Week 4 – Adding and Subtracting Radical Expressions

A radical expression that contains positive terms and negative terms have ways to be further simplified.

When a radical in an expression like such is presented: √4a + √16a – √9a, a ≥ 0

This is a radical expression because there is no equal sign and it includes radicals.

Since this expression has + and – symbols, we are able to combine like terms.

Combining like terms is done when the index of the root (4∛8, 3 is the index) is the same. The coefficient (4∛8, 4 is the coefficient) is the number that changes when the terms are combined and the radicand (4∛8, 8 is the radicand) stays the same.

Lastly, we need an understanding of simplifying radicals to help us simplify terms to create the same radicand and combine them.

√4a can be simplified to 2√a because √4 = 2

√16a can be simplified to 4√a because √16 = 4

– √9a can be simplified to – 3√a because √9 = 3, 3 x – = -3

√4a + √16a – √9a

2√a +  4√a – 3√a

Since the index and the radicands are the same, we are able to combine them to simplify the expression

3√a, a ≥ 0

 

Another example would be: 5e√24e³ – 7√54e⁵ + e²√6e + 6e, e ≥ 0

Each term is first simplified by finding the perfect square inside the radical and removing it. When it is removed from the radical, it is also rooted by the index (2√12
2√4⋅3
2√4⋅√3
2(2)√3
4√3)

5e√24e³ – 7√54e⁵ + e²√6e + 6e, e ≥ 0

5e√4⋅6e²⁻¹ – 7√9⋅6e²⁺²⁺¹ + e²√6e + 6e

5(2)e¹⁺¹√6e – 7(3)√6e²⁺²⁺¹ + e²√6e + 6e

10e²√6e – 21e²√6e + e²√6e + 6e

Then, since the index and the radicand is the same, the terms can be combined and simplified

-10e²√6e + 6e, e ≥ 0

 

Week 3 – Absolute Values

The absolute value of a real number is defined as the principal square root of the square of a number.

The absolute value is the number of spaces from a number to 0.

The absolute value symbol is \mid x \mid where x = real number

The principal square root is when the square root is given and the answer is positive. For example √25 = 5

So when we look for an absolute value of a number (that number being the one inside the absolute value symbol), we look for the number squared and then square rooted.

For Example : The absolute value of $latex \mid 5 \mid$

\mid 5 \mid = √5²

√25

5

Therefore the absolute value of the number 5 is 5

 

 

Week 2 – Geometric Sequence

This week we learned how to determine t_n in a geometric sequence.

Using the equation t_n = t_1r^n-1, we can find any term in the geometric sequence.

 

For example: The geometric sequence is 2, 8, 32…

 

We can find r, which is the variable for the ratio, by taking the second term and dividing it by the first.

Therefore \frac{8}{2} is the ratio, or 4

Now we know that t_1 or the first term is 2

 

So now we can fill in the equation. Let’s find the 7th term.

t_7 = 24^7-1

t_7 = 24^6

t_7 = 2(4096)

t_7 = 8192

 

This equation is very helpful to finding the t_n but it can also be helpful if you need to find another variable and you already know two.

Imagine we already knew t_7 and we want to find the first term. Using the equation, we can solve for t_1

8192 = t_14^7-1

8192 = t_14^6

8192 = t_14096

Now you divide both sides by 4096

2 = t_1

 

 

 

1.2 L’immigration Chinoise et la Perspectique Historique

Voici l’enregistrement de Nighina et Clare. Nighina a choisi de présenter la perspective un professeur européen dans une des écoles affectées et Clare a répondu avec la perspective d’une enfante chinoise. Ensuite, Clare a choisi de présenter la perspective d’une enfante chinoise et Nighina a répondu avec des questions d’une perspective d’un professeur européen.

Week 1 – My Arithmetic Sequence

Arithmetic Sequence 2, 9, 16, 23, 30…

Using these five terms, we are going to solve for  t_{50}, t_{n}, and S_{50}

 

The formula t_{n} = t_{1} + (n – 1)d will help us solve for t_{50}. d stands for the difference which is 7. We then are able to fill the formula in.

t_{n} = t1 + (n – 1)d

t_{50} = 2 + (50 – 1)7

t_{50} = 2 + 49(7)

t_{50} = 2 + 343

t_{50} = 345

 

To find the general form of t_{n} we use the same formula from above and just fill in what we know which is t_{1} and d.

t_{n} = t_{1} + (n – 1)d

t_{n} = 2 + (n – 1)7

Then you distribute the 7.

t_{n} = 2 + 7n – 7

t_{n} = 7n – 5

 

Finding s50 uses the formula \frac{n}{2} (t1 + tn).

s50 = \frac{50}{2} (2 + 345)

s50 = 25(347)

s50 = 8675