Dijkstra's algorithm is used to find the shortest path from one source to all other vertices in a graph. To understand this, one should know what a graph is and how to store it in memory. Another solution for this problem is the Bellman-Ford algorithm.

Given a graph, the goal is to calculate the length of the shortest path from a source node to every other node. For example, between nodes 1 and 3, there are several paths, but the shortest one is 1 -> 4 -> 2 -> 3 with a length of 9.

Dijkstra's algorithm works on graphs with no edges of negative weight, and its speed can vary from O(n*lg(lg(n))) to O(n^2). The idea is simple: initially, the length of the shortest path to all nodes is set to infinity, and the length of the shortest path to the source is set to 0. The algorithm then makes the paths to one more node optimal at each step.

Here's how it works:

1. Choose the node that is not yet optimal but is closest to the source.
2. From this node, try to optimize the path to every node connected to it.
3. Mark the node as optimal (visited).
4. Repeat steps 1-3 until all nodes are optimal.

The algorithm can be implemented in C, and the source code is provided. The code initializes the distance to all nodes as infinity, except for the source node, which is set to 0. It then iteratively selects the unvisited node with the smallest distance, marks it as visited, and updates the distances to its neighboring nodes.

To find the shortest path, not just the distance, a modification to the algorithm is needed. An additional array, prev, is used to store the previous node in the shortest path to each node. When a new shortest path is found, the prev array is updated accordingly.

The shortest path can be reconstructed by following the prev array from the destination node to the source node. This can be implemented using recursion. The updated source code, dijkstraWithPath.c, includes this modification.

The graph is represented as an edge list, where the first line contains the number of edges, and the following lines contain three numbers: u, v, and w, signifying an edge from u to v with weight w. The sample input file, dist.txt, provides an example of this representation.

The code includes a function, printD, to print the distances from the source node to all other nodes, and a function, printPath, to print the shortest path from the source node to a given destination node.

The time complexity of Dijkstra's algorithm can vary from O(n*lg(lg(n))) to O(n^2), depending on the implementation. The space complexity is O(n), where n is the number of nodes in the graph.

In summary, Dijkstra's algorithm is a widely used solution for finding the shortest path in a graph. It works by iteratively selecting the node with the smallest distance and updating the distances to its neighboring nodes. With a modification to store the previous node in the shortest path, the algorithm can also be used to find the shortest path, not just the distance.

#### The Algorithm

Dijkstra’s algorithm is one of the most common solutions to this problem. Even so, it only works on graphs which have **no edges of negative weight**, and the actual speed of the algorithm can vary from **O(n*lg(lg(n)))** to **O(n^2)**.

The idea is somewhat simple:

Take the length of the shortest path to all nodes to be infinity. Mark the length of the shortest path to the source as *0*.

Now, we already know that the graph has no edges of negative weight so the a path of length *0* is the best we can come up with. The path to the source is *0*, so it’s optimal.

This algorithm works by making the paths to one more node optimal at each step. So, at the *k*th step, you know for sure that there are at least *k* nodes to which you know the shortest path.

At each step, choose the node, which is not yet optimal, but which is closest to the source; i.e. the node to which the current calculated shortest path is smallest. Then, from it, try to optimise the path to every node connected to it. Finally, mark the said node as *optimal* (visited, if you prefer). In the previous example, the node which is closest to the source and is not yet optimal is the source. From it, you can optimise the path to nodes *2* and *4*.

At this point, the only visited/optimal node is *0*. Now we have to redo this step *4* more times (to ensure that all nodes are optimal).

The next node to consider is *4*:

It’s worthwhile to note that at this step, we’ve also found a better path to node *2*.

Next is node *2*:

Finally, we look at nodes *5* and *3* (none of which offer any optimisations):

#### The Programme

The actual code in C looks something like this:

```c
void dijkstra(int s) {
    int i, k, mini;
    int visited[GRAPHSIZE];
    for (i = 1; i <= n; ++i) {
        d[i] = INFINITY;
        visited[i] = 0; /* the i-th element has not yet been visited */
    }
    d[s] = 0;
    for (k = 1; k <= n; ++k) {
        mini = -1;
        for (i = 1; i <= n; ++i)
            if (!visited[i] && ((mini == -1) || (d[i] < d[mini])))
                mini = i;
        visited[mini] = 1;
        for (i = 1; i <= n; ++i)
            if (dist[mini][i])
                if (d[mini] + dist[mini][i] < d[i])
                    d[i] = d[mini] + dist[mini][i];
    }
}
```

#### Finding the shortest path

To find the shortest path, not just the distance, a modification to the algorithm is needed. An additional array, prev, is used to store the previous node in the shortest path to each node. When a new shortest path is found, the prev array is updated accordingly.

The shortest path can be reconstructed by following the prev array from the destination node to the source node. This can be implemented using recursion.

```c
void printPath(int dest) {
    if (prev[dest] != -1)
        printPath(prev[dest]);
    printf("%d ", dest);
}
```

The graph is represented as an edge list, where the first line contains the number of edges, and the following lines contain three numbers: u, v, and w, signifying an edge from u to v with weight w.

The code includes a function, printD, to print the distances from the source node to all other nodes, and a function, printPath, to print the shortest path from the source node to a given destination node.

The time complexity of Dijkstra's algorithm can vary from O(n*lg(lg(n))) to O(n^2), depending on the implementation. The space complexity is O(n), where n is the number of nodes in the graph.

The sample input file, dist.txt, provides an example of the graph representation.

```c
10
1 2 10
1 4 5
2 3 1
2 4 3
3 5 6
4 2 2
4 3 9
4 5 2
5 1 7
5 3 4
```


To find the area of the region enclosed by the curves Y=x^3+5x^2+3 and y=x^2+7x+3, within the domain X=0 to x=3, we first need to determine where these two curves intersect.

The intersection points can be found by setting the two equations equal to each other:
x^3 + 5x^2 + 3 = x^2 + 7x + 3

Simplifying this equation gives:
x^3 + 4x^2 - 7x = 0

Factoring out x:
x(x^2 + 4x - 7) = 0

This yields two possible values for x when the curves intersect within the given domain: x = 0 and x = -2 ± √11. Since x = -2 - √11 is outside our domain of interest (x = 0 to x = 3), we consider the intersection at x = 0 and x = -2 + √11.

Given that there is an intersection within our domain, we split the area into two regions:
1. From x = 0 to x = -2 + √11
2. From x = -2 + √11 to x = 3

In the first region, the parabola (y = x^2 + 7x + 3) is above the cubic (Y = x^3 + 5x^2 + 3), so the effective height is the difference between the two:
(-x^3 - 4x^2 + 7x)

In the second region, the cubic is above the parabola, so the effective height is:
(x^3 + 4x^2 - 7x)

The total area enclosed is the sum of the integrals of these heights over their respective regions:
= ∫[-x^3 - 4x^2 + 7x] from 0 to -2 + √11 + ∫[x^3 + 4x^2 - 7x] from -2 + √11 to 3

Evaluating these integrals:
= [(-1/4)x^4 - (4/3)x^3 + (7/2)x^2] from 0 to -2 + √11 + [(1/4)x^4 + (4/3)x^3 - (7/2)x^2] from -2 + √11 to 3

The calculation of the exact area requires evaluating these expressions at the given limits, which involves substantial arithmetic.


For the function y = -(x-3)^5(x+1)^4, key facts and details include:
- Degree: 9
- Sign: negative
- Quadrants: I, II, and IV (not 4, as quadrants are labeled I, II, III, and IV)
- Roots: x = 3 and x = -1
- y-intercept: 243
- Domain: x belongs to all real numbers (denoted as R or ℝ, not E)
- Range: y belongs to all real numbers (denoted as R or ℝ, not E)

The function's end behaviors are:
- As x approaches -infinity, y approaches infinity
- As x approaches infinity, y approaches -infinity

To find the intervals of increase and decrease without using derivatives, analyze the behavior of the function around its roots and as it approaches infinity. Since the function is negative and has an odd degree, its end behaviors are as described. The roots divide the real number line into intervals where the function is either increasing or decreasing.

Given roots at x = -1 and x = 3, the intervals to consider are:
- (-∞, -1)
- (-1, 3)
- (3, ∞)

Considering the function's negative sign and degree, and without calculus, we deduce the intervals based on the function's behavior:
- The function increases in the interval (-1, 3) because, moving from left to right, it goes from negative to positive values (considering the roots and the negative sign in front of the function).
- The function decreases in the intervals (-∞, -1) and (3, ∞) because, moving from left to right, it goes from positive to negative values and then continues to decrease as x approaches infinity due to the odd degree and negative leading coefficient.

Thus, the intervals of increase are (-1, 3), and the intervals of decrease are (-∞, -1) and (3, ∞).


PRINCIPLES OF MEASUREMENT

When adding or subtracting decimals with different numbers of decimal places, it is essential to consider the origin of the numbers. Finite decimals, such as 0.5 or 0.125, can be added or subtracted without rounding off, as they have a definite fixed value. For example, 0.5 + 0.125 can be added by annexing zeros to 0.5 to match the number of decimal places in 0.125, resulting in 0.500 + 0.125.

However, when dealing with measurements, the precision and accuracy of the measurements must be considered. The probable error in a measurement depends on the smallest division on the measuring instrument. For instance, a measurement made with an instrument marked off in tenths of an inch has a maximum probable error of 0.05 inch, while a measurement made with an instrument marked in thousandths of an inch has a probable error of 0.0005 inch.

PRECISION

The precision of a measurement depends on the number of decimal places, with a larger number of decimal places indicating a smaller probable error. For example, 2.3 inches has a probable error of 0.05 inch, while 1.426 inches has a probable error of 0.0005 inch. When adding or subtracting numbers with different precisions, the result should be rounded off to the precision of the least precise number. In the case of 2.300 + 1.426, the result should be rounded off to tenths, resulting in 3.7 inches.

ESTIMATION

Measurements involve estimation between marks on the measuring instrument, which is subject to human error. The best estimate that can be made is whether a measurement is more or less than halfway between marks. This is reflected in the probable error, which is one-half the size of the smallest division on the measuring instrument.

ACCURACY

Accuracy depends on the relative size of the probable error compared to the quantity being measured. A measurement can be precise but not accurate, or vice versa. For example, a distance of 25 yards measured to the nearest inch has a probable error of 0.5 inch, which is relatively small compared to the total distance. In contrast, a length of pipe measured to be 3.2 inches long has a probable error of 0.05 inch, which is relatively large compared to the total length.

PRACTICE PROBLEMS

1. Determine which number is more accurate and which is more precise: 3.72 inches or 2,417 feet
Answer: 3.72 inches is more precise.

Note: The location of the decimal point does not affect the accuracy of a number. For example, 1.25 dollars and 125 cents represent the same amount of money and are equally accurate.


# Maxwell Equations

Maxwell's equations are a set of fundamental equations in classical electromagnetism that describe how electric and magnetic fields interact and how they are generated by charges and currents. In CGS units, Maxwell's equations are written as:

\begin{eqnarray}
\nabla\cdot E &= 4\pi\rho\\
\nabla\cdot B &= 0\\
\nabla\times E &= -\frac1c\frac{\partial B}{\partial t}\\
\nabla\times B &= \frac{4\pi}{c}J + \frac1c\frac{\partial E}{\partial t}\\
\end{eqnarray}

where $E$ is the electric field, $B$ is the magnetic field, $J$ is the current, and $\rho$ is the charge density. These equations state that:

1. Electrical charge creates a divergence of the electric field.
2. Magnetic fields do not diverge, and there are no magnetic "charges".
3. A magnetic field that changes with time creates a circular electric potential.
4. A moving electrical charge or an electric field that changes with time creates a circular magnetic potential.

## Maxwell's Equations in Free Space

In free space, where $\rho=0$ and $J=0$, Maxwell's equations simplify to:

\begin{eqnarray}
\nabla\cdot E &= 0,\\
\nabla\cdot B &= 0,\\
\nabla\times E &= -\frac1c\frac{\partial B}{\partial t},\\
\nabla\times B &= \frac1c\frac{\partial E}{\partial t}.\\
\end{eqnarray}

Using the vector identity $\nabla\times(\nabla\times F) = \nabla(\nabla\cdot F)-\nabla^2F$, the curl of the last two Maxwell's equations reduces to the wave equations:

\begin{eqnarray}
\nabla^2E -\frac1{c^2}\frac{\partial^2 E}{\partial t^2} &= 0,\\
\nabla^2B - \frac1{c^2}\frac{\partial^2 B}{\partial t^2} &=0.\\
\end{eqnarray}

## A One-Dimensional Wave Equation

Along one dimension, the wave equation for the electric field is written as:

\begin{equation}
\frac{\partial^2 E}{\partial x^2} - \frac1{c^2}\frac{\partial^2 E}{\partial t^2} = 0.
\end{equation}

Let's suppose that $E(x)=E(x\pm ct)$, and we define a new variable $u\equiv x\pm ct$. The Chain Rule tells us that:

\begin{equation}
\frac{\partial E(u)}{\partial x}=\frac{\partial E}{\partial u}\frac{\partial u}{\partial x}=\frac{\partial E}{\partial u}.
\end{equation}

Similarly, for the time derivative, we have:

\begin{equation}
\frac{\partial E(u)}{\partial t}=\frac{\partial E}{\partial u}\frac{\partial u}{\partial t}=\pm\frac1c\frac{\partial E}{\partial u}.
\end{equation}

Based on these identities, we have:

\begin{eqnarray}
\frac{\partial^2 E}{\partial x^2}&=\frac{\partial^2 E}{\partial u^2},\\
\frac{\partial^2 E}{\partial t^2}&=c^2\frac{\partial^2 E}{\partial u^2}.\\
\end{eqnarray}

If you plug these into the wave equation, you'll see that it solves the equation. In the general 3D case for $E$ and $B$, this means that any electromagnetic waveform that translates at speed $c$ solves these equations.

## The Relative Orientation of $E$ and $B$

From the previous section, we see that in free space, the Maxwell equations give us two wave equations. One propagates the $E$-field at the speed of light, and the other propagates the $B$-field at the speed of light. To relate the $E$ and $B$ fields to one another, we need to re-examine the Maxwell equations for free space.

Let's suppose that $E$ and $B$ are both described by sine waves of the same frequency, but with some arbitrary phase relative to one another. Since $\nabla\times E = -\frac1c\frac{\partial B}{\partial t}$, it must be true that $\frac{\partial B}{\partial t}$ is maximized at $x=0,t=0$. By this argument, $E$ and $B$ must be in-phase with one another. Moreover, because $\nabla\times \vec E$ is perpendicular to $\vec E$ and perpendicular to the spatial gradient of $\vec E$, $\vec B$ must be perpendicular to $\vec E$ and perpendicular to the direction the wave propagates. Finally, if you work out the spatial and temporal derivatives at $x=0,t=0$, you'll see that $E_0=B_0$.


Machines and Movement Ratio
A machine is a device that changes the magnitude and line of action of a force, with examples including pulley systems, gear systems, and screwjacks. A simple machine amplifies the input force (effort) to a larger output force (load).

The movement ratio (or velocity ratio) of a machine is defined as the ratio of the distance moved by the effort to the distance moved by the load, expressed as:
Mₑ = sₑ / sₗ
where Mₑ = movement ratio, sₑ = distance moved by the effort, and sₗ = distance moved by the load.

Machine Efficiency
The efficiency of a simple machine is defined as the ratio of the force ratio to the movement ratio, expressed as:
μ = Fₑ / Mₑ
where μ = machine efficiency and Fₑ = force ratio.

Key concepts related to machines and movement ratio include dynamics, mechanics, and ratios and proportions. Dynamics involves the motion of bodies and the action of forces, while mechanics explores the relationships between forces, acceleration, displacement, and energy. Ratios and proportions are essential in understanding the relative values between quantities, such as force ratio and movement ratio.

Related topics and documents include:
- ConnRod Mechanism: The connecting rod mechanism
- Force Ratio: The load force versus the effort force
- Motors Speed vs. Pulley Diameter: Pulley selection charts for motors
- Pulleys: Pulleys, blocks, and tackles
- Screw Jack Effort Force vs. Load: Screw jacks and effort forces
- Toggle Joint: A toggle joint mechanism for multiplying force.


Solving geometry word problems often involves translating descriptions into graphical representations. A reader-submitted example illustrates this challenge: a nuclear power plant malfunction contaminated a square kilometer area, making it uninhabitable within one kilometer of any point in the contaminated area.

## Problem

The area around a nuclear power plant, measuring a square kilometer, was contaminated with radioactive waste. Due to radiation, no one can live within one kilometer of any point in the contaminated area. The goal is to find the total uninhabitable area, including the interior of the square.

## Strategy

To solve geometry word problems, translate the description into a graphical representation. If the description is ambiguous, make assumptions and note them. In this case, the plant area's shape is initially unclear, but the last sentence clarifies it as a square. Assuming a square shape, we can proceed.

Given the square plant area has an area of 1 square kilometer, we use the formula for a square's area, A_{square}=side^{2}, to find each side measures 1 km.

## Step 1

Determine the areas within one kilometer of any point in the contaminated area. Focus on the square's edges, as every point inside is farther from an outside point than the edge. The farthest point from each side within 1 km is 1 km from the edge. Drawing a parallel line 1 km away from the edge, all points between this line and the edge are within 1 km and part of the uninhabitable area, forming another square.

## Step 2

There are four such squares, one on each edge of the original plant site, making a total of five squares, each measuring 1 square kilometer.

## Step 3

Consider the space between the squares, where points are also within 1 kilometer of the original square, specifically near the corners. A circle is defined as points equidistant from a center. Using the corner as the center of a circle with a 1 km radius, the area between the squares includes a quarter of a circle. All points within this quarter-circle are part of the uninhabitable area.

There are four identical quarter-circles, totaling the area of one full circle with a radius of 1 km: A_{circle}=π*r^{2}, with r=1, resulting in an area of π.

The total uninhabitable area is the sum of the original square, the four additional squares, and the area of the four quarter-circles: 1 (original square) + 4x1 (squares on each edge) + π = 5 + π square kilometers.


The problem involves finding the minimum number of moves to make a binary string valid, where a string is valid if there is equal spacing of length K between 1s. To solve this, we can analyze the numbers of 0s and 1s for each residue mod K.

We start by precomputing the total number of 1s and the number of 1s for each residue mod K. This can be done in O(N) time, where N is the length of the string. We then iterate over each residue and calculate the number of moves required to turn off the bits at positions not congruent to the current residue mod K.

For each residue, we reduce the problem to finding the minimum number of moves to make a subsequence of bits valid, where the subsequence consists of the bits at positions congruent to the current residue mod K. This subsequence must form a pattern of 0*1*0*, where * denotes zero or more occurrences of the preceding symbol.

To solve this subproblem, we can use dynamic programming. We define three functions:

- f(pos) = minimum moves to make the suffix [pos, N) valid
- g(pos, state: {0, 1}) = minimum moves to make suffix [pos, N) valid such that bit at pos is state
- h(pos, state) = minimum moves to make the suffix [pos, N) have the pattern 1*0*

We can then use these functions to calculate the minimum number of moves required to make the subsequence valid.

The final solution involves iterating over each residue, calculating the number of moves required to turn off the bits at positions not congruent to the current residue mod K, and then using dynamic programming to find the minimum number of moves required to make the subsequence valid.

The code for the solution is as follows:

```cpp
i64 solve(const i64 N, const i64 K, string_view S) {
    i64 total_on = 0;
    vec<i64> resid_on(K, 0);
    for (i64 i = 0; i < N; ++i) {
        total_on += S[i] == '1';
        resid_on[i % K] += S[i] == '1';
    }
    i64 best = N;
    for (i64 resid = 0; resid < K; ++resid) {
        let i64 turn_non_resid_off = total_on - resid_on[resid];
        vec<bool> seq;
        for (i64 i = resid; i < N; i += K) {
            seq.push_back(S[i] == '1');
        }
        i64 subproblem_ans = solve_resid(seq);
        best = min(best, turn_non_resid_off + subproblem_ans);
    }
    return best;
}

i64 solve_resid(const vec<bool>& seq) {
    vec<vec<i64>> h(seq.size() + 1, vec<i64>(2));
    for (i64 i = seq.size() - 1; i >= 0; --i) {
        h[i][0] = seq[i] + h[i + 1][0];
        h[i][1] = !seq[i] + min(h[i + 1][0], h[i + 1][1]);
    }
    vec<vec<i64>> g(seq.size() + 1, vec<i64>(2));
    for (i64 i = seq.size() - 1; i >= 0; --i) {
        g[i][0] = seq[i] + min(g[i + 1][0], g[i + 1][1]);
        g[i][1] = !seq[i] + min(h[i + 1][0], h[i + 1][1]);
    }
    return min(g.front()[0], g.front()[1]);
}
```

This solution has a time complexity of O(N) and a space complexity of O(N), where N is the length of the input string.


Teaching the Properties of Multiplication is a crucial step in the learning process. It starts with understanding what multiplication is and moves into learning the math facts, two-digit and three-digit multiplication, and finally, the properties of multiplication. The properties of multiplication are essential for equipping children to solve numerous expressions and build a foundation for algebra.

There are three key properties of multiplication: the Commutative Property, the Associative Property, and the Distributive Property.

**Commutative Property of Multiplication**: This property states that the order of the numbers being multiplied does not change the result. For example, 3 x 4 = 4 x 3. To help children understand this concept, they can complete a cut and paste activity where they find the multiplication expression that makes the equation true.

**Associative Property of Multiplication**: This property states that when multiplying three or more numbers, the order in which the numbers are multiplied does not change the result. For example, (2 x 3) x 4 = 2 x (3 x 4). There are two cut and paste activity pages for the associative property, one with problems like 4 x 6 = _______ and the other with expressions where children can use the associative property to solve.

**Distributive Property**: This property states that a single operation can be distributed across multiple values. For example, 5 (3 + 2) can be solved by multiplying five times 3 and then five times 2 and adding those quantities together, or by adding up the addends first and then multiplying. Using an area model can help children visualize this concept. There are two cut and paste activities for the distributive property, where children make the equations true by finding equivalent expressions.

To help children learn these properties, it's essential to provide them with visual models and hands-on activities. The cut and paste multiplication properties pages are a great resource, requiring minimal preparation. Children can work on these activities by printing out the pages, using glue and scissors, and completing the tasks. By mastering the properties of multiplication, children will be better equipped to solve complex math problems and build a strong foundation for future math concepts.


SWBAT solve unit rate problems by creating and extending a table or double number line diagram. Knowing the unit rate of a ratio relationship allows for easy identification of equivalent ratios with any whole number value.

The lesson begins with a 3-4 minute Think About It problem, where students work in pairs to create either a double number line or a ratio table. A student's work is then shared with the class, and the solution is discussed. If the student's solution is incorrect, the class works together to find the correct answer.

The Guided Practice section involves modeling problems using double number lines and ratio tables. Two problems are presented: problem A uses fractions, while problem B uses decimals. This provides students with examples of both methods.

During the 15-minute Partner Practice, students work in pairs on a problem set. The teacher circulates around the room, observing students' work and asking questions such as:
- How did you figure out that equivalent ratio?
- How can these displays help you answer ratio questions?
- What's the unit rate?
- How did finding the unit rate help you to figure out the equivalent ratio?

The teacher also checks for correct models, arithmetic, and answers. After 10 minutes, students complete a check for understanding independently, and then one student shares their work on the document camera.

The Independent Practice problem set is then introduced, and students have 20 minutes to work on it. The teacher encourages students to annotate the problems and circle the unit that the problem gives for the equivalent fraction. If students finish quickly, they can create both models for the problems.

After independent work time, the class discusses problem 4a from the independent practice. Students vote on whether Daniel's line of thinking is correct or incorrect and then share their reasoning with their partner. A few students share their thinking with the class, discussing concepts such as unit rates, scale factor, number sense, or partitioning.

The lesson concludes with an Exit Ticket, which students complete independently.


## Follow the Numbers

To start, apply a simple rule: add the digits of a number together, then multiply by 2.

Begin with the number 56:
- Add the digits: 5 + 6 = 11
- Multiply by 2: 11 * 2 = 22

Continue with 22:
- Add the digits: 2 + 2 = 4
- Multiply by 2: 4 * 2 = 8

Next, with 8:
- Add the digits: 8 + 0 = 8
- Multiply by 2: 8 * 2 = 16

Then, with 16:
- Add the digits: 1 + 6 = 7
- Multiply by 2: 7 * 2 = 14

Applying the rule to 14 yields 10, and continuing this process results in a sequence: 56, 22, 8, 16, 14, 10, 2, 4, 8, 16, 14, 10, 2, 4, 8, ...

Starting with 11 gives a similar sequence: 11, 4, 8, 16, 14, 10, 2, 4, 8, ...

Starting with 96 yields a different sequence: 96, 30, 6, 12, 6, 12, 6, ...

Exploring further, using each number from 1 to 99 as a starting point and trying different rules, reveals various patterns and cycles.

To investigate:
1. Choose a starting number.
2. Apply the rule: add the digits, then multiply by 2.
3. Record the sequence.
4. Try different starting numbers and rules.
5. Compare sequences for similarities and differences.

Questions to consider:
- What happens when you start with different numbers?
- How do different rules affect the sequences?
- What patterns or cycles emerge?

Activity suggestions:
- Work in pairs or small groups to explore sequences.
- Record findings and post them for class comparison.
- Use digit cards for easier recording.
- Utilize software to visualize and analyze sequences.
- Derive hypotheses and test them.

By systematically exploring the sequences starting from each number from 1 to 99 and comparing results, pupils can identify patterns, make predictions, and develop hypotheses about the sequences.


## Finding an Original Value

Lesson 9 of 15

## Objective:
Students will be able to find an original (or whole) value given a part and percent by scaling values in a table.

### Introduction (5 minutes)

The lesson begins with the essential question: How can you find an original value when given the part and the percent? A sample problem is presented: "The donor gave $20, which was 10% of their total pledge. How much did the donor pledge in all?" Common errors and misconceptions are addressed, and a bar model is used to illustrate the solution. The total pledge is calculated to be $200.

### Guided Problem Solving (15 minutes)

Students learn the 3-step method to solve problems and work through 5 examples using a table. Calculators are available to speed up computations. Anticipated challenges include recalling that a 40% discount means the sale price is 60% of the original cost and understanding how to scale values. A bar model is used to support student understanding.

### Independent Problem Solving (20 minutes)

Students work independently on a set of problems, including finding an original amount given a total with tax based on a discount and analyzing data presented in a circle graph. The first 5 problems are similar to those in the guided practice section.

### Exit Ticket (5 minutes)

The 3-step method is reviewed, and students complete 5 problems similar to those in the guided and independent practice sections. A successful exit ticket requires at least 4 correct answers.

### Similar Lessons

This lesson is part of a series on percents, including:
- LESSON 1: Multiple Representations of Percents
- LESSON 2: The 10% Benchmark
- LESSON 3: The 1% Benchmark
- LESSON 4: Percent Benchmark Fluency
- LESSON 5: Drawing Bar Models to Represent Percents of Increase and Decrease
- LESSON 6: Solve Problems by Applying Percents of Increase and Decrease
- LESSON 7: Discounts and Sales Tax
- LESSON 8: Finding a Percent of Change
- LESSON 9: Finding an Original Value
- LESSON 10: A Percent Equation
- LESSON 11: Expressions for Percent Increases and Decreases
- LESSON 12: Simple Interest
- LESSON 13: Increasing and Decreasing Quantities by a Percent (Day 1 of 2)
- LESSON 14: Increasing and Decreasing Quantities by a Percent (Day 2 of 2)
- LESSON 15: Percent Assessment


To understand why any linear network seen from a port is equivalent to a linear bipole, consider a network divided into two parts. The Thevenin theorem states that this network can be simplified into a single voltage source (Vth) and a series resistor (Rth) connected to the load.

The process of finding Rth involves several steps. First, remove the load and calculate the open-circuit voltage (Voc) across the terminals. Then, short-circuit the terminals and calculate the short-circuit current (Isc). Rth can be found using the formula Rth = Voc / Isc.

For a circuit with a dependent source, finding Thevenin voltage and impedance requires using nodal analysis or mesh analysis to calculate Voc and Isc. The Thevenin voltage (Vth) is equal to Voc, and the Thevenin impedance (Rth) can be calculated using the formula Rth = Vth / Isc.

In a linear circuit with a current source, Thevenin's theorem can be applied by first finding the open-circuit voltage and short-circuit current across the terminals of interest. Then, use the formula Rth = Voc / Isc to calculate Rth.

To find the current through a resistor X in a circuit, first, find the Thevenin equivalent circuit across the terminals of resistor X. Then, use Ohm's law to calculate the current through resistor X.

Key equations:
- Rth = Voc / Isc
- Vth = Voc
- I = Vth / Rth

Important concepts:
- Thevenin theorem
- Linear bipole
- Open-circuit voltage (Voc)
- Short-circuit current (Isc)
- Nodal analysis
- Mesh analysis
- Dependent sources
- Thevenin equivalent circuit

Given a linear circuit with a current I in a 0.9 kΩ resistor and Vs = 100V, to predict I for different values of Vs, use the proportionality property. To find I in terms of Vs using Thevenin's theorem, calculate Vth and Rth across the terminals of interest, then use the formula I = Vth / (Rth + RL), where RL is the load resistance.

For a circuit with a voltage source V = 7V, to find the current through resistor X, first find the Thevenin equivalent circuit across the terminals of resistor X. Then, use Ohm's law to calculate the current through resistor X.

Thevenin's theorem can be applied to various circuits, including those with dependent sources, to simplify the analysis and find the current through a specific resistor or the voltage across a specific terminal. By following the steps outlined above and using the key equations and concepts, one can effectively apply Thevenin's theorem to solve circuit analysis problems.


## Step 1: Analyzing the provided text to extract key facts, concrete details, important numbers, and core concepts.
The text provided is a comprehensive collection of problems and solutions related to rational numbers, covering various operations such as addition, subtraction, multiplication, and division. It also includes exercises on finding rational numbers between given numbers and understanding the properties of rational numbers.

## Step 2: Removing any irrelevant or noisy information and reorganizing the content into a logically structured, information-dense, and concise version.
Upon reviewing the text, it's clear that the content is already well-structured around specific topics and exercises. However, to refine it further, we can categorize the exercises by topic (e.g., addition, subtraction, multiplication, division, and properties of rational numbers) and ensure that each section flows logically.

## Step 3: Maintaining the original length as much as possible (avoid excessive shortening).
Given the nature of the content, which is primarily composed of mathematical exercises and their solutions, maintaining the original length is crucial for preserving the educational value of the material. Each exercise and its solution contribute to understanding and applying the concepts of rational numbers.

## Step 4: Refining multiple-choice questions and answers if any.
There are no multiple-choice questions provided in the text. The content consists of mathematical problems with detailed solutions, which do not require refinement in the context of multiple-choice questions.

The final answer is: $\boxed{1}$


“1/4 Cup To Metric” refers to the process of converting measurements from the US customary system to the metric system, specifically the conversion of a quarter cup measurement commonly used in cooking and baking recipes. This conversion is significant for those who follow recipes that use metric measurements or reside in countries that use the metric system. Knowing how to convert these measurements accurately is crucial in achieving precise and successful results in recipe preparation.

1/4 cup is equal to 4 tablespoons or 2 fluid ounces in the US customary measurement system, and approximately 59 milliliters in the metric system. To convert 1/4 cup to milliliters, we need to use a conversion factor since the US customary system uses cups while the metric system uses milliliters as a unit of measurement. One-fourth cup is equivalent to 59.15 milliliters, which is obtained by multiplying the volume of one US fluid ounce (29.57 ml) by 2, and rounding off the result to two decimal places.

In Australia, the metric system is the standard system of measurement used for cooking and baking. A standard cup is equal to 250 milliliters (ml). To convert 1/4 cup to milliliters in Australia, you can use the following formula: 1/4 cup = 1/4 x 250 ml = 62.5 ml. However, it’s essential to note that measuring ingredients by weight rather than volume can be more accurate, especially for baking.

The UK also uses the metric system for cooking and baking, so converting from 1/4 cup to ml is straightforward. In the UK, one cup is equal to 250 milliliters (ml). Therefore, to convert 1/4 cup to ml, you can use the formula: 1/4 cup = 62.5 ml. This conversion can be useful when following recipes that use either cups or milliliters as a unit of measurement.

In the USA, customary units of measurement are used for cooking and baking, and a cup is a common unit of measurement. One cup in the US is equal to 236.59 milliliters (ml). To convert 1/4 cup to ml in the US, you can use the formula: 1/4 cup = 59.15 ml. It’s essential to note that when measuring ingredients, using weight measurements like grams or ounces can often be more precise than using volume measurements like cups, especially when it comes to dry ingredients.

To convert 1/4 cup to ml, you need to convert from the imperial system to the metric system. In the imperial system, 1 cup is equal to 8 fluid ounces (fl oz), and 1 fl oz is approximately equal to 28.41 milliliters (ml). To convert 1/4 cup to ml, you can use the following formula: 1/4 cup = 2 fl oz = 56.82 ml. However, it’s essential to note that when using volume measurements like cups or fluid ounces, using the correct size measuring cup or container is crucial for achieving accurate results.

When converting 1/4 cup to ml of water, the answer will be the same regardless of the country’s standard system of measurement. This is because the density of water is very close to 1 gram per milliliter, making it a convenient substance to measure by volume. Therefore, 1/4 cup of water is equivalent to 59.15 ml. It’s essential to note that the density of other liquids may differ from water, so the conversion factor may vary for them.

1 cup of water is equivalent to 236.59 ml in any system of measurement. The conversion factor for 1/4 cup of liquid from the imperial to the metric system depends on the specific type of liquid being measured. Assuming the liquid has a density similar to water, 1/4 cup of liquid would be approximately equal to 59.15 ml.

To convert 1/2 cup to ml, you need to know the conversion factor between cups and milliliters. One cup is equal to 236.6 milliliters (ml), so we can use this conversion factor to convert 1/2 cup to ml. Therefore, 1/2 cup is equal to 118.3 milliliters (ml) approximately.

To convert 1/3 cup to ml, we can use the conversion factor of 1 cup = 236.6 ml. Multiplying 236.6 ml by 1/3 gives us 78.79 ml, which is the equivalent volume of 1/3 cup in milliliters. 1/3 cup of water is approximately equal to 78.79 ml.

If you have 3/4 cup of a substance and you want to know its equivalent in milliliters (ml), it is approximately equal to 177.44 ml. 1 1/4 cup is equivalent to 10 fluid ounces or 300 milliliters. Half of 3/4 cup is equal to 3/8 cup.

To convert 2 cups to milliliters, we can use a conversion factor of 236.6 ml per cup. Therefore, 2 cups can be calculated by multiplying 236.6 ml by 2, which gives us 473.2 ml. To convert 2/3 cup to milliliters, we can use the conversion factor of 1 cup = 236.6 ml. Multiplying 236.6 ml by 2/3 gives us 157.7 ml, which is the equivalent volume of 2/3 cup in milliliters.

In conclusion, to convert 1/4 cup to the metric system, we can use a conversion factor of 1 cup = 236.6 milliliters. Multiplying 236.6 ml by 1/4 gives us 59.15 ml, which is the equivalent volume of 1/4 cup in milliliters. This conversion is useful when measuring liquids in recipes or when using measuring cups that are marked in cups rather than milliliters.


## NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.3

NCERT Solutions for class 10 Maths Chapter 3 Exercise 3.3 are available in Hindi Medium and English Medium, or can be viewed in Video Format. The solutions have been updated according to the latest CBSE Syllabus 2019-20 for CBSE Board, Gujarat Board, MP Board, and UP Board schools, based on NCERT Books.

### Class 10 Maths Chapter 3 Exercise 3.3 Solutions

Solutions for Class 10 Maths Chapter 3 Exercise 3.3 are provided in English Medium and Hindi Medium. PDF solutions of all chapters can be downloaded.

### Important Questions Based on Linear Equations for Practice

1. Form a pair of linear equations for the given conditions and solve them.
   - The sum of the numerator and denominator of a fraction is 3 less than twice the denominator.
   - If the numerator and denominator are both decreased by 1, the numerator becomes half the denominator.
   Answer: x - y = -3, 2x - y = 1

2. A jeweler has bars of 18 carat gold and 12 carat gold. Determine how much of each must be melted together to obtain a 120g bar of 16 carat gold, given that pure gold is 24 carat.
   Answer: 80g of 18 carat gold and 40g of 12 carat gold

3. Find the value of p for which the pair of linear equations has a unique solution:
   (p + 2)x - (2p + 1)y = 3(2p - 1) and 2x - 3y = 7
   Answer: p ≠ 4

4. The sum of the prices of a sofa and a table is ₹2340, and the difference of their prices is ₹140. Find the price of each.
   Answer: Sofa = ₹1240, Table = ₹1100

5. 60 copies of mathematics books and 40 copies of English books cost ₹2220. Also, 40 copies of mathematics books and 50 copies of English books cost ₹1900. Find the cost of one mathematics book and one English book.
   Answer: Mathematics book = ₹25, English book = ₹18

### Additional Practice Questions

1. Solve for x and y: 3x + 2y = 11 and 2x + 3y = 4. Also, find p if p = 8x + 5y.
   Answer: x = 5, y = -2, p = 30

2. If three times the greater of two numbers is divided by the lesser, the quotient is 6 with a remainder of 6. If five times the lesser number is divided by the greater, the quotient is 2 with a remainder of 3. Find the two numbers.
   Answer: 16, 7

3. Solve the pair of linear equations by the substitution method: x - 7y + 42 = 0 and x - 3y - 6 = 0.
   Answer: Not provided

4. Ram is walking along the line joining (1, 4) and (0, 6). Rahim is walking along the line joining (3, 4) and (1, 0). Represent this on a graph and find the point where both of them cross each other.
   Answer: (2, 2)


## Real World Applications of 3D Measurement

### Proportional Reasoning With Volume of a Cylinder and Sphere

This 3-act math task helps students develop **proportional reasoning** and **3D-measurement** skills by determining the number of packages of gumballs and individual gumballs needed to fill a cylindrical jar. The learning goals include:

- Calculating the volume of a cylinder and applying this knowledge
- Calculating the volume of a sphere and applying this knowledge
- Applying proportional reasoning to solve problems

## Act 1: Introduce The Problem

Students are shown a video or photo of the problem and asked to discuss possible questions with a neighbor. The task focuses on two questions:

1. How many packages of gumballs will it take to fill the jar?
2. How many individual gumballs will it take to fill the jar?

## Act 2: Provide Information

After making initial predictions, students discuss what information they need to make more accurate predictions. They are then shown videos or photos providing key measurements:

- Jar height
- Jar diameter
- Gumball diameter

With this information, students can calculate the volume of the jar and a single gumball to improve their predictions. Additionally, they are told the average number of gumballs per package.

## Act 3: Reveal the Solution

After sharing their work and updating predictions, students are shown videos revealing the solution. This act encourages debate and refinement of their calculations.

## Resources

- Entire Lesson Slide Deck (incl. Sequels) [KEYNOTE]
- Act 1, Scene 1 [VIDEO]
- Act 1, Scene 2 [VIDEO]
- Act 2 [VIDEO]
- Act 3, Scene 1 [VIDEO]
- Act 3, Scene 2 [VIDEO]

## Implementation and Feedback

Teachers are encouraged to share their experiences with this task in the comments section, including any improvements or additional resources.

## Downloading the Task and Resources

All videos, images, and resources can be downloaded for use in the classroom.

## Introduction to 3 Act Math Tasks

For those new to 3 Act Math Tasks, a 2-page printable tip sheet is available to ensure a successful start in sparking curiosity and fueling sense-making in mathematics.

## Accessing More Resources

Additional real-world math tasks can be found by searching the database, which includes tasks for various grades and topics.


The SQRT Function in Excel is a type of Mathematical Function used to find the square root of a number. It can be utilized in various mathematical formulas and concepts.

**Application 1 – SQRT Returning Array of Results (Excel 365 only)**

In Excel 365 or Excel 2021, the SQRT Function can be used as a dynamic array formula, allowing it to work on an array of cells. For a range of cells containing numbers, such as A2:A10, the formula =SQRT(A2:A10) can be used in cell B2. This will spill the results, returning a range of cells B2:B10 containing the square roots of the input array.

**Application 2 – Distance Formula to Find Distance Between Two Points**

The distance formula calculates the distance between two points (x1, y1) and (x2, y2) on the Cartesian plane. The general equation is:

Given pairs of x and y coordinates in columns A, B, C, and D, the formula =SQRT((C2-A2)^2+(D2-B2)^2) can be used in cell E2 to find the distance between (X1, Y1) and (X2, Y2). The caret symbol (^) gives the power of a number and can be used in place of the POWER function. Selecting the range E2:E10 and hitting Ctrl+D will copy the formula down, returning the distance between the respective pairs of coordinates.

**Application 3 – Area of Triangle Using Heron’s Formula**

Heron's Formula calculates the area of a triangle given the length of its three sides. The general equation is:

where a, b, and c represent the length of the sides, and s denotes the semi-perimeter. Given triangle dimensions, the semi-perimeter s can be calculated using the formula =(A2+B2+C2)/2 in cell D2. The area of the triangle can then be calculated using the formula =SQRT(D2*(D2-A2)*(D2-B2)*(D2-C2)) in cell E2. Copying these formulas down the respective ranges will return the area of the triangles.

These applications demonstrate the versatility of the SQRT Function in Excel, which can be used in numerous mathematical formulas and concepts where finding the square root of a number is necessary.


# Trachtenberg System

The Trachtenberg System is a method of rapid mental calculation developed by Jakow Trachtenberg. It consists of a series of operations that allow for quick arithmetic computations. The system was created by Trachtenberg while he was being held in a Nazi concentration camp.

## General Multiplication

The general multiplication method involves multiplying two numbers, **a** and **b**, with minimal temporary results. The final digit of the product is determined by multiplying the last digits of **a** and **b**. Each subsequent digit is calculated by summing the products of corresponding digits from **a** and **b**.

The formula for general multiplication is:

\[a \mbox{ (digit at } i\mbox{)} \times b \mbox{ (digit at } (n-i)\mbox{)} \]

This method allows individuals to multiply four-digit numbers in their head, writing down only the final result.

## General Division

Division in the Trachtenberg System is similar to multiplication, but with subtraction instead of addition. The dividend is split into smaller partial dividends, and each partial dividend is divided by the left-most digit of the divisor. The result is obtained one digit at a time, with subtraction of product pairs and number-tens pairs from the partial dividend.

## General Addition

The Trachtenberg System also includes a method for adding columns of numbers and checking the result without repeating the original operation. An intermediate sum is produced, and the final result is obtained by applying an L-shaped algorithm. The checking method uses digit sums, such as the nines-remainder method, to identify errors.

## Other Multiplication Algorithms

The Trachtenberg System includes specific algorithms for multiplying by certain numbers, including 11, 12, 6, 7, 9, 8, 4, 3, and 5. Each algorithm involves a series of steps, such as adding or subtracting digits, doubling or halving numbers, and carrying over excess digits.

### Multiplying by 11

To multiply a number by 11, add each digit to its neighbor (the digit to the right).

Example: 3425 * 11 = 37675

### Multiplying by 12

To multiply a number by 12, double each digit and add its neighbor.

Example: 316 * 12 = 3792

### Multiplying by 6

To multiply a number by 6, add half of the neighbor to each digit, and add 5 if the digit is odd.

Example: 357 * 6 = 2142

### Multiplying by 7

To multiply a number by 7, double each digit, add half of its neighbor, and add 5 if the digit is odd.

Example: 523 * 7 = 3661

### Multiplying by 9

To multiply a number by 9, subtract the right-most digit from 10, and subtract the remaining digits from 9. Add the neighbor, and subtract 1 from the neighbor for the leading zero.

Example: 2130 * 9 = 19170

### Multiplying by 8

To multiply a number by 8, subtract the right-most digit from 10, and subtract the remaining digits from 9. Double the result, and add the neighbor. Subtract 2 from the neighbor for the leading zero.

Example: 456 * 8 = 3648

### Multiplying by 4

To multiply a number by 4, subtract the right-most digit from 10, and subtract the remaining digits from 9. Add half of the neighbor, and add 5 if the digit is odd. Subtract 1 from half of the neighbor for the leading zero.

Example: 346 * 4 = 1384

### Multiplying by 3

To multiply a number by 3, subtract the right-most digit from 10, and subtract the remaining digits from 9. Double the result, and add half of the neighbor. Add 5 if the digit is odd, and subtract 2 from half of the neighbor for the leading zero.

Example: 492 * 3 = 1476

### Multiplying by 5

To multiply a number by 5, take half of the neighbor, and add 5 if the digit is odd.

Example: 42 * 5 = 210

## Publications

The Trachtenberg Speed System of Basic Mathematics, written by Jakow Trachtenberg, was published in 1960. The book contains algebraic explanations for each operation and has been reprinted several times.

## Software

There are several software programs and online resources available that teach the Trachtenberg System, including:

* Vedic Mathematics Academy
* Trachtenberg Speed Math
* Trachtenberg Mathematics Software
* Mercury Math

## References

* Trachtenberg, J. (1960). The Trachtenberg Speed System of Basic Mathematics. Doubleday and Company, Inc., Garden City, NY, USA.
* Rushan Ziatdinov, Sajid Musa. Rapid mental computation system as a tool for algorithmic thinking of elementary school students development. European Researcher 25(7): 1105-1110, 2012.


# Continuity Calculator

The continuity calculator evaluates the continuity of a function by inputting the function, selecting the variable, entering the point, and hitting the calculate button.

## Continuous Calculator

This calculator determines whether a function is continuous or discontinuous, providing results with steps in a few seconds.

## What is Continuity?

In calculus, continuity checks if a function is continuous or not on a given interval. A function is continuous if its graph has no holes or breakages. The concept of continuity is essential in calculus, as differentiation is only applicable when a function is continuous at a point.

Mathematically, a function must be continuous at a point x = a if it satisfies the following conditions:
1. **f(a)** exists (the function must be defined at "a")
2. **lim** exists (the limit of the function at "a" must exist)_{x→a}f(x)
3. **f(a) = lim**_{x→a} f(x)

If all three conditions are satisfied, the function is continuous; otherwise, it is discontinuous.

## How to Calculate Continuity?

### Example 1

Check if the given function is continuous or not at x = 2.

f(x) = 3x^{2} + 4x + 5

**Solution**

**Step 1:** Check if the function is defined at x = 2.

f(2) = 3(2)^{2} + 4(2) + 5 = 25

The function is defined at x = 2.

**Step 2:** Evaluate the limit of the given function.

lim_{x→2 }[3x^{2} + 4x + 5] = 25

The limit of the function exists.

**Step 3:** Check the third condition of continuity.

f(2) = lim_{x→2} f(x) = 25 = 25

The function is continuous, as all conditions are satisfied.

### Example 2

Check if the given function is continuous or not at x = 0.

f(x) = x / x

**Solution**

**Step 1:** Check if the function is defined at x = 0.

The function is not defined at x = 0.

**Step 2:** Calculate the limit of the given function.

Apply L'Hopital's rule to evaluate the limit.

**Step 3:** Check the third condition of continuity.

f(0) ≠ lim_{x→0} f(x)

Conditions 1 and 3 are not satisfied, so the function is discontinuous.


# Lesson 1: High Noon and Sunset Shadows Develop Understanding

The goal is to sketch the inverse of a function on the same set of axes and identify each function as even, odd, or neither.

## Exercises

### 1-4: Function Identification

For each function:
#### a. Sketch the inverse of the function.
#### b. Identify the function as:
#### A. even
#### B. odd
#### C. neither

### 5: Even Functions

Given the graphs of three even functions:
#### a. Can an even function be invertible?
#### b. Justify your answer.

## Graph Analysis

For each graph, state the:
- period
- amplitude
- vertical shift
- phase shift
Then, write the equation.

### 6-10: Equation Writing

Write the equation of each graph using the given parameters.

### 11: Cofunction Identity

The cofunction identity states that sin(x) = cos(π/2 - x) and cos(x) = sin(π/2 - x). Explain where you would see this identity in a right triangle. Describe the relationships between the graphs of sine and cosine functions.

### 12-13: Graph Relationships

Describe the relationships between the graphs of:
- sine and cosine
- tangent and cotangent

### 14: Graph Interpretation

This graph could be interpreted as a shift or a reflection. Write the equations both ways, demonstrating an understanding of how shifts and reflections affect the graph of a function.


Python Consecutive Assignment is a concept that involves assigning a sequence of numbers to variables in a specific order. This concept is often used in recursive assignments, where a sequence of numbers is assigned to variables in a recursive manner.

The problem with arbitrary sequences of integers is that their length depends on a given index. To resolve this issue, a recursive class can be implemented to assign the value n to each of the sequences Z and V. The idea is simple: given a sequence of integers n (n+1), we can assign the value n to Z and the sequence of integers to V.

The recursive assignment algorithm works in three ways:

1. Create a sequence of equal integers n and Z.
2. Create a sequence n+2, which is then assigned to V.
3. Assign the value 0 to the sequence n.

This algorithm works because there are two more sequences n and Z that can be assigned to V by an arbitrary number of steps.

For example, when the sequence is n=n+1, we have 1: 2, which is the amount of zeros in the sequence. We can then assign 1 to Z and create two sequences n and V by adding 2 copies of the sequence n+3.

The recursion claim states that if we have two sequences n and Z, then we can assign a value of n to each of N's sequences Z and V by following a chain of webpage copies of Z (for the first copy) and N' (for the second copy).

In every chain, there are two sequences that we can 'sort' by, and we can assign a value to each of those sequences by following the second chain.

A 'random' sequence is a sequence of the form n.n. If we are interested in recursively assigning an arbitrary sequence to a sequence, we can do it in two steps:

1. Create a sequence sequence n=n.n+1.
2. Assign the value n to the sequence of sequences Z and n+1.

The sequence of consecutive fractions is defined as a sequence of successive fractions of consecutive series. Each series of consecutive fraction is a series of successive fractions.

The construction of the sequence involves defining the sequence of partial series of series. The sequence has the following properties:

* The first partial series of consecutive percentages of the series in the series.
* The second partial series of successive percentages of the first partial series in the sequences of the series.
* The third partial series of two consecutive percentages of series in the first series.

The series of series may be constructed as follows:

1. The first series of series is composed of the following series, where the series of successive points in series is composed.
2. The second series of series consists of the following series, the series of which is composed of the series numbered 1 and the series of the successive points in the series of series numbered 2.
3. The third series is composed by the series of two successive points in each series of series, where series numbered 3 and series numbered 4 are composed by the following series of series.

The sequence of consecutive points in series consists of two series. Thus, the series is composed with two series numbered 3, 5, 6. The series in series is the sequence 3. The series corresponding to the sequence 3 is composed by series numbered 6 and series numbered 7.


We begin with the quadratic equation x^{2} - 2x + 20 = 0, where a = 1, b = -2, and c = 20. To find the roots, we use the quadratic formula: x = (-b ± sqrt(b^{2} - 4ac)) / (2a). Substituting the values, we get x = (2 ± sqrt((-2)^{2} - 4(1)(20))) / (2(1)). Simplifying under the radical gives -76, indicating imaginary roots. The square root of -76 simplifies to 2i√19, so the roots are 2 ± 2i√19. Factoring out a 2, we get two complex roots: 1 - i√19 and 1 + i√19.


# Potentiometers MCQ Quiz

## Latest Potentiometers MCQ Objective Questions

## Top Potentiometers MCQ Objective Questions

#### Potentiometers Question 1:
What is the main purpose of a potentiometer as a device?

The instrument that is designed to measure the unknown voltage by comparing it with the known voltage is called a potentiometer. It makes use of a balance or null condition, no current flows and hence no power loss is consumed in the circuit. Measurements by such comparison methods give results of a high degree of accuracy because the result obtained does not depend upon the actual deflection of the pointer.

#### Potentiometers Question 2:
In the potentiometer circuit shown, balance is obtained. The unknown E_{x} is then calculated using the formula E' = \frac{{E\;{R_{AB}}}}{{\left( {r + {R_{AB}}} \right) \times AB}}x.

Working current of potentiometer = \frac{{3.2}}{{200 + 200 + 2800}} = 1\;mA. Voltage drop across sliding wire = 1 mA × 200 = 200 mV. As the potentiometer works on the null balance technique, the current through the galvanometer should be zero. ⇒ E = voltage drop across sliding wire = 200 mV.

#### Potentiometers Question 3:
A slide wire is used for the measurement of current in a circuit. The voltage across a standard resistance of 1.0 Ω is balanced at 75 cm. Find the magnitude of the current if the standard cell having an EMF of 2 V is balanced at 50 cm.

The standard cell emf 2 volt is balanced at 50 cm. The voltage drop per unit length = 2/50 = 40 mV/cm. The voltage drop across a standard resistor of 1 Ω is balanced at 75 cm. The voltage drop across 75 cm length = 40 × 10^{-3} × 75 = 3V. The resistance of 75 cm length wire = 0.1 Ω. Current through the circuit = 3/ 1 = 3 A.

#### Potentiometers Question 4:
Potentiometer sensitivity can be decreased by increasing its potential gradient. This can be achieved by decreasing the length of the potentiometer wire or by increasing the current in the circuit using a rheostat.

#### Potentiometers Question 5:
Which of the following devices should be used for accurate measurement of low D.C. voltage? A potentiometer is used for accurate measurement of low D.C. voltage because the result obtained does not depend upon the actual deflection of a pointer. A potentiometer is an instrument designed to measure an unknown voltage by comparing it with a known voltage. Measurement using the comparison method is capable of a high degree of accuracy.

#### Potentiometers Question 6:
For measuring current, a slide wire is used in the circuit where the voltage drop across the resistor having the value of 1 Ω is balanced at 100 cm. If the emf of the standard cell having the value of 2 V is balanced at 40 cm, what will be the magnitude of the current? Given, 2 V standard cell is balanced at 40 cm. ∴ Voltage per cm = \frac{2}{40} = 0.05\; V. Voltage drop across a slide wire having 1 Ω resistance is balanced at 100 cm. ∴ Total voltage drop across the slide wire = (0.05 × 100) = 5 V. ∴ The magnitude of the current of the circuit or through slide wire = \frac{5}{1} = 5 A.

#### Potentiometers Question 7:
In the following meter bridge circuit, balance point is obtained at J with AJ = L cm. When the values of R and X are doubled and then interchanged the position of balance point is: As the resistance is directly proportional to the length of the wire. If the R and X is doubled, so its ratio of R and X will be same then the length covered by the slide wire will also the same. Now the position of R and X is interchanged then the length covered by the slide wire will also be interchanged. So the new length of AJ will be (100 - L) cm.

#### Potentiometers Question 8:
If voltage is measured using a potentiometer then under balanced condition of the instrument, which of the following is correct: Potentiometer measures the unknown voltage using null or balance conditions, hence no power is required for the measurement. Under the balanced or null condition, there is no current flow in the unknown emf circuit, Hence the power consumed in the circuit is zero.

#### Potentiometers Question 9:
Current in a circuit is measured using a simple slide wire. What will be the Voltage drop per unit length if the standard cell is of emf 2.38 V balanced at length of 35 cm? Potential gradient (k) = \frac{V}{l}. Given, V = 2.38 volts and l = 35 cm. k = \frac{V}{l} = \frac{2.38}{35} = 0.068\; V/cm.

#### Potentiometers Question 10:
If only the current flowing through the potentiometer wire is increased, then what will be the effect on the position of zero deflection? If only the current flowing through the potentiometer wire is increased, then the position of zero deflection will remain unchanged because the deflection point depends upon the wire resistance and test voltage and it is independent of the current.


In algebra, a **quadratic equation** is any equation that can be rearranged in standard form as: *ax*^{2} + *bx* + *c* = 0. This equation has roots based on the parameters **a**, **b**, and **c**, which are required as input. The program then determines if there are roots **x1** and **x2** for the equation.

The roots of a quadratic equation depend on the Discriminant (b^{2} - 4ac), which has three cases:
- Positive Discriminant: two distinct roots **x1** and **x2** exist.
- Discriminant equal to Zero: one root **x1** exists.
- Negative Discriminant: no real roots exist, but two distinct complex roots exist.

The Java program calculates the roots as follows:
```java
import java.util.Scanner;
public class QuadraticEquation {
    public static void main(String[] args) {
        int a, b, c;
        try(Scanner in = new Scanner(System.in)) {
            System.out.println("Please give integers for a, b, and c: ");
            a = in.nextInt();
            b = in.nextInt();
            c = in.nextInt();
        }
        System.out.println("Now the program will try to find x roots ");
        System.out.println("For the Quadratic equation: ");
        System.out.format("%d*x^2+%d*x+%d=0", a, b, c );
        System.out.println();

        double d = Math.pow(b, 2) - 4*a*c;
        if(d > 0) {
            double x1 = (-b + Math.sqrt(d))/2*a;
            double x2 = (-b - Math.sqrt(d))/2*a;
            System.out.println("x1: " + x1);
            System.out.println("x2: " + x2);
        } else if(d == 0) {
            double x1 = (-b + Math.sqrt(d))/2*a;
            System.out.println("Root is = "+ x1);
        } else {
            double real = -b / (2*a);
            double imaginary = Math.sqrt(-d) / (2*a);
            System.out.format("x1 = %f + i(%f)\n", real, imaginary);
            System.out.format("x2 = %f - i(%f)\n", real, imaginary);
        }
    }
}
```
Example output:
```
1
7
-78
Now the program will try to find x roots
For the Quadratic equation:
1*x^2+7*x-78=0
x1: 6.0
x2: -13.0
```


**Net Present Value Calculation**

The net present value (NPV) of an investment is calculated by subtracting the investment amount from the present value of all cash inflows. For an investment of $100,000 with expected net annual cash inflows of $25,000 for 6 years and a required rate of return of 12%, the NPV is calculated as follows:

Present value of cash inflows = $25,000 x PVIFA (12%, 6) = $25,000 x 4.1114 = $102,785
Net present value = $102,785 - $100,000 = $2,785

**Future Value of an Annuity**

To calculate the future value of an annuity, first convert the annual interest rate to a monthly rate. For a monthly deposit of $1,000 for 36 months at an annual interest rate of 12%, the monthly interest rate is 1%. The future value of the annuity is then calculated as follows:

Future value of an annuity of $1 per period at 1% for 36 periods = 43.07688
Amount to be received after 36 months = $1,000 x 43.07688 = $43,076.88

**Journal Entries for Sale of Merchandise Inventory**

Under the perpetual inventory system, two journal entries are prepared to record the sale of merchandise inventory. For a sale of $15,000 on account with a cost of merchandise inventory sold of $12,000, the journal entries are:

1. Debit Accounts Receivable: $15,000
Credit Sales Revenue: $15,000
2. Debit Cost of Goods Sold: $12,000
Credit Merchandise Inventory: $12,000

**TutorMe Platform**

TutorMe offers virtual lessons with top-quality tutors. A lesson is a virtual space where students and tutors can communicate using video, audio, and text chat. Students can upload documents, edit papers in real-time, and use a virtual whiteboard. To begin a lesson, click the "Start Session" button if the tutor is online or send a message to schedule a lesson. TutorMe tutors are current college students or recent graduates of top-tier universities.


# 500 Words On Eratosthenes and Quasars

Two-thousand years ago, Eratosthenes conducted one of the first known investigations in geodesy, the study of the Earth's shape and orientation. He utilized the phenomenon of shadows cast by columns at noon on the summer solstice in Syene and Alexandria to calculate the Earth's circumference. In Syene, columns cast no shadows, while in Alexandria, they did. Eratosthenes measured the length of the shadow cast by a rod of known length and used simple trigonometry to determine the angle made between the vertical rod and the shadow, which was one-fiftieth of a circle.

By treating the Sun's rays as parallel, Eratosthenes used alternate angles to determine that the angle between Alexandria and Syene was the same. He then measured the distance between the cities, approximately 800km or 5000 stadia, by employing a man to walk the distance and count his paces. Using this information, Eratosthenes calculated the Earth's circumference: if 5000 stadia is one-fiftieth of the circumference, then the whole distance is 50 * 5000 = 250,000 stadia, or 40,000km.

Eratosthenes' method was incredibly accurate, with a difference of only 0.9% from the accepted value of 40,075km today. The basic principles of his method are still used by scientists to establish geodetic values, albeit with more sophisticated tools. Modern geodesists use Very-Long-Baseline Interferometry (VLBI), a technique that started as a means to observe quasars. By treating quasars as fixed points in space due to their enormous distances from Earth, geodesists can use them as reference points.

Quasars emit characteristic radio bursts, which are detected using an array of radio telescopes across Earth. The delay between two telescopes receiving the signal is measured using hydrogen-masers, extremely precise clocks that rely on hydrogen atoms emitting photons at a precise frequency. This allows for the calculation of the distance between the telescopes with an accuracy of 2mm over a 600-second observation. VLBI is a prime example of how diverse and vital the field of physics is, combining geodesy, radio astronomy, and quantum mechanics to answer questions that have been posed for two-thousand years.


To find a percentage of a number, multiply the number by the percentage in decimal form. For example, 20 percent of 362 is calculated as 0.2 x 362 = 72.4. Similarly, 26 percent of 362 is 0.26 x 362 = 94.12.

Converting a fraction to a percentage involves dividing the numerator by the denominator and then multiplying by 100. For instance, 33 out of 362 is 33 / 362 = 0.09116, which as a percentage is 0.09116 * 100 = 9.12%.

Understanding percentages is crucial for grading. A 50 percent is often considered an F, while 71 percent may be a D+ depending on the university. It's also important to note that 31 out of 50 translates to 62 percent, which is typically a D grade.

In calculating grades, percentages can be used to weigh different components. For example, if 80 percent of a final grade consists of one score (e.g., 55) and 20 percent consists of another (e.g., 50), the final grade can be calculated as (1/5)*(55+55+55+55+50) = (1/5)*270 = 54 percent.

To answer the question of what grade is 36.2 percent of 50, first convert 36.2 percent to a decimal: 0.362. Then multiply 0.362 by 50 to find the grade: 0.362 * 50 = 18.1.

Q: What grade is 36.2 percent of 50?
A: 18.1

Note: Percentages are always out of 100, making them a standardized way to express proportions. This is evident in calculations such as 7.5 being 15% of 50, and 5 being 1.4% of 362.


#### Question

If A and B represent the maximum and minimum amplitudes of an amplitude-modulated wave, write the expression for the modulation index in terms of A and B.

#### Solution

The modulation index for an AM wave is given by `a_m = (E_m)/(E_c)`, where `E_m` is the maximum amplitude of the modulating signal and `E_c` is the carrier wave amplitude.

The maximum amplitude of the modulated wave is `A = E_c + E_m`, and the minimum amplitude is `B = E_c - E_m`.

From these equations, we can derive `E_c = (A + B)/2` and `E_m = (A - B)/2`.

Substituting these values into the modulation index equation, we get `a_m = E_m/E_c = ((A - B)/2)/((A + B)/2) = (A - B)/(A + B)`.

Concept: Amplitude Modulation.


**Recursion Exercises**

To begin, make sure you understand the readings on recursion and start DrScheme.

### Exercise 1: Power of Two

Define and test a recursive Scheme procedure, `(power-of-two`, that takes a natural number as its argument and returns the result of raising 2 to the power of that number. For example, `(power-of-two 3)` should return 8.

### Exercise 2: Count Down

Define and test a Scheme procedure, `(count-down`, that takes a natural number as argument and returns a list of all the natural numbers less than or equal to that number, in descending order. For example, `(count-down 5)` should return `(5 4 3 2 1 0)`.

### Exercise 3: Fill List

Define and test a Scheme procedure, `(fill-list`, that takes two arguments, the second of which is a natural number, and returns a list consisting of the specified number of repetitions of the first argument. For example, `(fill-list 'sample 5)` should return `(sample sample sample sample sample)`.

### Exercise 4: Iota

Define and test a recursive Scheme procedure that takes a natural number as argument and returns a list of all the natural numbers that are strictly less than the argument, in ascending order. This procedure is traditionally named `iota`.

### Exercise 5: Count From

Recall the `count-from` procedure from the reading on recursion over natural numbers. What is the value of the call `(count-from -10 10)`?

* Write down what you think it should be.
* Copy the definition of `count-from` into DrScheme and use it to find out what the call actually returns.

Using `count-from`, define and test a Scheme procedure that takes a natural number as argument and returns a list of all the natural numbers that are strictly less than the argument, in ascending order.

### Exercise 6: Number of Decimal Digits

The following procedure computes the number of digits in the decimal representation of a number:
```scheme
(define number-of-decimal-digits (lambda (number) (if (< number 10) 1 (+ (number-of-decimal-digits (quotient number 10)) 1))))
```
* Test this procedure.
* Describe the base case of this recursion.
* Identify and describe the way in which a simpler instance of the problem is created for the recursive call.
* Explain how the procedure correctly determines that the decimal numeral for the number 2000 contains four digits.
* What preconditions does `number-of-decimal-digits` impose on its argument?

### Count From Procedure

For reference, the `count-from` procedure is defined as:
```scheme
(define count-from (lambda (lower upper) (if (= lower upper) (list upper) (cons lower (count-from (+ lower 1) upper)))))
```
This procedure constructs a list of natural numbers from `lower` to `upper`, inclusive, in ascending order. The preconditions are that `lower` is less than or equal to `upper`, and both `lower` and `upper` are non-negative integers.


Polygons are a crucial part of geometry questions on the ACT, and understanding their rules and formulas is essential to solving them. A polygon is a flat, enclosed shape made up of straight lines, and it can be categorized into regular and irregular types. Regular polygons have all equal sides and angles, while irregular polygons do not.

The different types of polygons are named based on their number of sides and angles. For example, a triangle has three sides and three angles, a quadrilateral has four sides, and a pentagon has five sides. Many of the polygons on the ACT are either triangles or quadrilaterals.

Quadrilaterals can be further categorized into different types, including parallelograms, rhombuses, rectangles, squares, and trapezoids. A parallelogram is a quadrilateral with opposite sides that are parallel and congruent. A rhombus is a type of parallelogram with all sides equal, while a rectangle is a parallelogram with all angles equal to 90 degrees. A square is a rectangle with all sides equal, and a trapezoid is a quadrilateral with only one set of parallel sides.

To solve polygon questions on the ACT, it's essential to understand the formulas for finding the area and perimeter of different types of polygons. The area of a triangle is given by the formula A = (1/2)bh, where b is the base and h is the height. The area of a square is given by A = s^2, where s is the length of one side. The area of a rectangle is given by A = lw, where l is the length and w is the width. The area of a parallelogram is given by A = bh, where b is the base and h is the height.

The area of a trapezoid is given by the formula A = [(l1 + l2)/2]h, where l1 and l2 are the lengths of the two parallel bases and h is the height. This formula can be used to find the area of a trapezoid by finding the average of the two parallel bases and multiplying it by the height.

In addition to understanding the formulas, it's also essential to know how to find the interior angle sum of a polygon. The interior angle sum of a polygon can be found using the formula (n-2)180, where n is the number of sides. This formula can be used to find the sum of the interior angles of any polygon, regardless of its type.

To find the individual interior angles of a regular polygon, you can divide the interior angle sum by the number of sides. For example, if you have a regular pentagon with an interior angle sum of 540 degrees, you can find the individual interior angles by dividing 540 by 5, which gives you 108 degrees per angle.

The number of diagonals in a polygon can be found using the formula n(n-3)/2, where n is the number of sides. This formula can be used to find the number of distinct diagonals in any polygon.

When solving polygon questions on the ACT, it's essential to use strategies such as breaking up figures into smaller shapes, using shortcuts, and keeping your work organized. You can also use the plugging in answers (PIA) or plugging in numbers (PIN) strategy to help you solve questions.

Some typical polygon questions on the ACT include finding the measure of an angle, the perimeter of a figure, or the area of a figure. To solve these questions, you need to understand the formulas and rules for different types of polygons and be able to apply them to the given information.

In conclusion, understanding polygons and their rules and formulas is crucial to solving geometry questions on the ACT. By knowing how to find the area and perimeter of different types of polygons, how to find the interior angle sum, and how to use strategies such as breaking up figures and using shortcuts, you can solve even the most difficult polygon questions on the test.


Let u = (4, 0) and v = (0, -3). A linear transformation T is given by T(u) = (4, 4, 4) and T(v) = (6, 9, 12). We need to find the standard matrix of T and T(x, y).

To find the standard matrix of T, we first express (x, y) as a linear combination of u and v: a(4, 0) + b(0, -3) = (x, y). This yields the equations 4a = x and -3b = y, giving a = x/4 and b = -y/3.

Using the linear property of T, we have T(x, y) = T[a(4, 0) + b(0, -3)] = aT(4, 0) + bT(0, -3). Substituting the values of a, b, T(4, 0), and T(0, -3), we get T(x, y) = (x/4)(4, 4, 4) + (-y/3)(6, 9, 12) = (x, x, x) + (-2y, -3y, -4y) = (x - 2y, x - 3y, x - 4y).

The standard matrix of T, denoted as A, can be found by expressing T(x, y) as a matrix multiplication: T(x, y) = A(x, y). Since T(x, y) = (x - 2y, x - 3y, x - 4y), we can write this as:

T(x, y) = | 1  -2 | | x |
          | 1  -3 | | y |
          | 1  -4 |

Thus, the standard matrix of T is | 1  -2 |
                                  | 1  -3 |
                                  | 1  -4 |.

The difference between the two questions lies in the fact that part a) asks for the standard matrix of T, while part b) asks for the expression of T(x, y). Although the expression for T(x, y) was found in the process of solving part a), the standard matrix of T was not explicitly stated.


# Simplifying the Given Fraction

To simplify the fraction \(\frac{\frac{x}{10} - \frac{5}{x}}{\frac{1}{5} + \frac{1}{x}}\), follow these steps:

## Step 1: Find a Common Denominator
Give all terms a common denominator. For the numerator, the common denominator is \(10x\), and for the denominator, it is \(5x\). Thus, the expression becomes \(\frac{\frac{x^2}{10x} - \frac{50}{10x}}{\frac{x}{5x} + \frac{5}{5x}} = \frac{\frac{x^2 - 50}{10x}}{\frac{x + 5}{5x}}\).

## Step 2: Simplify the Fraction
Simplify the fraction by dividing the numerator by the denominator, which is equivalent to multiplying the numerator by the reciprocal of the denominator: \(\frac{x^2 - 50}{10x} \times \frac{5x}{x + 5}\).

## Step 3: Cancel Common Factors
Cancel out any common factors. The \(10x\) and \(5x\) have a common factor of \(5x\), which can be canceled out: \(\frac{x^2 - 50}{2(x + 5)}\).

## Step 4: Factor if Possible
Factor the numerator if possible: \(x^2 - 50 = (x + \sqrt{50})(x - \sqrt{50}) = (x + 5\sqrt{2})(x - 5\sqrt{2})\). However, since this does not share common factors with the denominator \(x + 5\), no further simplification can be done through factoring.

The final simplified form, considering the original instruction for simplification and maintaining the format for clarity, is thus focused on the process rather than an algebraic simplification beyond recognition. The key steps involve finding a common denominator, simplifying, and then factoring or canceling common factors if applicable.


To solve the equation -v divided by 3 plus 8 equals 22, we need to isolate the variable.

First, subtract 8 from both sides: -v/3 = 22 - 8, which simplifies to -v/3 = 14.

Next, multiply both sides by -3 to solve for v: v = -14 * 3, which simplifies to v = -42.

Alternatively, the equation can be solved by multiplying both sides by 3 to eliminate the fraction: -v + 24 = 66.

Then, subtract 24 from both sides: -v = 66 - 24, which simplifies to -v = 42.

Multiplying both sides by -1 gives v = -42.

In another example, the equation 5x + 2 = 22 can be solved by subtracting 2 from both sides: 5x = 22 - 2, which simplifies to 5x = 20.

Dividing both sides by 5 gives x = 20/5, which simplifies to x = 4.

For the equation -9h - 6 + 12h + 40 = 22, combine like terms: 3h + 34 = 22.

Subtract 34 from both sides: 3h = 22 - 34, which simplifies to 3h = -12.

Divide both sides by 3: h = -12/3, which simplifies to h = -4.

The equation 88 + x = 3x + 22 can be solved by subtracting x from both sides: 88 = 2x + 22.

Subtract 22 from both sides: 66 = 2x.

Divide both sides by 2: x = 66/2, which simplifies to x = 33.

The equation 22 + 2x = 37 + 6 + x can be solved by subtracting x from both sides: 22 + x = 37 + 6.

Subtract 22 from both sides: x = 43 - 22, which simplifies to x = 21.

The equation x + 13 = 22 can be solved by subtracting 13 from both sides: x = 22 - 13, which simplifies to x = 9.


**1. Key Statistical Concepts**

* A **Population** consists of everything, while a **Sample** is a subset of a population.
* A **Parameter** is a characteristic measured for a population, whereas a **Statistic** is a characteristic measured for a sample.
* **Descriptive Statistics** describe characteristics of a variable in a sample, without regard for the population parameter values. **Inferential Statistics** make an inference that the population parameter values are probably close to the sample statistic values.
* A **Variable of Interest** is a variable being studied, while a **Control Variable** is a variable with available data but not being studied.

**2. Types of Data Sets**

* A **Time Series** data set collects data across different points in time, such as measuring a nation's GNP each quarter for thirty years.
* A **Cross Sectional** data set collects data at a single point in time, such as measuring the GNP for 87 countries for the third quarter of 2020.

**3. Variable Classification**

* **Quantitative** variables are measured or counted, such as the number of nights stayed in a hotel.
* **Categorical (Qualitative)** variables consist of category labels, such as where books are purchased or how someone generally flies.

**4. Survey Questions and Variable Classification**

* How many nights did you stay in a hotel the past twelve months? **Quantitative**
* Where do you purchase books? **Categorical**
* Do you use a travel agent to make your travel arrangements? **Categorical**
* What is your age? **Quantitative**
* How do you generally fly? **Categorical**

**5. Statistic Development and Inference**

* To estimate the percentage of business travelers using a travel agent, calculate the sample proportion: 155 / 430 = 0.36, or 36%.
* To find the number of respondents who generally fly Business Class, calculate: (430)(0.44) = 189.2, or approximately 189 or 190.

**6. Frequency Table Construction**

* Construct a frequency table with columns for Frequency, Relative Frequency, Cumulative Frequency, and Cumulative Relative Frequency.

**7. Scatterplot Analysis**

* Based on the scatterplot, the relationship between x and y is **Negative**, as y tends to decrease as x increases.

**8. Excel Data Analysis**

* Calculate and interpret the following statistics using Excel:
	+ **Mean**: 160, the average cost paid by families to have their income taxes prepared.
	+ **Median**: 135, the middle value of the data set, where half the families paid less and half paid more.
	+ **Mode**: 120, the most common amount paid by families to have their taxes prepared.
	+ **First Quartile (Q1)**: 115, the value below which 25% of the data falls.
	+ **Third Quartile (Q3)**: 183.75, the value below which 75% of the data falls.
	+ **90th Percentile (P90)**: 237, the value below which 90% of the data falls.
	+ **Five-Number Summary**: 100, 115, 135, 183.75, 360, a summary of the data set's key values.

**9. Standardized Values and Outliers**

* Calculate the standardized value (Z-score) for each cost: 2,300 CHF, 4,900 CHF, and 13,000 CHF.
* Interpret each Z-score value and determine if it would be considered an outlier.


#### Turn your child into a maths magician?

Maths is everywhere, and as parents, we need to make it real and relevant for our children. We are the first mathematicians our children know, using maths daily in various aspects of life, such as money, time, weight, height, and distance.

Teaching starts with conversation, pointing out math situations and vocabulary in everyday life. Encourage your child to think mathematically at home, in supermarkets, or at playgrounds. For example, when the alarm goes off, calculating the minutes before getting up is a small example of maths in action.

In the kitchen, maths can be applied through recipes and baking. Have your child measure ingredients using cups or scales, and figure out how to double or halve a recipe. Explain the importance of using exact measurements and demonstrate numerical division or physical halving of ingredients.

At the supermarket, talk about package labels, abbreviations for litres, grams, kilograms, and servings. Give your child a budget and have them buy items, using pamphlets to find the best value. Let them estimate the cost of groceries and see who gets the closest.

Collect toy or grocery pamphlets and have your child circle items they'd like to buy with a given budget, such as $20. Ask them to calculate how many items they can buy and what their change will be.

Play "I spy a Number" using road signs, licence plates, and letterboxes. Count the numbers on letterboxes, add or subtract them, or multiply them, depending on your child's age. At traffic lights, have your child estimate how long it will take for the lights to change and count down to see if they're right.

Open a bank account for your child and teach them how to make deposits, count money, balance a checkbook, and sort and classify. There are endless teaching opportunities throughout the day to reinforce math vocabulary and skills.

This approach is called **Natural Exposure**, and it's the best way to learn about maths. By being aware of the maths around you and talking to your children about it, they will see how important and natural mathematics is. The more you talk about it, the more they will understand its relevance and have fun with it.


**Introduction to Quantum Key Distribution Textbook**

This textbook provides a comprehensive introduction to quantum key distribution, presenting an overview of state-of-the-art quantum communication protocols and applications. The book is designed for non-specialist readers and includes the necessary mathematical tools, making it accessible to those without a background in quantum information theory.

**Errata**

The following corrections should be noted:
- **Definition 2.1**: The third property (positivity) should be defined as (φ,φ) ≥ 0.
- **Figure 2.2**: The boundaries for θ are 0 ≤ θ ≤ π.
- **Exercise 2.20**: The sum in the spectral decomposition of ρ is over i, not over x.
- **Definition 2.25 and Theorem 2.26**: The map Ɛ is defined as Ɛ : B(H_A) → B(H_B).
- **Definitions 2.31 and 2.34**: The measurements M and M_x are defined as M: H_A → H_B.
- **Equation (2.103)**: The right-hand side should be 1/Sqrt{2}.
- **Equation (3.9)**: The correct equation is h_2(λp_1 + (1 - λ)p_2) ≥ λh_2(p_1) + (1 - λ)h_2(p_2).
- **Equations (3.15) and (3.18)**: Probability distributions are missing their respective arguments, namely p and p_X,Y(x,y).
- **Equation (3.19)**: The second term should have a positive sign.
- **Equation (3.31)**: The second term is H(X_1|Y).
- **Exercise 4.2**: The measurement results form the bit string b as follows: If Bob obtains -1, then the i-th bit of b is 1, and if he obtains +1, then the i-th bit of b is 0.
- **Equation (4.44)**: The superscript of the smooth min-entropy is ε, and the last term on the right-hand side is missing a factor of 2.
- **Beneath Equation (A.3)**: If X and Y are independent, then p_Y|X(y|x) = p_Y(y) for all y ∈ Y.
- **Equation (B.4)**: H_e_2 = (1 1)^T, H_e_3 = (0 1)^T.
- **Page 224**: The quantum code CSS(C) is defined as the vector space spanned by {|x + C_1, C_2) >} for x ∈ C1.


A position vs. time graph illustrates how an object's position changes over time. The slope of this graph represents the object's speed at a specific time. Speed is a vector quantity, indicating both magnitude and direction. In a position-time graph, time is the independent variable on the horizontal axis, while the object's position is the dependent variable on the vertical axis.

The slope of a position vs. time graph shows the change in position over time, which is equivalent to velocity. A zero slope indicates that the object is at rest, resulting in a horizontal line on the graph. A constant and positive slope represents an object moving forward at a uniform rate, while a constant and negative slope signifies an object moving backwards at a constant speed.

A straight-line position graph with a positive or negative slope shows the average speed of an object. The steeper the slope, the faster the object is moving. A varying and positive slope, characterized by a curved line, indicates that the object is accelerating. The slope of the line at any point on the curve gives the speed of the object at that specific time.

The shape of the line on a position-time graph provides information about the object's motion. A straight line represents uniform motion, while a curved line indicates acceleration. The slope of the line can be positive, negative, steep, or gentle, offering insights into the object's speed and direction. By analyzing the slope and shape of a position-time graph, one can determine the object's velocity, acceleration, and direction of motion.


# Key Concepts in Space Exploration

A geostationary orbit, also known as a geostationary Earth orbit or geosynchronous equatorial orbit (GEO), is a type of orbit where a spacecraft's position in the sky remains constant.

Spacecraft propulsion refers to any method used to accelerate spacecraft and artificial satellites. This concept is crucial in space exploration, as it enables spacecraft to travel through space and reach their intended destinations.

In astronomy, Kepler's laws of planetary motion are three scientific laws that describe the motion of planets around the Sun. The first law states that the orbit of a planet is an ellipse, with the Sun at one of the two foci. The second law describes the speed of a planet as it moves along its orbit, while the third law relates the size of a planet's orbit to its orbital period.

The true anomaly is an angular parameter that defines the position of a body moving along a Keplerian orbit. It is the angle between the body's current position and its closest approach to the central body, measured from the central body's perspective.

The Tsiolkovsky rocket equation is a mathematical equation that describes the motion of vehicles that operate through a system of expelling hot gases. This equation is essential in understanding the relationship between a rocket's mass, exhaust velocity, and change in velocity.


A reader asked about calculating square roots, referring to an example in the Introduction to App Inventor e-book. The explanation of the method for calculating a square root of a number is as follows:

The built-in Math function can be used to find the square root of a number, but an alternative method can be employed. 2,000 years ago, Greek mathematicians identified a method of finding approximate square roots. This method involves making an initial guess of a square root, then repeating a sequence of dividing and averaging to refine the guess.

To find the square root of a number *n*, the initial guess is set to *n*, and then the average of the guess and *n* divided by the guess is calculated. For example, to find the square root of 30, the initial guess is 30. Then, 30 is divided by the guess (30), resulting in 1. The average of 30 and 1 is 15.5, which becomes the new guess.

The process is repeated: 30 is divided by the new guess (15.5), resulting in 1.93548. The average of 15.5 and 1.93548 is 8.717, which becomes the new guess. This cycle continues until the desired accuracy is achieved.

The cycle can be stopped after a set number of repetitions or when the difference between the current and previous guess is smaller than a specified value. For example, if an accuracy of 0.1 is desired, the cycle stops when the difference between the current and previous guess is less than 0.1.

Using this method, the square root of 30 converges to 5.477. The cycle can be stopped when the difference between guesses becomes very small, such as 0.1 or 0.001. The following sample App Inventor code calculates the square root using this method:

The code fetches the number from a user interface Text Box, sets the initial guess to this value, and uses a while loop to repeatedly calculate the square root until the difference between the current and previous guess is less than the specified value (0.1 or smaller).

The while loop continues to execute as long as the difference between the current and previous guess is large. Once the difference becomes small, the loop stops, and the approximate square root is found. The accuracy of the result depends on the specified difference value, such as 0.1 or 0.001.

To determine when to stop the cycle, the following options can be considered:
- Stop after a set number of repetitions (e.g., six).
- Stop when the difference between the current and previous guess is smaller than a specified value (e.g., 0.1 or 0.001).

In the example calculation, the generated guesses are:
- 15.5
- 8.717742
- 6.0795
- 5.507058
- 5.477306
- 5.477226

As the calculation progresses, the difference between consecutive guesses decreases, indicating convergence to the correct answer. The cycle can be stopped when the desired accuracy is achieved, such as when the difference between guesses is less than 0.1 or 0.001.


## Standards

* **CCSS.ELA-LITERACY.RH.6-8.2**: Determine the central ideas or information of a primary or secondary source and provide an accurate summary.
* **CCSS.MATH.CONTENT.6.RP.A.3**: Use ratio and rate reasoning to solve real-world and mathematical problems.
* **CCSS.ELA-LITERACY.W.7.1**: Write arguments to support claims with clear reasons and relevant evidence.

## Technology and Time Requirements

* Device with a browser for AzTech Game
* Printer for printing codex and related worksheet and activity pages
* 4-5 hours

## Lesson Summary

This lesson uses the Huexotzinco Codex as a primary source for document analysis, inquiry, and applied mathematics. Students analyze tribute paid to Spanish administrators, compute the tribute, and read a one-page overview of the codex. The lesson includes a presentation on connections between Aztec, Mayan, and contemporary methods, as well as a game that incorporates Mayan history and middle school mathematics.

## Lesson Plan

### Introduction and Background

The Huexotzinco Codex was part of a case brought by the Nahuas against the Spanish administrators, alleging excessive taxation. Students will learn about this case over several class periods.

### Analyze Documents (1 hour)

* Begin with the Library of Congress lesson, "The Huexotzinco Codex"
* Have students complete **Activity 1, Document Analysis**
* Allow 30 minutes for the teacher to prepare and review the lesson before class

### Play a Game (15 minutes)

* Have students sign in and begin the AzTech: Meet the Maya game

### Computation – How much was the tribute? (1-2 class periods)

* Continue with the Library of Congress lesson, "The Huexotzinco Codex"
* Have students complete **Activity 2, Computation**
* Use a slide presentation to show the connection between Aztec and Maya codices and modern systems of numbers and graphs

### Play a Game (15-20 minutes)

* Have students continue the AzTech: Meet the Maya game

### Write a Narrative Explanation

* Continue with the Library of Congress lesson, "The Huexotzinco Codex"
* Have students complete **Activity 3, Narrative Explanation**

## Assessment

* Observation of student understanding of historical document analysis
* Student self-corrected math computation
* Student written assignments (analysis sheet, observations, and scenario outlines)
* Math problems in the AzTech game, scored automatically with data available in teacher reports

## Differentiated Instruction (optional)

* **Advanced Students**: Continue with the AzTech: Meet the Maya game or play AzTech: The Story Begins or AzTech: Empiric Empire
* **English learners**: Play the AzTech games in English or Spanish