Monthly Archives: February 2018

Week 4 – Addition, Subtraction, Multiplication and Division of Radicals

In Grade 9, we looked at the concept of grouping like terms.

Grade 9; 5x + 6y – 3x + 9y + x

x = 3x

y= 15y

Now in Grade 11, we now group radicals.

Grade 11; 4\sqrt[3]{9}, 6\sqrt[3]{9}, 15\sqrt[3]{9}

= 25\sqrt[3]{9}.

Rule

Radicand and index must be the same.

Subtraction

\sqrt{160} – \sqrt{40}

= \sqrt{16\cdot10} – \sqrt{4\cdot10}

= 4\sqrt{10} – 2\sqrt{10}

= 2\sqrt{10}

 

Division and Multiplication Rule

The denominator CANNOT be a radical

 

Rationalizing the Denominator

\sqrt{5} /    x    \sqrt{6} =  \sqrt{30} /   ——->       \sqrt{30} /

\sqrt{6}       x    \sqrt{6} =  \sqrt{36}      ——->        6

 

Rationalizing the denominator is a way of cancelling out the radical so the denominator is not a radical.

1.) Recognize that the denominator is a radical

2.) Multiply both the numerator and the denominator (what you do the bottom you got to do the top)

3.) Simplify to find the new values

4.) See if there are any perfect squares (if the index is for example 3, then look for perfect cubes)

5.)Simplify the new values

6.) MAKE SURE A RADICAL IS NOT IN THE DENOMINATOR PLACE.

Week 3 – Absolute Value, Roots

This week we learned some new concepts and reviewed concepts from grade 10.

2.1 Absolute Value – The principal square root of the square root of a number.

ex. \mid{-7}\mid      absolute value = 7

the absolute value is the amount that the given number is different from 0.

Note – The radicand can only be positive

\mid{7-16}\mid

7 – 16 = -23

absolute value = 23

-23 & 23, are both 23 digits away from 0.

2.2 Roots and Radicals

Perfect Squares 4,9,16,25,36,49,64,81,100

Perfect Cubes 8,27,64,125

Simplify

\sqrt [3]{40}

=\sqrt[3]{8\cdot5}

=2\sqrt[3]{5}

 

 

 

 

Week 2 – Geometric Sequences

This week in math we saw some questions that displayed percentages.

For example. 

Billy a wanted to estimate the value of his car (New) after several years of use. The car was worth $34 000 new and depreciates in value by 10% each year. What will be the value of his car at the end of 1 year of use and 5 years of use?  

Analysing the Question.

Value of New Car = 34 000

t_{1} = 34 000 x  100% – 10% (Starting value x years of depreciation)

t_{1} = 34 000 x .90 = 21 420
de·pre·ci·ate
dəˈprēSHēˌāt/
verb
  1. diminish in value over a period of time.

5th year 

Formula = t_{5} = ar^{n-1}

Take t_{1} r ^{n-1}

34 000 (.90) ^ 5-1 = 20 075.66