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Another example of a decimal number is 23.025. In this example, 23 is the whole number, while 025 is the fraction. In this instance, the 025 means 25/1000 or 25 parts out of 1000. As a rule, the number of digits after the decimal place denotes the number of zeroes used in the denominator of the fraction. So one zero would indicate tenths, while two indicates hundredths.

The cube root of a decimal number is a number that, when multiplied by itself three times, will give the result in the decimal number. For example, the cube root of 8 would be 2 because 2 * 2 * 2 = 8.

Perfect cubes are numbers whose cube roots are whole numbers. Eight is a perfect cube because its cube root, 2, is a whole number. Examples of perfect cubes are as follows:

- 23 = 8
- 33 = 27
- 43 = 64
- 53 = 125
- 63 = 216
- 73 = 343
- 83 = 512
- 93 = 729
- 103 = 1000

Let’s start with the cube roots of non-fractional decimal numbers.

Find the cube root of 216.0.

The number 216 is a whole number. The first step is to express the number as a product of its prime factors. When the number is expressed as a product of prime factors, the factors can then be grouped, and the cube root picked from the groups. Let’s see how this works.

The first number to start with is 2. It's the smallest prime number. Let’s check if 2 is a factor of 216. If the last digit of an integer is either zero or an even number, then the integer is divisible by 2: 216 / 2 = 108.

Our example, 216, ends with 6, which is even. So 2 is a prime factor. The result, 108, is also divisible by 2: 108 / 2 = 54. Also, 54 is divisible by 2: 54 / 2 = 27.

Notice that 27 ends with 7, which is not even, so it is not divisible by 2. However, 27 is divisible by 3: 27 / 3 = 9. Then, 9 is also divisible by 3: 9 / 3 = 3. And 3 is also divisible by 3: 3 / 3 = 1. We stop the division when we arrive at 1.

Express the number whose cube root you intend to find as a product of prime factors:

216 = 2 * 2 * 2 * 3 * 3 * 3

Pick the prime factors that occur up to three times. From the above prime factors, 2 appears three times, and 3 appears three times, so we pick 2 and 3:

2 * 3 = 6

The product of the numbers we have picked is 6, so the cube root of 216 is 6.

Find the cube root of 3375. We will follow the same steps as in the above example.

Express 3375 as a product of prime factors. 3375 ends with 5, which is an odd number, so 2 is not a factor of 3375. Let us check for 3. To check if 3 is a factor of a number, sum the digits of the number and divide the sum by 3. If the remainder is zero, then 3 is a factor of the number. The sum of the digits of 3375 is 18. 18 / 3 is 6. So 3 is a factor of 3375: 3375 / 3 = 1125

Using the rule we established earlier, 1125 is divisible by 3: 1125 / 3 = 375

375 is also divisible by 3: 375 / 3 = 125.

But 125 is no longer divisible by 3. Let’s check the next prime number, 5. All integers that end with 5 or 0 are divisible by 5: 125 / 5 = 25

25 is also divisible by 5: 25 / 5 = 5. And 5 is also divisible by 5: 5 / 5 = 1. Since the result is now 1, we will stop the division.

The next step is to express 3375 as a product of prime factors

3375 = 3 * 3 * 3 * 5 * 5 * 5

We will now pick the prime factors that appear 3 times. 3 and 5 have appeared 3 times each.

3 * 5 = 15

The cube root of 3375 is 15.

This refers to numbers that are not whole numbers and are expressed in decimal fractions.

What is the cube root of 0.343?

0.343 has a whole number part, which is 0, and the fractional part, which is 343.

The first step is to express the number as a fraction with a numerator and a denominator.

0.343 = 343 / 1000

The numerator is 343, and the denominator is 1000

The next step is to find the cube root of both the numerator and the denominator using the method of factorization explained in the examples above.

Let's look at the prime factors of 343, which is not divisible by 2, 3, or 5. It is divisible by 7

- 343 / 7 = 49
- 49 / 7 = 7
- 7 / 7 = 1
- 343 = 7 * 7 * 7

For the denominator, we already know that 1000 = 10 * 10 * 10

Pick the factors that occur three times: 7 for the numerator and 10 for the denominator. Hence the cube root of 343 / 1000 = 7 / 10. The cube root of 0.343 is the same as the cube root of 343/1000 = 7/10 = 0.7.

So the cube root of 0.343 = 0.7.

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**Regular** refers to a regular polygon with a unique set of attributes, not the typical definition of "normal." Whenever you see "regular" in mathematics, it is about a **polygon**.

- All sides are equal in length
- All angles are equal in degree

Typically, people think of squares. They are not wrong, but there are other polygons than can be regular: triangles, pentagons, and hexagons, for example. There are also **regular polyhedrons.** These are three-dimensional polygons such as cubes, tetrahedrons, and octahedrons. A polyhedron is regular when all its faces are regular polygons.

**Mean** is another way of writing "average." When paired with **geometry**, it clues us in that we are working in geometric dimensions as opposed to arithmetic. You will be finding the **geometric mean**.

For this particular topic, we will learn how to find the geometric mean of an **irregular polygon** and an **irregular polyhedron**. This way, you will better understand what makes a regular polygon and how to form one from an irregular polygon using geometric mean.

Unless specifically asked on a test, quiz, or by a teacher, different real-life situations require you to use the geometric mean. When it comes to mean, however, there are two different options: arithmetic and geometric. Understanding when to use the arithmetic mean can significantly help you figure out when to use geometric mean.

When finding the arithmetic mean, we are finding the average on a number line. This number line is generated with a low number, high number, and frequently, other values that fall in between. These numbers are added up and divided by the total amount of numbers on the line.

The formula for arithmetic mean is (*A+B)/2*.

The underlying question we are asking ourselves when finding the arithmetic mean is what number would each value (*A* and *B*) have to be to add up to the original total of *A+B*?

**Example: **

*6+12=18*

*18/2=9*

To prove our question, we take the mean and add it twice to make our original total before division.

*9+9=18*

The geometric mean is used to calculate growth. We would use geometric mean not when something is added to find a total, but rather multiplied to find a product.

The standard formula for geometric mean is *√(AB)*.

In other words, we are finding the square root when provided with two values, the cubed root when dealing with three values, etc. We can also apply the same question used for the arithmetic mean but replacing addition terms with multiplication terms.

Geometric mean is commonly used to calculate investment growth. Say you have investments that grow by three different values over three consecutive years. Your investments changed according to different interest rates (multiplication). You now want to replicate that exact pattern of growth. You'll have to find out what the interest rate would have to be every year to recreate this pattern using the geometric mean.

We won't be going into investment calculations, though, so let's jump back into geometry.

Remember, our base formula for geometric mean is *√(AB)*. It may help to draw out these problems on a piece of paper to visualize it better.

A rectangle is an irregular polygon -- its lengths are different. You want to figure out what the lengths of a rectangle's sides need to be to generate the same area as its regular square counterpart.

This question can be presented to you as: **What is the geometric mean of 4 and 16?**

On your rectangle, label *Length A (height) = 4*. Label *Length B (width) = 16*.

Plug these values into the formula.

*√(4*16)*

Solve the multiplication within the formula. You can also think of this as finding the area of your rectangle.

*√(64)*

Find the square root of 64.

*√64=√8*

Your geometric mean is 8! This means that to change your rectangle to a square in which the area remains the same, all four sides must be equal to 8. Let's prove this. Find the area of a square where lengths *A* and *B* are equal to 8.

*8*8=64*

The value 8, when multiplied twice, results in the original product of the formula.

*4*16=8*8*

Finding the geometric mean of a polyhedron involves adding a third number to the mix. Instead of dealing with the square root, we will be finding the cubed root.

The formula for geometric mean involving three numbers is *3√(ABC)*.

Let's start with an irregular polyhedron: a box. A box is a rectangle with an added dimension. Do you remember the formula for finding the volume (area in three dimensions)? It is *length x width x height*. You'll need this to properly label the box.

**What is the geometric mean of 1, 7, and 49?**

On your rectangle, label *Length A (length) = 1*. Label *Length B (width) = 7*. Label *Length C (height) = 49*.

Input these numbers in your formula.

*3√(1*7*49)*

Solve for volume.

*3√(343)*

Now, find the cubed root of 343.

*3√7*

To prove this, let's see if our irregular polyhedron's geometric mean will give us a regular polyhedron (a cube).

*7*7*7=343*

Therefore, *1*7*49=7*7*7.*

When put into terms of investment calculations, finding the geometric mean is a little daunting. It's better to practice with polygons and polyhedrons first and then advance to the larger concepts. Geometric means are nothing to be afraid of, though! Nothing in math should be feared.

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For example:

- -4 is an integer.
- 0 is an integer.
- 5.3 is not an integer.

It is worth noting that some people learn differently when it comes to whole numbers. Some do not believe that 0 is a whole number and that whole numbers automatically include negatives. If this criterion applies to you, simply adjust the definition of an integer provided here.

It’s that easy! It stays that easy when finding them in algebraic equations, too. This is because integers are simply elements added to mathematical equations. Because they are so identifiable, there’s no question when you come across one -- except maybe when it’s hidden in a variable.

An algebraic equation incorporates more advanced elements than your basic arithmetic equation. Also referred to as algebraic expressions, these make use of variables, integers and other numbers, and at least one algebraic operation.

**Example:**

*2x + 3y(15-3) = 30*

The variables in the equation are *x* and *y*. **Variables **stand in place of numbers for which we must solve. We do not yet know if the variable is an integer or not. Remember, integers do not include fractions and decimals.

The **coefficient** -- the number by which we are multiplying our variable -- for term *2x* is 2. For *3y* it is 3.

Our **algebraic operations** include addition (+), subtraction (-), and multiplication (*3y(15-3)...*).

However, we can identify integers in our equation right away. These are 2, 3, 15, and 30.

There’s only one way to determine if a variable is an integer or not, and that’s to solve the equation. Not every problem you’ll solve will provide you with variable values. In fact, you’ll most likely be creating your own algebraic equations where you need to solve for a variable in real life. This is especially true if calculating profit. You can use any variable to represent your value.

In this case, *a=$12,000 (total revenue)*,* b= $10,000.45 (total expense)*, and *c=profit*.

*a - b = c*

Insert your values into the profit equation above.

*$12,000 - $10,000.45 = c*

Solve for *c*.

*$1,999.55 = c*

In this case, *c* is not representing an integer. If your values are not whole numbers and include parts -- if there are cents with your dollars -- unless the parts negate each other, your profit will not be an integer.

Because we have to solve to find a variable that could be hiding an integer, let’s learn a simple way to solve for the variable. This equation will use like terms with one variable.

*14x + 13(7*6) - 6x = 610*

First, let’s combine the like terms in this equation: *14x *and *-6x*. Only when a variable is the same can you combine them.

*8x + 13(7*6) = 610*

Following Order of Operations, we solve what is in parentheses first, followed by multiplication.

*8x + 13(42) = 610*

*8x + 546 = 610*

Next, we need to start isolating the variable or removing the remaining numbers and the coefficient from the left side of the equation. The rule goes that anything done to one side of the equation must also be done to the other. In this case, we will subtract 546 from both sides.

*8x = 64*

Finally, to fully isolate the variable, divide *8x* by *8* and *64* by *8*.

*x=8*

According to this answer, our variable is an integer. Let’s prove our equation by placing the variable in the equation and solving for the result, 610.

*14(8) + 13(7*6) - 6(8) = 610*

*112 + 13(42) - 48*

*112 + 546 - 48*

*658 - 48 = 610*

Our integer works!

Finding integers in algebraic equations is all about knowing what is considered an integer or not. It also relies on your ability to solve for variables. Again, never assume variables are integers. There are plenty of cases when they aren’t.

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A mathematical equation, also known as a statement, shows that two separate expressions are equal. The two expressions are usually connected by the "equal-to" (=) sign. You can find the expressions in a mathematical statement by looking at the left-hand side and the right-hand side of the equation, using the = sign as the point of reference.

**Example**: *2x + 3 = 11*

Can you identify the expression(s)? Can you locate the equal-to sign?

It can be helpful to view equations as a scale in which weights are used to keep it balanced. The scale is balanced when what is placed on the right-hand side is equal in weight to what is on the left-hand side. Solving an equation can also be thought of in terms of adding weight to the left or right-hand side of the scale to balance it out.

Finding the solution to the equation involves manipulating the weights. Whatever is done on the right-hand side of the scale must also be done on the left-hand side to keep the scale balanced.

Your tools for this happen to be, most commonly, addition, subtraction, multiplication, or division.

The most popular equations are algebraic equations, in which letters are used as variables. Variables are placeholders that can assume any value. They can also be referred to as *unknowns*. You must solve for the correct value of the variable to make the equation true.

Let's look at an example.

*4x + 2x = 12*

The letter *x* is the *variable*. You must find the value of *x.* Finding this value will make the right-hand side of the equation equal to the left-hand side. Let's solve this together.

In this case, since both variables are the same, you can add the values on the left-hand side.

*6x = 12*

Now, we divide both sides by the value of '6', which will help us solve for x. Consider this to be a way to unravel the equation. Also, remember that whatever is done to one side must be done to the other.

*x = 2*

In this case, the value of *x* happens to be 2.

Go back into the original equation and substitute 2 in place of *x.*

*(4 x 2) + (2 x 2) = 12*

*8 + 4 = 12*

*12 = 12.*

We have successfully balanced the scale!

The beauty of mathematics is that it provides a logical and sequential means of solving problems. Sometimes, the word problems that we encounter can be solved if they are represented in mathematical notation. A great way to do this by writing the problems as algebraic equations. Once the problem is represented as an algebraic equation, it can then be solved.

The first key lesson is to use letters to represent the unknown quantities of the problem.

*Jack left his house in the morning with $12. On his way to school, he bought ice cream. He had $8 remaining when he reached school. How much did he pay for the ice cream?*

The amount Jack paid for the ice cream is unknown. This amount will be represented as *x.* We know that if we subtract the amount he paid for the ice cream from the money he left home with, we will have the amount of money he reached school with.

**Step 1**: Write the equation based on the relationships in the problem.

*12 – x = 8*

We can then find the solution to the equation by solving for *x*.

**Step 2**: Collect like terms by taking *x* to the right-hand side and bringing 8 to the left hand.

*12 – 8 = x*

Note that when a variable crosses the equality sign, the sign of the variable or number changes. Positive 8 turns into negative 8.

**Step 3:** Solve for *x*.

Follow operational rules to solve for the variable.

*x = 4*.

Jack paid $4 for the ice cream.

*Twice a number is 30.*

This equation is written as:

*2x = 30.*

We know that a number multiplied by two ("twice") equals 30. The unknown number is *x. *

*Bob’s age 5 years ago was half the age he will be in 8 years. How old is Bob now?*

This one is a bit trickier—the important thing when writing equations from word problems is to take it step-by-step. What we want to know is Bob’s age. It will be represented by *x.*

Bob’s age 5 years ago was *x – 5.*

Bob’s age 8 years from now is *x + 8.*

The problem states that his age 5 years ago was half the age he will be in 8 years.

*x – 5 = ½(x + 8)*

The equation also means “his age 5 years ago was half the age he will be in 8 years.”

The key to forming equations from word problems is by identifying the relationships between the various quantities in the problem. Your equation then brings together the relationships into a single math representation.

The key to writing your own equation in algebra is learning how to solve them. Like learning any language, this means you have to practice consistently. Try turning word problems into equations first. Then go ahead and make your own equation from scratch to solve. You may find that it ends up being great fun.

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There are some common methods used to simplify algebraic equations which are as follows:

- Combining like terms
- Factoring
- Expansion of equations that is the Distributive Property
- Multiplication or division of terms

We shall go through each of these methods step by step to truly understand how to make an algebraic equation simpler to solve.

First, we'll need to combine the like terms in the equation. This simply means putting together whichever terms are the same to shorten the equation.

For example:

17x + 3y – 9x would be simplified to 8x + 3y because we combined the x terms.

Or 3a -5b + 3ab + 7a = 22, which would be simplified to: 10a – 5b +3ab = 22.

Essentially, look for any like terms like a, y, e, or with any other unknown coefficients.

Secondly, we have the factoring method. This one is trickier as it involves a different knowledge of algebra operations, and it is also not technically simplified. The process only applies as a simplification method if you are solving long, combined algebraic equations. Simply put, we take a long equation and simplify it by factoring it back into a shorter equation. This makes it easier to use in other mathematical calculations and also easier to write out.

For example:

X2 - 2x - 3 would be simplified to:

(x - 3) (x + 1)

I know this one seems counterproductive, especially in the light of the next simplification methods. Still, it helps when dealing with larger equations in algebra and longer mathematical questions that have complex algebra potions. Knowing that one or two sub-equations in the lager equation can be easily divided into smaller parts helps one easily calculate the solutions for the larger equation.

The next method is also known as the Distributive Property. It is where you expand the equation by removing brackets and make it into a longer but easier to manipulate.

For Example:

5b (b – 6) + 4 = 10 would be simplified to

5b2 – 30b + 4 = 10 which is much easier to work with.

Finally, we have the multiplication and division of the terms. This simply means multiply what can be multiplied or divide what can be divided. This ensures you finally work with the equation with the smallest possible coefficients.

One example is this: (4x – 12y) ÷ 4 + 3 = 0

It would be simplified to this: (x – 3y) + 3 = 0

A second example would be this: 10a * 2a ÷ 4a = 20

It would be simplified to this: 20a2 ÷ 4a = 20

You will use all the above methods at some point or another as you solve equations in algebra. You often use them together and apply them to broader topics in mathematics that involve some algebra.

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So the square of the imaginary unit would be -1. Here's an example: *j2* = -1.

The square of an imaginary number, say *bj*, is (*bj*)2 = -*b*2. An imaginary number can be added to a real number to form another complex number. For example, *a *+ *bj* is a complex number with *a* as the *real part* of the complex number and *b* as the *imaginary part* of the complex number.

Complex numbers are sometimes represented using the Cartesian plane. The x-axis represents the real part, with the imaginary part on the y-axis. From this representation, the magnitude of a complex number is defined as the point on the Cartesian plane where the real and the imaginary parts intersect.

Care must be taken when handling imaginary numbers expressed in the form of square roots of negative numbers. For example:

Sqrt(-6) = sqrt(-1) * sqrt(6)

= sqrt(6)*j*.

However, this does not apply to the square root of the following,

Sqrt(-4 * -3) = sqrt(12)

And not sqrt(-4) * sqrt(-3) = 2*j * sqrt(3)j*

So when the negative signs can be neutralized before taking the square root, it becomes wrong to simplify to an imaginary number.

The nature of problems solved these days has increased the chances of encountering complex numbers in solutions. And since imaginary numbers are not physically real numbers, simplifying them is important if you want to work with them. We'll consider the various ways you can simplify imaginary numbers.

The imaginary unit, *j,* is the square root of -1. Hence the square of the imaginary unit is -1. This follows that:

*j0 =*1*j1*=*j**j2 =*-1*j3 = j2*x*j =*-1 x*j*= -*j**j4 = j2*x*j2 =*-1 x -1 = 1*j5 = j4*x*j =*1 x*j = j**j6 = j4*x*j2 =*1 x -1 = -1

Understanding the powers of the imaginary unit is essential in understanding imaginary numbers. Following the examples above, it can be seen that there is a pattern for the powers of the imaginary unit. It always simplifies to -1, -*j*, 1, or *j*. A simple shortcut to simplify an imaginary unit raised to a power is to divide the power by 4 and then raise the imaginary unit to the power of the reminder.

For example: to simplify *j*23, first divide 23 by 4.

23/4 = 5 remainder 3. So *j23 =* *j3 = -j *…… as already shown above.

Consider another example

(*2j)6* = 26 x *j6* = 64 x -1 = -64

Simply put, a conjugate is when you switch the sign between the two units in an equation. The conjugate of a complex number would be another complex number that also had a real part, imaginary part, the same magnitude. However, it has the opposite sign from the imaginary unit.

For example, if *x* and *y* are real numbers, then given a complex number, *z = x + yj*, the complex conjugate of *z* is *x – yj.*

Complex conjugates are very important in complex numbers because the product of complex conjugates is a real number of the form *x2* + *y2*. They are important in finding the roots of polynomials.

To illustrate the concept further, let us evaluate the product of two complex conjugates.

(*x + yj*)(*x – yj*)

= x2 – xy*j* + xy*j* – y2*j2*

= x2 – (-y2)

= x2 + y2

**Example**

*Simplify the expression* 2 / (1 + 3*j*)

The above expression is a complex fraction where the denominator is a complex number. As it is, we can't simplify it any further except if we rationalized the denominator. The concept of conjugates would come in handy in this situation.

When dealing with fractions, if the numerator and denominator are the same, the fraction is equal to 1.

Hence (1 – 3*j*) / (1 – 3*j*) = 1

Also, when a fraction is multiplied by 1, the fraction is unchanged. So we will multiply the complex fraction 2 / (1 + 3*j*) by (1 – 3*j*) / (1 – 3*j*) where (1 – 3*j*) is the complex conjugate of (1 + 3*j*).

(2 / (1 + 3*j*)) * ((1 – 3*j*) / (1 – 3*j*))

= 2(1 – 3*j*) / (1 + 3*j*)(1 – 3*j*)

The denominator of the fraction is now the product of two conjugates. As stated earlier, the product of the two conjugates will simplify to the sum of two squares.

Hence

2(1 - 3*j*) / (1 + 3*j*)(1 – 3*j*) = 2(1 - 3*j*) / (12 + 32)

= 2(1 - 3*j*) / (1 + 9)

= (2 - 6*j*) / 10

= 0.2 - 0.6*j*

We've been able to simplify the fraction by applying the complex conjugate of the denominator.

Complex numbers can also be written in polar form. The earlier form of *x* + y*j* is the rectangular form of complex numbers. Given a complex number z = x + y*j*, then the complex number can be written as z = r(cos(*n*) + *j*sin(*n*))

Where r = sqrt(x2 + y2)

*n* = arctan (y/x).

De Moivre’s theorem states that r(cos(*n*) + *j*sin(*n*))p = rp(cos(p*n*) + *j*sin(p*n*))

Here's an example that can help explain this theory.

*Given z = 1 + j, find z2*

Let us convert the complex number to polar form.

r = sqrt(12 + 12) = sqrt (2)

*n* = arctan (1/1) = 45o

So *z* in polar form is z = sqrt(2)(cos(45) + *j*sin(45)).

Z2 = sqrt(2)(cos(45) + *j*sin(45))2.

You can see what happens when we apply De Moivre’s theorem:

sqrt(2)(cos(45) + *j*sin(45))2 = (sqrt(2))2(cos(2 x 45) + *j*sin(2 x 45))

= 2(cos(90) + *j*sin(90))

= 2(0 + *j*) or = 2*j*

So z2 = (1 + *j*)2 = 2*j*

You can verify the answer by expanding the complex number in rectangular form.

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Ratios express the quantitative relationship between values by showing how many times one is contained in another. Percentages are similar to ratios, but they're expressed as a fraction of 100.

Percentage point, on the other hand, is a unit of measurement that represents the difference between two percentages.Percentage point is often represented as "pp" or "p.p." In other books, it is written in full as “percentage points” or “percent points” to avoid confusing it with “pages.”

Percentage point is used in a wide range of applications, ranging from probability studies to fault/ risk analysis to descriptive statistics to investment predictions. Many statisticians and analysts prefer the use of percentage points to percentages when comparing amounts. Some examples to illustrate the percentage point calculations are given below:

In 2018, at St. Andrew’s Catholic Boys’ School, forty percent (40%) of students passed the A levels mock examinations that were administered by the state. In 2019, in the same school, sixty-five percent (65%) of students passed the same examinations, and in 2020, 69% were successful. What is the change in the examination success rates between 2018 and 2019?

To calculate the percentage point, follow these steps:

**Step 1**: Highlight the percentages you need to compare. In this case, we have 40% (2018) and 65% (2019)

**Step 2**: Subtract the smaller percentage from the larger percentage

*Change in exam success rates = 65 - 40 = 25*

**Step 3**: Add your unit (percentage points): The difference in exam success rates between 2018 and 2019 is 25 percentage points.

In the example above, we can say that there was an increase in the success rate by 25 pp.

In texts and even in presentations, percentage and percentage points are often interchanged. However, using one in place of the other can give the information being conveyed a completely different meaning.

To illustrate this, imagine if the government increases taxes from 4% to 6%, that’s a 50 percent tax increase but a 2-percentage point tax increase. Example 2 shows how similar calculations can be done.

In a survey of road accidents in Durban, South Africa, it was discovered that during the rainy season of 2006, eighty percent (80%) of car accidents involved men. The same survey showed that in the dry season in 2007, that figure dropped to sixty percent (60%). Compare the changes in percentage and percentage points.

**Step 1**: Highlight the percentages you need to compare. In this case, we have 80% (rainy season) and 60% (dry season).

**Step 2** (For the percentage): Subtract the smaller percentage by the larger percentage and divide the quotient by the larger percentage:

*Percentage change = ((80 - 60)/80) * 100 *

This gives us: *(20/80) * 100 = 25*

**Step 3**: Add your unit (%). There was a 25% decrease in the number of accidents involving men during the dry season.

**Step 4**: (For the percentage point): Subtract the smaller percentage from the larger percentage

*Change in number of accidents involving men = 80 – 60 = 20*

**Step 5**: Add your unit (percentage points): The difference between the accidents involving men in the rainy season and the dry season is 20 pp.

From the results above, we can see that the percentage difference is 25%, while the percentage point is 20 pp.

To derive percentage points from ratios, the rations have to be converted to percentages, which can be subtracted to obtain the percentage points. This is further explained in the example below.

In January 2020, the P.E teacher, Mr. Percy, started a Yoga class for teenagers aged between 12 and 17. The ratio of boys to girls in the class was 3:7. The total number of children that signed up was 20. Unfortunately, schools had to be closed down due to a pandemic.

After the pandemic, only half of the students that signed up initially returned. Of this lot, the ratio of boys to the total number was 2:5. Find the changes in the number of boys and girls using percentage points.

**Step 1**: Convert the ratios to percentages

*Combining the ratio, we have 3 + 7 = 10*

Therefore, there are *(3/10) * 100% boys and (7/10) * 100% girls.*

This gives us 30% boys and 70% girls.

If the ratio of boys to the total number is 2:5, then the percentage of boys is given as *(2/5) * 100% = 40%*

Therefore, there are *40% guys left and 60% girls left*

**Step 2: **Highlight the percentages you need to compare. In this case, we have the following:

- For boys: 30% (Before the pandemic) and 40% (After the pandemic)
- For girls: 70% (Before the pandemic) and 60% (After the pandemic)

**Step 3**: Subtract the smaller percentage from the larger percentage

- For boys:
*Change in the percentage of boys = 40 – 30 = 10 (increase)* - For girls:
*Change in the percentage of girls = 70 -60 = 10 (decrease)*

**Step 4**: Add your unit (percentage points): The difference in the percentage of boys at the Yoga class before and after the pandemic is 10 pp (increase).

The difference in the percentage of girls at the Yoga class before and after the pandemic is 10 pp (decrease).

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For others, it’s merely a source of confusion or unnecessary stress. Due to their fundamental application to calculus, a misunderstanding of derivatives can also lead to unnecessarily lower grades and stressed students.

When read properly, this article can alleviate some of your concerns with a proper explanation of derivatives and their applications. It is important to note that these are general overviews, and watching video examples on specific rules or methods can allow you to apply what you’ve learned more efficiently.

Before finding the derivative, it will be helpful to define and thoroughly understand what a derivative is. Simply put, the derivative is the slope. More specifically, it is the slope of the tangent line at a given point in a function. To make this more understandable, let’s look at the function f(x) = x^2 at the point (1, 1) on a graphing calculator.

The function is graphed as a U-shaped parabola, and at the point where x=1, we can draw a tangent line. This means that at (1, 1), we can draw a line that touches only this point and is below the curve on either side of this same point. The slope of this line (which is 2) is actually the derivative at that given point.

To actually find this value of the derivative, there are two main methods.

This method relates to a conceptual understanding of the derivative. If you let the x-axis difference between two points on a curve equal h, this definition of the derivative can be derived and explained in further detail. For the purposes of this explanation, things are simplified with a statement of the formula.

f’(x) = dy/dx = lim as h→0 of [f(x+h) - f(x)] / h

It is really a representation of 'rise over run' or the slope between two points, where the x-axis value between the two points is a, and the distance between the two points is approaching 0.

Example: Find the derivative of f(x) = x^2 at (3, 9).

f’(3) = dy/dx= lim as h→0 of [f(3+h) - f(3)] / h = lim as h→0 of [(3+h)^2 - 9] / h

f’(3) = (9 + 6h + h^2 - 9) / h = 6

This method is a lot more methodical, and can be used more generally to find the slope at any given point. It does, however, require understanding of several different rules which are listed below.

dy/dx = nx^(n-1)

This rule finds the derivative of an exponential function.

Example: dy/dx = 4x^3

dy/dx[f(x) x g(x)] = f(x)g’(x) + f’(x)g(x)

This rule finds the derivative of two multiplied functions.

Example: dy/dx[(3x^2)(x^4)] = (3x^2)(4x^3) + (x^4)(6x) = 12x^5 + 6x^5

dy/dx[f(x)/g(x)] = [g(x)f'(x) - f(x)g'(x)] / [g(x)]^2

This rule finds the derivative of divided functions.

Example: dy/dx = [(3x^2)(4x^3)-(x^4)(6x)]/(3x2)^2 = (2x^5)/(3x^4)

dy/dx[(f(g(x))] = f’(g(x))g’(x)

This rule finds the derivative of two functions where one is within the other. It is frequently forgotten and takes practice and consciousness to remember to add it on.

Example: f(x) = x^4 g(x) = 3x^2

dy/dx[(f(g(x))] = f’(3x^2) x 6x

dy/dx[cx^n] = cnx^(n-1)

The derivative of the constant multiple is always just the constant multiple.

Example: dy/dx[6x^3] = 6 x 3x^2 = 18x^2

After establishing how to find the first derivative, the second derivative comes fairly easily. The second derivative is simply the derivative of that initial derivative. For example, if the function was f(x), and you took the derivative, the first derivative would be f’(x). The second derivative would be the derivative of f’(x), and it would be written as f’’(x).

Curvature can actually be determined through the use of the second derivative.

When the second derivative is a positive number, the curvature of the graph is concave up, or in a u-shape. When the second derivative is a negative number, the curvature of the graph is concave down or in an n-shape.

It’s easier to understand this through an example. The applications of derivatives are often seen through physics, and as such, considering a function as a model of distance or displacement can be extremely helpful.

If there is a function graphing the distance of a car in meters over time in seconds, the speed of the car is going to be distance over time or the slope of that function at any given point. You will come to realize that the speed of this car is essentially the first derivative.

Now let’s consider the second derivative. Since the first derivative models how fast the function is changing, the second derivative models how fast the first derivative is changing.

An easier way to look at this is with the acceleration of the car. The acceleration of the car shows how fast the speed or the first derivative of the car is changing. Acceleration is, therefore, a good example of the second derivative.

When acceleration is positive, this means that the speed at which the car is increasing speed is increasing. On a graph of the distance, this appears in the u-shape, which we can describe as the concave up curvature.

When acceleration is negative, this means that the speed at which the car is increasing speed is decreasing. On a graph representing the distance traveled, this would instead appear as an n-shape, which represents the concave down curvature.

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Regardless, an operator in algebraic terms tells us what type of calculation to perform on an expression involving two numbers.

Algebra is a branch of mathematics involving the study and manipulation of symbols using defined laws. These laws apply to almost all branches of mathematics.

Algebra's unique feature is that it uses letters and other symbols to represent numbers. These are called **variables**. The use of variables permits the assignment of different numerical values depending on the situation.

For example, in the expression, *x* + 2, the letter *x* can assume any value. Let's assign it the value of 5. The sum of the expression is, therefore, 7.

The operator in this expression is “+”. It defines how *x* and 2 should be manipulated. In this case, we add the variable with 2.

There are situations in algebra when more than one operator is used in an expression. We must now determine which operation to do first.

Here is an example of this instance:

*x + y * 6*

Let's now assume* x* = 2 and *y* = 3. Your possible solutions could be:

Case 1: *2 + 3 * 6 = 5 * 6 = 30*

Case 2: * 2 + 3 * 6 = 2 + 18 = 20*

The math in both cases is correct, but according to **Operator Precedence**, also known as **Order of Operations**, only one case is the right answer.

If we used the above mathematical expression to represent a process or some other phenomenon, another person who uses the expression might arrive at a different result. Without a law to the expression, it opens it up to inconsistent usage.

To resolve this problem, mathematicians all over the world agreed on which operators should have precedence over others.

Anything in parentheses and brackets is calculated first, regardless of the operator inside. Exponents and square roots are calculated second. Multiplication and division hold precedence over addition and subtraction.

If you are in the US, you can remember this order by the acronym **PEMDAS **(Parentheses, Exponents, Multiplication, Division, Addition, Subtraction).

In other English speaking countries, including Nigeria, they use the variation **BODMAS** (Bracket Of Division, Multiplication, Addition, and Subtraction).

Convention also states that if operators are of the same precedence -- as is the case with multiplication/division and addition/subtraction -- then the expression should be evaluated from left to right.

The rules of Operator Precedence apply even if variables are used instead of numbers.

Step 1: 6 + 3 * 4 + 8

Step 2: 6 + 12 + 8 = 26

Multiplication was done before addition. Note on the second step of the evaluation that addition was the only operator. Based on the commutative nature of addition (6 + 3 = 3 + 6 = 9), an expression involving only addition can be done in any order.

Step 1: 2 * 6 / 3 – 3

Step 2: 12 / 3 – 3

Step 3: 4 – 3 = 1

Both multiplication and division are at the same precedence, so the expression is evaluated from left to right.

Step 1: 2 * 2^3 + 11

Step 2: 2 * 8 + 11

Step 3: 16 + 11 = 27

The exponential is evaluated first before the multiplication.

Step 1: (3 + 2) ^ 2

Step 2: 5 ^ 2 = 25

Here, brackets are used to break the convention by forcing the addition operator to be evaluated before the exponentiation.

Step 1: [5 – (8 - 5)] * 2

Step 2: (5 - 3) * 2

Step 3: 2 * 2 = 4

Where the brackets are nested, a different type of brackets may be used, such as square brackets or curly braces.

There are some situations when you may be unsure of the order in which you should solve a multi-operator equation. Despite your knowledge of operator precedence, when you're given a lot of data, it can throw you off. Here are some common instances you may encounter:

3 ^ 2 ^ 2 = 3 ^ 4 = 81.

In cascading exponentials, the cascaded powers are evaluated first.

100 / 10 / 2

The correct order of division is from left to right.

10 / 2 = 5

If done the other way, it will be 100 / 5 = 20.

Division is not commutative. However, multiplication is commutative and will yield the same answer regardless of the direction of operation.

There are cases when multiplication is not explicit, such as 2 / 3*x. *This expression should be evaluated as 2 / (3*x*). The implied multiplication in the denominator takes precedence over the division.

Knowing the correct order of operator precedence is important in various fields such as computer programming, physics, and all areas of mathematics. Fortunately, it is easy to learn this useful tool.

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The simple answer is: yes you can get negative numbers out of square roots. In fact, should you wish to find the square root of any positive real numbers, you will get two results: the positive and negative versions of the same number.

Consider the following:

16 = 4 * 4 = (-4) * (-4)

In the equation above, you can either multiply 4 by itself or multiply (-4) by itself to get the result of 16. Thus, the square root of 16 would be:

The √ symbol is called the **radical symbol**, while the number or expression inside the symbol—in this case, 16—is called a **radicand**.

Why would we need a ± symbol in front of 4? Well, as we have discussed before, the square roots of 16 can be either 4 or (-4).

Most people would simply write this equation simply as . While it is technically true and there's nothing wrong with it, it doesn’t tell the whole story.

Instead, you can also write the equation in such a way that it explicitly indicates that you want both the positive and negative square root’s results:

This way, other people can easily tell that the one who writes the equation wishes to have positive and negative numbers as the result.

As you may know, 16 is a **perfect square**. Perfect squares are radicands in the form of an **integer**, or a whole number, that has a square root of another integer. In the example above, 16 is a perfect square because it has the number 4 as its square root.

Positive real numbers are not always perfect squares. There are also other numbers such as 3, 5, or 13 that are referred to as **imperfect squares**. If a radicand is not a perfect square, then the square root of the radicand won’t result is an integer. Take a look at the equation below.

5 is not a perfect square. Therefore, its square root won’t be an integer. Additionally, the square root is not even a rational number. **Rational numbers** are numbers that can be expressed as fractions composed of two integers, e.g. 7/2, 50/4, and 100/3.

The square root of 5 is an **irrational number** since it can’t be expressed as fractions. The numbers right of the decimal 2.236067 … would continue on endlessly without any repeating pattern. Still, both irrational and rational numbers are part of real numbers, meaning they have tangible values and exist on the number line.

We mentioned earlier that any positive real numbers have two square roots, the positive one and the negative one. What about negative numbers and zero?

For zero, it only has one square root, which is itself, 0.

On the other hand, negative numbers don’t have any real square roots. Any real number—whether it’s positive or negative—that is multiplied by itself is always equal to a positive number, except for 0. Instead, the square root of all negative numbers is an **imaginary number**.

or

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By definition, the square root of (-1) is *i*, which is an imaginary unit. As a side note, imaginary numbers do not have a tangible value. They are not part of **real numbers** in the sense that they can’t be quantified on the number line. However, they are still used in math and the study of sciences including quantum mechanics, electricity, and more.

To get a better understanding, let’s take a look at an example. For instance, let’s say we want to identify the square root of (-9), what would it be?

Therefore, is equal to 3*i*. This result can also be written as

How about the square root of (-3)?

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