completed on the assignment so far.
By contrast, percent complete indicates the status of the duration for the task.

The task percent complete can be calculated either on an average of completed tasks or on work time completed.
For the average, it is calculated by dividing the number of the completed work segments by the total number of segments in the work (tasks completed / total # of tasks).

if a task does not have a Complete Date, Task % Complete = 0%.
if a task has a Complete Date, task % Complete = 100%.
Summary/Parent Task/Project % Complete value is calculated as follows (immediate child tasks only):
(Total number of completed active tasks/total number of open active tasks) * 100.

This budget planning represents the following percentage of the job:
60/2088 = .0287 = 2.87%.
(This can be rounded to 3% of the job.)
The following tables indicate percentages for tasks that are performed for various periods of time.
If you have questions, contact your Human Resources (HR) Office.

Let's see how to calculate the percentage of task status in excel with example.
Look the image above here number of tasks completed is 3 among 11 projects tasks, so the calculation of percentage will be = (required part/total value)*100 = (3/11)*100 = 27% (value decimal omitted).

This site is great and I really appreciate those of you helping those of us who are technically deficient.
=COUNTA(B5:B9)/5.
The cells reflect the following:
A B.
Task 1 check box.
Task 2 check box.
Task 3 check box.
Task 4 check box.
Task 5 check box.

For changing only one task's percentage completed in Outlook, please do as follows:
1. In the Tasks view, double click to open the specified task whose percentage completed you want to change.
2. Please change the percentage completed in the % Complete box.
See screenshot:
3. Click the File > Save & Close button to save the change.

How Calculated
When a task is first created, the percent work complete is zero percent.
If you enter actual work for the task, Project calculates percent work complete as follows:
Percent Work Complete = (Actual Work / Work) * 100.
If you type a value in the % Work Complete field, Project automatically calculates actual work and remaining work.
Similarly, entering a value in the Actual Work or Remaining Work field automatically recalculates the other fields.


# Smartermaths 2022 HSC Revision Series

## 2022 HSC Comprehensive Revision Series

The 2022 Comprehension Revision Series is available earlier than usual, providing teachers with flexibility in its use. This series is slightly longer than the 2021 version, offering students around 13-14 hours of past exam questions, covering key areas identified. It serves as a roadmap for the HSC exam, with recommended use for motivated students aiming for a Band 5 or 6 result.

The 2022 Revision Series release dates are:
- STANDARD 2: now available
- ADVANCED: 70% of topics now available
- EXTENSION 1: ready mid-next week

The series can be accessed via a provided link.

## Structure of the HSC Revision Series

Dividing revision time between topics is crucial, especially as the HSC exam approaches. The series' design is based on past HSC exam contributions, with adjustments made for new content, growing importance, and regularly tested concepts. Question selection aims to expose students to common types, untested concepts, and complex examples, including multiple-choice and longer-answer questions.

Before starting the series, students should review topic basics, especially in areas where they are less confident. However, time management is essential to avoid dedicating too much time to a single area. This series complements, rather than replaces, the need to complete past HSC papers in their entirety. Students will benefit from question overlap, promoting speed and improving exam performance if they prioritize finishing practice exams with 15-20 minutes to spare for review.


**Unit 8: Polygons and Quadrilaterals**

A parallelogram is a quadrilateral with two pairs of parallel sides. In geometry, a closed plane figure formed by 3 or more segments is considered a polygon.

The curriculum is divided into units, including:
- Unit 2: Reasoning and Proof
- A unit on Polygons and Quadrilaterals, which covers topics such as:
  - Properties of quadrilaterals
  - Types of quadrilaterals, including kites and parallelograms
  - Sum of interior angles of a quadrilateral
  - Number of sides in different polygons (e.g., heptagon has 7 sides, decagon has 10 sides)

To determine the total number of sides in a shape, add the number of sides of both shapes together. A quadrilateral has four sides, and a parallelogram is a specific type of quadrilateral with two pairs of parallel sides.

Key concepts include:
- A quadrilateral with four congruent sides and four right angles
- Properties of convex and concave polygons
- Identifying polygons and their properties

**Multiple Choice Questions:**
1. How many sides are in a quadrilateral polygon?
2. What is the sum of the interior angles of a quadrilateral?
3. What is the definition of a parallelogram?
   A) A quadrilateral with two pairs of parallel sides
   B) A quadrilateral with four congruent sides
   C) A quadrilateral with four right angles
   D) A quadrilateral with two pairs of congruent sides

**Answers:**
1. 4
2. 360 degrees
3. A) A quadrilateral with two pairs of parallel sides


The triangle is defined by the formula T(n, k) = n*(n+9) + k + 5, with initial values T(0, 0) = 5 and T(1, 0) = 15. The triangle begins as:
5;
15, 16;
27, 28, 29;
41, 42, 43, 44;
57, 58, 59, 60, 61;
75, 76, 77, 78, 79, 80;

The formula can be expressed in Mathematica as: T[n_, k_]:= If[n==0 && k==0, 5, If[k==0 && n==1, 15, n*(n+9) +k +5]];
and in Sage as: @CachedFunction def T(n, k): if (n==0 and k==0): return 5 elif (k==0 and n==1): return 15 else: return n*(n + 9) + k + 5.

Some values in the triangle are:
5, 15, 16, 27, 28, 29, 41, 42, 43, 44, 57, 58, 59, 60, 61, 75, 76, 77, 78, 79, 80, 95, 96, 97, 98, 99, 100, 101, 117, 118, 119, 120, 121, 122, 123, 124, 141, 142, 143, 144, 145, 146, 147, 148, 149, 167, 168, 169, 170, 171, 172, 173, 174, 175, 176, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205.

The status of this entry is approved.


Two boats left the harbor at the same time, traveling at different speeds and bearings. The first boat traveled at 10 km/h on a bearing of N47°E, while the second boat traveled at 8 km/h on a bearing of N79°E. To find the distance between the boats after 78 minutes, we can use trigonometry.

First, we need to convert the speeds from km/h to km/min: 1 km/h = 0.01667 km/min. Then, we can calculate the distance traveled by each boat:
- First boat: 10 km/h * 0.01667 km/min = 0.1667 km/min * 78 min = 13 km
- Second boat: 8 km/h * 0.01667 km/min = 0.1333 km/min * 78 min = 10.4 km

We can visualize the situation using a graph with the y-axis as North and the x-axis as East. The two bearings form a triangle with angles A, B, and C. The angle between the two bearings is A = 79° - 47° = 32°.

Using the law of cosines, we can find the distance between the boats (side a):
a^2 = b^2 + c^2 - 2bc cos A
where b = 10.4 km (second boat's bearing) and c = 13 km (first boat's bearing).

Plugging in the values, we get:
a^2 = 10.4^2 + 13^2 - 2(10.4)(13)cos 32°
a^2 = 108.16 + 169 - 270.4 * 0.8
a^2 = 277.16 - 216.32
a^2 = 60.8 km
a = √60.8 km ≈ 7.8 km

Therefore, the distance between the boats after 78 minutes is approximately 7.8 km.


## Class 7 Maths NCERT Solutions Chapter 4 Simple Equations InText Questions

### Try These (Page 78)

**Question 1.5**
The value of the expression (10y – 20) depends on the value of y. Verify this by giving different values to y and finding the value of (10y – 20) for each.

**Solution:**
The value of (10y – 20) changes with different values of y. To find a solution to 10y – 20 = 50, we test various values of y.

| y | 10y - 20 |
| --- | --- |
| 1 | 10*1 - 20 = -10 |
| 2 | 10*2 - 20 = 0 |
| 3 | 10*3 - 20 = 10 |
| 4 | 10*4 - 20 = 20 |
| 5 | 10*5 - 20 = 30 |
| 6 | 10*6 - 20 = 40 |
| 7 | 10*7 - 20 = 50 |

At y = 7, 10y – 20 = 50. Thus, y = 7 is a solution to the equation 10y – 20 = 50.

### Try These (Page 80)

**Question 1**
Write another form for each of the given equations.

**Solution:**
Other forms of the equations are:
- p multiplied by 5 gives 20
- 7 added to thrice of n gives 1
- one-fifth of m less 2 gives 6

### Try These (Page 88)

**Question 1**
Start with x = 5 and create two different equations. Ask classmates to solve them and verify if they get x = 5 as the solution.

**Solution:**
1st equation:
Starting with x = 5,
Multiplying both sides by 5 gives 5x = 25,
Adding 5 to both sides gives 5x + 5 = 30.

2nd equation:
Starting with x = 5,
Multiplying both sides by 3 gives 3x = 15,
Subtracting 3 from both sides gives 3x – 3 = 12.

**Question 2**
Create two number puzzles, one with the solution 11 and the other with 100.

**Solution:**
1st puzzle (solution 11): Think of a number, multiply it by 3, and add 2. The result is 35. What's the number?
2nd puzzle (solution 100): Think of a number, divide it by 10, and subtract 5. The result is 5. What's the number?

### Try These (Page 90)

**Question 1**
When you multiply a number by 6 and subtract 5, you get 7. What's the number?

**Solution:**
Let the number be x.
6x - 5 = 7,
6x = 12,
x = 2.

**Question 2**
What number, when one-third of it is added to 5, gives 8?

**Solution:**
Let the number be y.
(1/3)y + 5 = 8,
(1/3)y = 3,
y = 9.

**Question 3**
There are two types of mango boxes. The larger box contains 4 more mangoes than 8 of the smaller boxes combined. Each larger box has 100 mangoes. Find the number of mangoes in a smaller box.

**Solution:**
Let x be the number of mangoes in a smaller box.
8x + 4 = 100 (since the larger box has 4 more mangoes than 8 smaller boxes),
8x = 96,
x = 12.

Thus, each smaller box contains 12 mangoes.


Most digital clocks run on a 12-hour cycle, displaying times from 1:00 to 12:59, with no distinction between AM and PM. Since there are 60 minutes in an hour, a digital clock can display 12*60 = 720 possible times.

A typical digital clock has four numbers, each using a seven-segment display, resulting in 4*7 = 28 individual LEDs. However, only 720 of the 2^28 (268,435,456) possible combinations are valid times, indicating redundancy.

Theoretically, only 10 LEDs are needed to display 1024 possible messages, which is sufficient for a digital clock. However, this solution is not elegant, as 720 is not a power of two, resulting in 304 invalid values.

A more efficient approach is to use a factorial number system, which is a mixed radix system. This enables the design of a triangular digital clock, where each row represents a different unit of time. The dots in each row are worth 360, 120, 30, 6, and 1 minutes, respectively.

To decode the time, add up the values of the dots in each row. For example, a dot in the first row is worth 360 minutes, while a dot in the second row is worth 120 minutes. This system allows for the representation of times between 0 and 720 using exactly 5 factoradic numbers.

A computer implementation of the triangular clock can be achieved using Xlib, a library for creating graphical user interfaces. The implementation involves drawing the dots in the clock and updating the display in real-time.

The idea of a binary clock is not new and has been explored in various forms. One approach is to represent the hour with four dots and the minute with six dots, allowing for a total of 11 LEDs. Another possibility is to develop a new binary-based system of time measurement, defining a new unit of time called the "tock," where there are exactly 131,072 tocks in a day.

This system would require 17 LEDs to express the current time to the nearest tock, which is more accurate than a standard digital clock. A cheaper model could use 11 LEDs, ignoring the last six LEDs, similar to how we usually ignore the seconds on a clock.

The concept of a digital clock has inspired various projects, including a website that allows users to share associations with the numbers on the clock. While the idea of a triangular clock may not be practical, it demonstrates the potential for innovative and efficient designs in time-keeping devices.

In conclusion, the design of a digital clock can be optimized using different number systems and representations. The use of a factorial number system and a triangular clock design provides an efficient and elegant solution, although it may not be practical for everyday use. The exploration of alternative time-keeping systems and devices continues to inspire new ideas and innovations.

What is the total number of possible times that a standard digital clock can display?
A) 720
B) 1024
C) 2^28
D) 268,435,456

Answer: A) 720

What is the minimum number of LEDs required to display 1024 possible messages?
A) 10
B) 11
C) 15
D) 28

Answer: A) 10

What is the value of a dot in the first row of the triangular clock?
A) 1 minute
B) 6 minutes
C) 30 minutes
D) 360 minutes

Answer: D) 360 minutes

How many LEDs are required to express the current time to the nearest "tock" in the new binary-based system of time measurement?
A) 11
B) 15
C) 17
D) 28

Answer: C) 17


## Introduction to Lean Body Mass Calculator

The lean body mass calculator determines LBM based on age, gender, weight, and height, using either English or metric measurements. This health tool evaluates a body parameter known as LBM, which is the lean body mass, through age, gender, weight, and height.

## How the Calculator Works

To make it easier to input data, users can choose between the English or metric tab, allowing for input in either kg and cm or lbs and ft. The result comprises the LBM value found by employing three different formulas, given in both measurement units. These formulas are applied to users aged 15 or above, while a specific formula is used for children.

### Formulas Used

- **Hume's LBM Formula**:
  - Male: Estimated LBM = 0.32810*Weight (in kg) + 0.33929*Height (in cm) - 29.5336
  - Female: Estimated LBM = 0.29569*Weight (in kg) + 0.41813*Height (in cm) - 43.2933

- **Boer's LBM Formula**:
  - Male: Estimated LBM = 0.407*Weight (in kg) + 0.267*Height (in cm) - 19.2
  - Female: Estimated LBM = 0.252*Weight (in kg) + 0.473*Height (in cm) - 48.3

- **James's LBM Formula**:
  - Male: Estimated LBM = 1.1*Weight (in kg) – 128*[Weight (in kg)/Height (in cm)]^{2}
  - Female: Estimated LBM = 1.07*Weight (in kg) – 148*[Weight (in kg)/Height (in cm)]^{2}

- **Peter's Formula for Children** (up to 14 years old):
  - Estimated LBM = 3.8*[0.0215*(Weight in kg^0.6469)*(Height in cm^0.7236)]

## Example Calculations

- **Adult Female** (aged 34, weighing 124lbs, height 5ft 4in):
  - By Hume’s Formula: 91.07 lbs / 41.31 kg
  - By Boer’s Formula: 94.28 lbs / 42.76 kg
  - By James’s Formula: 93.62 lbs / 42.46 kg

- **Child** (aged 10, weighing 20lbs, height 28.6 in):
  - By Peter’s Formula: 16.67 lbs / 7.56 kg

## Understanding Lean Body Mass

Lean body mass is the body mass resulting from subtracting body fat from total body weight, typically around 60 to 90% of the body mass. The equation is: Body mass = Lean body mass + Body fat. Thus, body fat percentage averages 10 to 40%. LBM accounts for organs, blood, bones, muscle, and skin.

## Importance of LBM

LBM is used as an index to analyze metabolic functions and in prescribing medication for metabolic disorders. It helps track weight loss, ensuring that muscle mass is not lost instead of the targeted fat.

## References

1. Hume R. (1966) Prediction of lean body mass from height and weight. J Clin Pathol; 19(4):389-91.
2. Fuchs RJ, Theis CF, Lancaster MC. (1978) A nomogram to predict lean body mass in men. Am J Clin Nutr. 31(4):673-8.
3. Boer P. (1984) Estimated lean body mass as an index for normalization of body fluid volumes in man.
4. Absalom AR, Mani V, DeSmet T, et al. Pharmacokinetic models for propofol-defining and illuminating the devil in the detail. Br J Anaesth 2009; 103:26-37
5. A. M. Peters, H. L. R. Snelling, D. M. Glass, N. J. Bird (2011) Estimation of lean body mass in children.


# Lesson 5.4 of Exploding Dots

The division problems done so far have worked out nicely. For example, 276 divided by 12 equals 23. This was visualized by drawing a picture of 276 and looking for groups of 12. The result showed two groups of ten and three groups of one, totaling 23 groups of 12.

Now, consider the problem 277 divided by 12. The picture of 277 would be two dots, seven dots, and one extra dot. When looking for groups of 12, the same groups as before are found, but with one extra dot remaining. This extra dot can be interpreted as a remainder.

The result of 277 divided by 12 can be expressed as 23 with a remainder of 1. Depending on the grade level or country, this might be written as "remainder 1" or "23 + 1/12". The long division approach with pictures is useful for handling remainders, as they become visible and easy to work with.

In this approach, remainders are straightforward to deal with, even when the division problem does not result in a whole number. The visual representation makes it clear what to do with remainders, making them easy to understand and work with. This method is helpful for division problems, especially when remainders are involved.


# CT: Morphism

In category theory, a morphism is a transformation from one object to another, either within a category or between categories, denoted as f: A → B, where f is a morphism from object A to object B. Morphisms can be categorized based on their behavior across every object into types such as isomorphism, epimorphism, and monomorphism. The morphism is central to category theory, describing relations between objects.

A category, which is fundamental to category theory, consists of a collection of objects and morphisms. Each object in a category has at least one 'identity morphism', and morphisms are represented as arrows pointing from one object to another. Objects serve as named points to provide context for the morphisms. Category theory focuses on the composition of morphisms within different categories and the various states that are possible.

Key concepts related to morphisms include:
- **Isomorphism**: A morphism that can be inverted.
- **Epimorphism** and **Monomorphism**: Types of morphisms categorized by their behavior.
- **Identity Morphism**: A morphism that leaves an object unchanged, a requirement for every object in a category.

Categories can be represented in various ways:
- A category with a single object is considered a monoid, where the object can have multiple morphisms, and the composition of these morphisms corresponds to an associative binary operation.
- Graphs can represent categories, with objects as nodes and morphisms as edges, although not every graph represents a category, every category can be described as a graph, and any graph can be made into a category.

Understanding morphisms and their properties is essential for exploring category theory further, including the study of specific types of categories, such as the empty category, and the application of category theory in representing and analyzing complex structures and relationships.


1. A survey of people in Kansas reported that 30% of people had been in a tornado. To estimate the percentage of people who have been in a tornado with 94% confidence and a margin of error of 4%, how many people must be surveyed? Round up to the next whole number.

2. In a sample of 1550 students, 785 brought their own lunch. What is the margin of error for the 99% confidence interval estimate for the true percentage of students who bring their lunch? Answer as a percent rounded to the nearest tenth.

3. Find the t(α/2) for an 80% confidence interval based on a sample size of 12. Use the t-distribution table to find this value to the thousandth decimal place.

4. To qualify for the Police Academy, candidates must score in the top 7% on the General Abilities Test, which has a mean of 350 and a standard deviation of 15. What is the lowest score that a candidate can have and still qualify? Round up to the nearest whole number.

5. Find the 2 z-scores in the standard normal distribution that define the middle 70% of the area under the curve. Round answers to the nearest hundredth decimal place and give in ascending order separated by a comma.

6. In a standard normal distribution, find the area under the curve for z-scores above -0.5. Round the answer to 4 decimal places.

7. Lengths of pregnancies are normally distributed with a mean of 265 days and a standard deviation of 15 days. What is the z-score for a pregnancy lasting 275 days? Round to the nearest hundredth.

8. Find the z-score in the standard normal distribution such that 25% of the area under the curve will be to the right of this z-score. Round the answer to the nearest hundredth decimal place.

9. The average beginning annual salary for teachers in California is $41,181 with a standard deviation of $725. 10% of starting teachers will make at most how much money per year? Round to the nearest dollar.

10. The amount of money a Harvard University student carries is normally distributed with a mean of $50 and a standard deviation of $10. What is the probability that a randomly selected student will have more than $75? Give the answer as a decimal rounded to 4 decimal places.

11. Women's heights are normally distributed with a mean of 64.5 inches and a standard deviation of 2.3 inches. What is the probability that a randomly selected woman's height will be between 64 and 65 inches? Give the answer as a decimal rounded to 4 places.

12. The average number of hours SJSU students spend working at a computer is 4.3 hours with a standard deviation of 0.8 hours. What percentage of students spend less than 5.5 hours per day working at a computer? Give the answer as a decimal rounded to four decimal places.

13. The average age of a vehicle in Santa Clara County is 9 years (108 months) with a standard deviation of 18 months. What percent of vehicles are less than 8 years old (96 months)? Give the answer as a decimal rounded to four decimal places.

14. The average time to respond to a roadside emergency is 25 minutes with a standard deviation of 4.5 minutes. For a sample of 10 emergency response calls, what is the probability that the average wait time is more than 28 minutes? Give the answer as a rounded 4-place decimal.

15a. The average number of hours SJSU students spend working at a computer is 4.3 hours with a standard deviation of 0.8 hours. What percentage of students spend less than 5.5 hours per day working at a computer? Give the answer as a decimal rounded to four decimal places.

15b. The average number of hours SJSU students spend working at a computer is 4.3 hours with a standard deviation of 0.8 hours. What is the probability that a sample of 16 students will spend an average of less than 4.5 hours per day working on a computer? Give the answer as a decimal rounded to four decimal places.

16. The average time to respond to a roadside emergency is 25 minutes with a standard deviation of 4.5 minutes. For a sample of 12 emergency response calls, what is the probability that the average wait time is between 24 and 27 minutes? Give the answer as a rounded 4-place decimal.

17. Women's heights have a mean of 64.5 inches and a standard deviation of 2.3 inches. What is the probability that a sample of 30 women will have an average height greater than 65 inches? Give the answer as a decimal rounded to 4 places.

18. A California State Lottery official wants to estimate the average amount spent on lottery tickets with 90% confidence. He samples 150 people with an average spend of $39.50 and a sample standard deviation of $3.50. What is the lower boundary of the confidence interval estimate to the nearest penny?

19. In a survey of 1500 men, 590 supported the death penalty. What is the best point estimate of the true population percentage of men supporting the death penalty? Answer as a percent rounded to the nearest tenth of a percent.

20. A statistics student wants to estimate the average age of students with 99% confidence and a margin of error of 1/4 year. The population standard deviation is estimated to be 3.5 years. How many people must she survey?

21. In a sample of 1550 students, 785 brought their own lunch. What is the margin of error for the 99% confidence interval estimate for the true percentage of students who bring their lunch? Answer as a percent rounded to the nearest tenth.

22. A political pollster wants to estimate the percentage of female voters who believe in a "Woman's Right to Choose" with 90% confidence and a margin of error of 10%. What is the minimum number of female voters to sample?

23. A researcher measures the tread depth of 18 randomly selected cars with a sample mean of 0.33 inches and a sample standard deviation of 0.07 inches. Find the 99% confidence interval estimate for the average tread depth.

24. The math scores for a sample of 100 eighth graders had a mean of 80 with a sample standard deviation of 9. What is the best point estimate for the population mean of the math scores?

25. A California State Lottery official wants to estimate the average amount spent on lottery tickets with 95% confidence. He samples 150 people with an average spend of $39.50 and a sample standard deviation of $3.50. What is the upper boundary of the confidence interval estimate to the nearest penny?

26. Find the Z(α/2) for an 80% confidence interval. Round the answer to the hundredths place.

27. Find the t(α/2) for an 80% confidence interval based on a sample size of 12. Use the t-distribution table to find this value to the thousandth decimal place.


To solve the inequality 6y - 10 - y > 14, we start by simplifying the left side.

Subtracting y from 6y gives 5y - 10 > 14.

Next, add 10 to both sides of the inequality to get 5y > 14 + 10.

Adding 14 and 10 yields 5y > 24.

Divide each term in 5y > 24 by 5 to solve for y, resulting in y > 24/5.

Simplifying 24/5 gives y > 4.8.

The solution can be expressed in different forms:

Inequality Form: y > 4.8

Interval Notation: (4.8, ∞)


## Composition of Functions

The composition of two functions f: A → B and g: B → C, denoted by gof, is defined as the function gof: A → C given by gof = g(f(x)) for all x ∈ A.

### Example 1

Given f(x) = (x + 3) and g(x) = x^2, we can find the compositions gof and fog as follows:
- gof = g(f(x)) = g(x + 3) = (x + 3)^2
- fog = f(g(x)) = f(x^2) = x^2 + 3

In this case, fog ≠ gof.

### Example 2

Consider f: N → N, g: N → N, and h: N → R defined as f(x) = 2x, g(y) = 3y + 4, and h(z) = sin z, for x, y, and z in N. We need to show that ho(gof) = (hog)of.

- ho(gof)(x) = h(gof(x)) = h(g(f(x))) = h(g(2x)) = h(3(2x) + 4) = h(6x + 4) = sin(6x + 4) for x in N.
- ((hog)of)(x) = (hog)(f(x)) = (hog)(2x) = h(g(2x)) = h(3(2x) + 4) = h(6x + 4) = sin(6x + 4) for x in N.

This demonstrates that ho(gof) = (hog)of.

### Example 3

Given f(x) = x^2 + 1 and g(x) = 5x + 3, find:
(a) (f o g)(x)
(b) (g o f)(x)

#### Solution

(a) (f o g)(x) = f(5x + 3) = (5x + 3)^2 + 1 = 25x^2 + 30x + 9 + 1 = 25x^2 + 30x + 10

(b) (g o f)(x) = g(x^2 + 1) = 5(x^2 + 1) + 3 = 5x^2 + 8

### Related Topics

For further study, explore the main pages of Class 12 subjects:
- Class 12 Maths
- Class 12 Physics
- Class 12 Chemistry
- Class 12 Biology


To take the square root of exponents, you divide the exponent by 2. The square root of a number is that number to the 1/2 power. For example, the square root of x cubed (x^3) is x^3 to the half power, which equals x^(3/2).

When finding the square root of a number with an even exponent, you divide the exponent by 2. For instance, the square root of x^10 is x^5, and the square root of x^7 is x^3.5, which can also be expressed as x^3 * x^0.5. The x^0.5 is equivalent to the square root of x.

When multiplying two values of the same base raised to different exponents, you add the exponents. When dividing them, you subtract the exponents. In the case of roots, the exponents are fractions, so x^(1/2) ÷ x^(1/3) = x^(1/2 - 1/3) = x^(3/6 - 2/6) = x^(1/6).

Squaring an expression and then taking the square root cancels each other out, except that the sign will be positive. For example, squaring 16x and then taking the square root results in 16x or -16x, depending on whether x is positive or negative.

The exponent for a square root is 0.5 or 1/2. To find the square root of a number using a calculator, raise the number to the power of 1/2. Roots in math are the opposite of exponents. If you square a number and then take the square root, you undo the operation. For instance, the square root of 81 is 9 because 9 squared is 81.

The square root of 729 is 27. As a product of its prime factors in exponents, it is 3^6. To find the square root of 5184, first find its prime factorization using exponents. Then, use the answer to calculate the square root.

A square root is a number raised to the exponent (power) 1/2. Fractional exponents follow the same algebra rules as integer exponents. The numerator of the exponent is the normal integer type of exponent degree, and the denominator is similar to the degree of the root.

The prime factorization of 10,000 is 2^4 * 5^4. The square root is 2^2 * 5^2, or 100. Exponents, such as 2 to the 3rd power or 4 to the 4th power, can be simplified using the rules of exponents.

A root is like a fractional power. For example, (x^(1/4))^(1/2) equals x^(1/8) or the eighth root of x. The cube root of the square root of π is the 6th root of π, which can be expressed as (π^(1/2))^(1/3) = π^(1/6).

Q: How do you take the square root of exponents?
A: You divide the exponent by 2.

Q: What is the prime factorization of 5184 using exponents?
A: First, find the prime factorization of 5184, which is 2^6 * 3^4. Then, the square root of 5184 is 2^3 * 3^2, or 36.

Q: What is the square root of x^7?
A: The square root of x^7 is x^3.5, which can also be expressed as x^3 * x^0.5.

Q: What is the cube root of the square root of π?
A: The cube root of the square root of π is the 6th root of π, which can be expressed as (π^(1/2))^(1/3) = π^(1/6).


# Post Hole Concrete Calculator

The post-hole concrete calculator is a tool designed to determine the volume of concrete mix required for a fence post hole. It provides the necessary calculations to ensure a stable and secure fence installation.

To use the calculator, simply input the depth and dimensions of the post, and it will calculate the required volume of concrete. If you know the total number of posts needed, you can also input that to obtain the total volume of concrete required.

The recommended post depth is at least 1/3 of the post length. Our fence calculator can help determine the number of posts and their length.

## How to Use the Concrete Post Calculator

Calculating the post-hole concrete mix is straightforward:

1. Input the depth and dimensions of the post to calculate the required concrete volume.
2. If you know the total number of posts, input that to get the total volume of concrete.

## Formulas for Calculating Concrete Volume

The volume of concrete required is calculated by subtracting the volume of the buried portion of the post from the hole volume.

### Volume Formula for Cuboid Shape Post

The volume of the buried portion of a cuboid post is calculated as:
Volume = width × thickness × depth

It is recommended to make the hole three times the width and three times the post thickness. Ensure that the width, thickness, and depth have the same units (e.g., inches or meters) to obtain the correct volume units.

### Volume Formula for Cylindrical Shape Post

The volume of the buried portion of a cylindrical post is calculated using the cylinder volume formula. It is recommended to use a hole with three times the post diameter. Ensure that the post diameter and depth have the same units to obtain the correct volume units.

## FAQ

### Can You Put Dry Concrete in a Post Hole?

Yes, you can put dry concrete in a post hole. The dry concrete will come into contact with the ground and collect water, which is necessary to bind the mortar.

### What Size Hole for a 4"x4" Fence Post Should I Use?

For a 4"x4" fence post, a 12"x12" hole size is recommended. If the post is 3 ft deep, this requires approximately 0.09877 cubic yards of concrete, equivalent to 240 lb of cement or three 94 lb bags.


The prompt is suitable for secondary school students aged 11 to 16 and involves placing a number in a larger circle and calculating the results of taking steps in two paths. To facilitate communication, the teacher may label the larger circle 'input' and the two others 'output A' and 'output B'. Students typically conclude that one output is always three more than the other, leading to a preliminary phase of inquiry.

The selection of a regulatory card often reveals students' mathematical thinking, with two common choices being to find more examples by experimenting with different pairs of operations or to prove the prompt is always true. The latter can lead to the development of algebra directly from a numerical case. For example, starting with the number 4, the top path gives 4 + 4 + 3 or 2 x 4 + 3, while the bottom path gives (4 + 3) + (4 + 3) or 2 x (4 + 3). The algebraic expressions are 2n + 3 and 2(n + 3), where n is the starting number. As 2(n + 3) expands to 2n + 6, it becomes clear that the difference between the outputs will always be three.

Students have shown enthusiasm for finding algebraic expressions and substituting into them to deduce the difference between outcomes for a particular starting number. The prompt can be modified in various ways, such as using three steps, three or four operations, or aiming for equal outputs. Students have also used other operations like division, squaring, and cubing.

The design of the prompt is crucial, with smaller numbers being less 'threatening' and allowing students to change the prompt to create more examples or test particular cases. A classroom display of the steps inquiry carried out by a year 8 mixed attainment class at Longhill High School, Brighton, UK, demonstrates the effectiveness of the prompt.

Guided posters, such as the one devised by Emma Morgan, can help students present their mathematical reasoning. The story of an inquiry, as presented in a grade 5/6 class at the Fred Varley Public School, Markham, Ontario, shows how students can explore the prompt in search of patterns and use algebraic expressions to explain their generalization.

Higher-order thinking is encouraged through the use of question stems, as seen in Rachel Mahoney's year 7 class at Carre's Grammar School, Sleaford, Lincolnshire, UK. Students' questions and conversations were fantastic, and the question stems aided higher-order thinking. Presenting inquiry, as demonstrated by Emma Morgan's students, can turn students into "active learners who are fearless and methodical when attacking a problem."

Creating examples, as seen in year 9 students at Holyport College, Berkshire, UK, involves asking questions and creating examples to test conjectures. Classroom inquiry, as demonstrated by year 8 students at The Bishop Wand Church of England School, Sunbury-on-Thames, UK, involves changing the operation and exploring permutations of three steps.

The inquiry in action, as seen in a year 7 class, starts slowly until one student declares "a number at the start would lead to two answers." The teacher introduces the terms input and outputs and invites the student to choose a number, leading to the development of algebraic reasoning. The class uses a sheet to create their own examples, finding that the outputs from each pair of steps are related in the same way.

Conjecturing, as seen in Amanda Klahn's grade 4 IB PYP class at the Western Academy of Beijing, China, involves making novel ideas, such as the conjecture that the amount added equals the difference between the outputs. This can be extended to the general case when one operation is 'multiply by b': b(n + a) - (bn + a) = a(b - 1).

Some possible multiple-choice questions based on the prompt are:
1. What is the difference between the two outputs in the prompt?
A) 1
B) 2
C) 3
D) 4

Answer: C) 3

2. What is the algebraic expression for the top path in the prompt?
A) 2n + 2
B) 2n + 3
C) 2n + 4
D) 2n + 5

Answer: B) 2n + 3

3. What is the effect of changing the number of operations in the prompt?
A) The difference between the outputs increases
B) The difference between the outputs decreases
C) The difference between the outputs remains the same
D) The difference between the outputs becomes unpredictable

Answer: C) The difference between the outputs remains the same

4. What is the purpose of using smaller numbers in the prompt?
A) To make the calculation more difficult
B) To make the calculation easier
C) To make the prompt more 'threatening'
D) To make the prompt less 'threatening'

Answer: D) To make the prompt less 'threatening'

5. What is the benefit of using guided posters in presenting inquiry?
A) To make the presentation more confusing
B) To make the presentation more complicated
C) To help students present their mathematical reasoning
D) To make the presentation more boring

Answer: C) To help students present their mathematical reasoning


This guide explains how to count blank or empty cells in a given range of cells using a formula in Excel 2013/2016. Excel offers several count functions to count cells that match specific criteria, such as containing a specific text string or letter. To count cells without data values, use the COUNTBLANK function.

To count empty cells in a selected range, such as A1:A6, use the formula: **=COUNTBLANK(A1:B6)**. Simply type the formula and adjust the cell range as needed. The formula returns the number of blank cells in the specified range.

The COUNTBLANK function counts cells without any value, excluding cells with text, numbers, errors, etc. Alternatively, the COUNTIF function can achieve the same result using the formula: **=COUNTIF(A1:B6,””)**. Both functions return the same result, which is the number of blank cells in the range.

### Related Functions

- **Excel COUNTIF function**: Counts cells in a range that meet a given criteria, including cells with numbers, dates, text values, blank, non-blanks, or specific characters. The syntax is: **=COUNTIF(range, criteria)**.
- **Excel COUNTBLANK function**: Counts the number of empty cells in a range of cells. The syntax is: **=COUNTBLANK(range)**. This function is used to get the number of blank cells in a given range in Excel.


FRACTIONS INTRODUCTION LESSON PLAN

The lesson plan covers key concepts and operations involving fractions. To introduce fractions, the instructor will:
1. Ask the class for examples of fractions and discuss the definitions and differences between the numerator and denominator.
2. Emphasize the importance of a common denominator when adding and subtracting fractions.
3. Explain the differences between proper, improper, and mixed fractions.

The instructor will then demonstrate:
4. Converting improper fractions to mixed fractions.
5. Converting mixed fractions to improper fractions.
6. Applying fractions to word problems using everyday examples.
7. Simplifying fractions and explaining its importance.
8. Reducing fractions and finding the Greatest Common Factor (GCF) within a fraction.

Assessment will be based on a worksheet completion, requiring a passing grade of 70% or greater. Supporting materials include:
- Teaching lesson pages
- Exercise worksheets

These materials will facilitate a comprehensive understanding of fractions and their applications.


**Respect Responsibility Readiness**

1. A rectangle's length is twice its width, and its perimeter is 138 feet. Find the length.
Let x = width, 2x = length. The perimeter equation is 2x + 2x + x + x = 138, which simplifies to 6x = 138. Solving for x, we get x = 23, and the length is 2x = 46.

2. A rectangle's length is 4 feet less than three times its width, and its perimeter is 224 feet. Find the absolute difference between the dimensions.
Let x = width, 3x - 4 = length. The perimeter equation is 2x + 2(3x - 4) = 224, which simplifies to 8x = 232. Solving for x, we get x = 29, and the length is 3x - 4 = 83.

3. In a triangle, the first angle is twice the second, and the third is 5 degrees larger than the first. Find the absolute difference between the two largest angles.
Let x = second angle, 2x = first angle, 2x + 5 = third angle. The sum of angles equation is x + 2x + (2x + 5) = 180, which simplifies to 5x + 5 = 180. Solving for x, we get x = 35, and the angles are 35, 70, and 75.

4. A rectangle's length is twice its width. If the length decreases by 5 feet and the width increases by 5 feet, the area increases by 75 square feet. Find the absolute difference between the length and width.
Let x = width, 2x = length. The area equation is x(2x) + 75 = (x + 5)(2x - 5). Solving for x, we get x = 20, and the length is 2x = 40.

5. A triangle's first side is 2 inches less than twice the second side, and the third side is 10 inches longer than the second side. The perimeter is 12 feet (144 inches). Find the absolute difference between the two largest sides.
Let x = second side, 2x - 2 = first side, x + 10 = third side. The perimeter equation is x + (2x - 2) + (x + 10) = 144, which simplifies to 4x + 8 = 144. Solving for x, we get x = 34, and the sides are 34, 66, and 44.

6. A rectangle's length is 10 feet more than twice its width, and its perimeter is 320 feet. Find the length and width.
Let x = width, 2x + 10 = length. The perimeter equation is 2x + 2(2x + 10) = 320, which simplifies to 6x + 20 = 320. Solving for x, we get x = 50, and the length is 2x + 10 = 110.

7. A triangle's second angle is 20 degrees greater than the first angle, and the third angle is twice the second. Find the angles.
Let x = first angle, x + 20 = second angle, 2(x + 20) = third angle. The sum of angles equation is x + (x + 20) + 2(x + 20) = 180, which simplifies to 4x + 60 = 180. Solving for x, we get x = 30, and the angles are 30, 50, and 100.

8. A farmer wants to fence a rectangular area with a length three times its width. The perimeter is 320 feet. Find the dimensions.
Let x = width, 3x = length. The perimeter equation is 2(3x) + 2x = 320, which simplifies to 8x = 320. Solving for x, we get x = 40, and the length is 3x = 120.

9. A triangle's first side is twice the second side, and the third side is 20 feet less than three times the second side. The perimeter is 106 feet. Find the sides.
Let x = second side, 2x = first side, 3x - 20 = third side. The perimeter equation is x + 2x + (3x - 20) = 106, which simplifies to 6x - 20 = 106. Solving for x, we get x = 21, and the sides are 21, 42, and 43.

10. A room's length is 8 feet more than twice its width. The perimeter is 124 feet. Find the dimensions.
Let x = width, 2x + 8 = length. The perimeter equation is 2x + 2(2x + 8) = 124, which simplifies to 6x + 16 = 124. Solving for x, we get x = 18, and the length is 2x + 8 = 44.

11. A triangular lawn has a perimeter of 42 yards. The first side is 5 yards less than the second, and the third side is 2 yards less than the first. Find the sides.
Let x = second side, x - 5 = first side, (x - 5) - 2 = third side. The perimeter equation is x + (x - 5) + (x - 5 - 2) = 42, which simplifies to 3x - 12 = 42. Solving for x, we get x = 18, and the sides are 13, 18, and 11.

12. A rectangular metal plate's length is five times its width. The perimeter is 72 meters. Find the dimensions.
Let x = width, 5x = length. The perimeter equation is 2(5x) + 2x = 72, which simplifies to 12x = 72. Solving for x, we get x = 6, and the length is 5x = 30.

13. A rectangular box of cereal has a volume of 252 cubic inches, a height of 12 inches, and a width of 7 inches. Find the depth.
Let x = depth. The volume equation is (7)(12)(x) = 252, which simplifies to 84x = 252. Solving for x, we get x = 3.

14. A rectangle's length is 8 feet more than its width. If the width increases by 4 feet and the length decreases by 5 feet, the area remains the same. Find the original dimensions.
Let x = width, x + 8 = length. The area equation is x(x + 8) = (x + 4)(x + 3). Solving for x, we get x = 11, and the length is x + 8 = 19.

15. A rectangle's length is 4 meters more than its width, and its area is 45 square meters. Find the dimensions.
Let x = width, x + 4 = length. The area equation is x(x + 4) = 45, which simplifies to x^2 + 4x - 45 = 0. Solving for x, we get x = 5, and the length is x + 4 = 9.

16. A rectangle's length is three times its width, and its area is 108 square centimeters. Find the dimensions.
Let x = width, 3x = length. The area equation is x(3x) = 108, which simplifies to 3x^2 = 108. Solving for x, we get x = 6, and the length is 3x = 18.

17. A photograph's length is 1 centimeter less than twice its width, and its area is 28 square centimeters. Find the dimensions.
Let x = width, 2x - 1 = length. The area equation is x(2x - 1) = 28, which simplifies to 2x^2 - x - 28 = 0. Solving for x, we get x = 3.5, and the length is 2x - 1 = 6.

18. A square field's length and width are increased by 3 meters and 2 meters, respectively. The new area is 90 square meters. Find the original side length.
Let x = original side length, (x + 2)(x + 3) = 90. The equation simplifies to x^2 + 5x - 84 = 0. Solving for x, we get x = 7.

19. A rectangular mural's length is 1 meter greater than its width, and its area is 20 square meters. Find the dimensions.
Let x = width, x + 1 = length. The area equation is x(x + 1) = 20, which simplifies to x^2 + x - 20 = 0. Solving for x, we get x = 4, and the length is x + 1 = 5.

20. A rectangle's length is 6 centimeters more than its width, and its area is 11 square centimeters. Find the dimensions.
Let x = width, x + 6 = length. The area equation is x(x + 6) = 11, which does not have a straightforward solution.

21. A rectangular garden's length is 4 meters more than its width, and its area is 77 square meters. Find the dimensions.
Let x = width, x + 4 = length. The area equation is x(x + 4) = 77, which simplifies to x^2 + 4x - 77 = 0. Solving for x, we get x = 7, and the length is x + 4 = 11.

22. A rectangular park's length is 2 kilometers more than twice its width, and its area is 112 square kilometers. Find the dimensions.
Let x = width, 2x + 2 = length. The area equation is x(2x + 2) = 112, which simplifies to 2x^2 + 2x - 112 = 0. Solving for x, we get x = 7, and the length is 2x + 2 = 16.