Tag Archives: Math11

Week 14 – Multiplying and Dividing Rational Expressions

In earlier previous grades we learned to solve and simplify rational and irrational expressions and equations. A lot of people panic when they see expressions that include fractions, myself included. But it doesn’t need to be hard.

First, vocabulary learned in previous years.

  • Denominator
  • Numerator
  • Multiplying the reciprocal
  • Simplifying
  • Equivalent expressions
  • Factoring

All this vocabulary and concepts are still useful. But now we are including fractions!

By now we have learned how to factor expressions, if not remember

Common

Difference of squares

Pattern

Easy

Ugly

Example;

There is a new concept that relates to previous a pre-calculus unit similar to restrictions, known as non-permissible values. Non-permissible are basically x values that make the expression untrue or make it equal to nothing. If you did 6/0, you wouldn’t get an answer

Week 12 – Solving Absolute Value Equations Algebraically

A couple of weeks ago we learned how to solve absolute value equations by using algebra.

 

Verifying is how you can check for extraneous roots (Roots that give you a different answer). Make sure you remember it’s in absolute value form, so there’s never any negative values, if there’s ever any negative  you know that there are no solutions. If the left side does not equal the right side it’s an extraneous root.

Week 13 – Graphing Reciprocal Quadratic Functions.

Last week we learned about how to graph reciprocal functions.

The steps to solve.

1.) Graph the parabola

2.) Locate the invariant points

3.) Find the asymptotes

 

Three types of Quadratics

Positive Slope, Positive y-intercept

 

It never touches (0,0)

so no invariant points

 

Positive slope (can be negative), y-intercept=0

The asymptotes are touching the parabola

Positive or Negative Slope, opposite y-intercept

 

 

 

 

Week 9 – Modelling Problems with the Quadratic Equation.

Find two integers with a sum of 36 and the greatest possible product.

Find the variables

I personally like to use the variable “x”, and since there are two integers that means there are two integers. That means the second variable is (36 – x). For example, if the first variable was 4 and using 36-x, would result in 36 – 4, that means the second variable would be 32.

Variables  

1st x= x

2nd x= 36-x

when it says the greatest possible product, that means that we’re looking for the maximum value.

We put the variables in the factored form.

(x) (36 – x)

1st x = 0

2nd x = 36

Now we need to find the vertex

(1st x) + (2nd x) / 2

0 + 36 /2= 18

(18,_)

Then we input the new “x” value

(x) (36-x)

(18) (36-18)

(18) (18)

=324

So the answer is (18,324).

 

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