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# Math Review

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on 17 June 2013

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#### Transcript of Math Review

Unit # 1: Linear Equations
We started this unit by reviewing order of operations.
The order of operations
Parenthesis
Exponents
Multiplication or Division
Example:
2+3*4
Since there are no parenthesis or exponents, we proceed to multiply 3*4
2+12
Then the last step is to add because there is no other operation left.
We learned how to solve one and two step equations.
What are Linear Equations?
Linear equations contain at least one variable (Variable can only be raised to the power of 1)
How to solve linear equations
6m+7=31
Divide both sides of the = sign by greatest common denominator (-7, to take care of the sign change) of both sides of = sign.
m=
x6
+7
31
-7
/6
24
4
You can use either method for one or two step equations
Visual Method (Two step equation)
Draw two or three boxes depending on the equation (One step= 2 Two step=3) and connect them using arrows top and bottom. Then write the answer on the last box. Write the mathematical operations the equation let you reach the answer on the top arrows. Then do the opposite operation of the top arrows on the bottom arrows. (X=/, +=-, and vise-versa) Then do the mathematical operation of the answer, and put each answer on each box respectively. The integer on the first box is your answer
Linear equations have countless real world applications. From maximizing your money on the grocery store, to making a rover land on Mars.
Unit # 2: Graphing
Example
One common example is, if we had a amount of dollars and we wished to convert that amount into Euros. We would use a linear equation to calculate the amount.
D= US dollars
E=Euros
D*0.7749=E
We started this unit by reviewing graphs and inequalities symbols.
Inequalities symbols:
< Greater than
> Less than
≥ Greater then or equal to
≤ Less than or equal to
≠ Not equal
Graphs:
Then we learned how to solve inequalities and plotting them.
Linear relationships
Graphing
Systems of equations
Exponents
Pythagoras Theorem

Solving Simple (linear) inequalities:
Equation:
Inequality:
j-51=-139
+51
+51
When solving simple inequalities, one starts with a linear equation.
j=-88
After we solve the equation, we proceed to change it into a inequality by inserting a inequality symbol.
j=-88
j-51

-139
+51
+51
Plotting inequalities on a Number line:
If the symbol is (≥ or ≤) then you fill in the dot ( ) on a number line
if the symbol is (> or <) then you do not fill in the dot ( ) on a number line.
.
j

-88
Example :
x≥2
Since the symbol is greater than or equal to, we would plot the inequality in a number line with a fill in dot.
Since the inequality is greater than or equal to two, we plot it in a number line with a arrow indicating that it is larger than two and the dot, indicates it is larger than or equal to two.
Afterward, we revised how to plot points on a graph and analyzing them.
Plotting points on a graph:
If you are plotting points, the first thing you must know is that they come in pairs. (x,y) These pairs correspond to different quadrants in a graph.
Examples:
1.(2,4)
2.(3,-2)
3.(-4,3)
4.(-3,-4)
.
.
.
.
1
2
3
4
Finally, we learned how to analyze them using many different forms.
Point-Slope form
There are many ways to represent gradient of two points. The simplest one is to compare the vertical distance over the horizontal distance by using the point-slope form. We rearrange both coordinates : (x1,y1), (x2, y2) into the formula below, to represent the gradient in vertical distance traveled over horizontal distance traveled.
x2-x1
y2-y1
Y=mx+b
In the y=mx+b, m is the slope and b is the y intercept. We can use these equations to represent graphs and also to create them.
Example: (describing a line)
Slope: 2/1
Y-int.: 2
y=2/1x+2
Example:(plotting a line)
y=3/1x-1
Slope: 3/1
Y-int.:-1
Graphs let us represent and analyze data. For example if we wished to know the fuel consumption over a amount of time, we could represent it by using any form we discussed earlier and use it so we can display it on a graph.
Unit # 3: Systems of equations
The first thing we learned was the different types of solutions.
Types of solutions
1 solution: When there is a point of interception
No solutions: Parallel lines
Infinite solutions: overlapping lines.
Then we learned the different methods to solve where are two lines going to intercept.
Trial and error method
On the trial en error method, we do a table with each equation, and we use trial and error to find the point of intersection.
Example:
x+y=4
x-2y=-2
Ax+By=C
This is called the standard (or general) form. In this form, A is equal to the slope of a line (multiplied to get a whole #), B is the # multiplied (to get the slope into a whole number) and C is equal to the y int. multiplied (by that whole #)
Example:
y =-2/3x+4
+2/3x +2/3x
2/3x+y=4
*3
2x+3y=12
After you do the table, you insert the values for X and Y to check which values follow the equations. In this case, it is (2,2), therefore, the lines intercept at (2,2)
Graphing Method
On the graphing method you can only use them on one solution problems. When using this method, we graph both lines from each equation and then we can check our answers.
2+2=4
2-2(2)= -2
Example:
A.y=2x+5
B.y=-x-1
A
y int. 5
y int.-1
B
point of interception

Checking: (replace x variable with the x of point of int.)
A.1=2(-2)+5
B.1=-4+5

Solution by substitution
When using solution by substitution, we can use them in all three types of solution problems. Since y is the subject of the y=mx+b we can compare both mx+b of both of the equation involved
Example:
y=-x+4
y=2x-8
-x+4=2x-8
+x +x
4=3x-8
+8 +8
12=3x
3
x=4
y=0
Solution by elimination
This method can also solve problems with all three types of solutions. The problem has to be in the form of Ax+By=C. However, the symbols of Ax or By have to have opposite symbols.
Example:
3x+2y =6
1x-2y =10
3x+2y=6
+ 1x-2y=10
4x+0=16
4x=16
4
x=4
3*4+2* =6
12+2* =6
-12 -12
2*-3=6
y=-3
Systems of equations play a huge role in real life. One example is if we wished to know how many people went to a even and we know the amount that was charged to adults and children, and the amount that was raised. We could figure out the amount of people that went using any of the methods I have explained.
Unit # 4: Exponents and Radicals
On this unit, the first thing we learned was the laws of exponents.
Laws of Exponents
1. When multiplying exponents, add the indecies. Ex: p
12
*p
3
= p
15
2. When dividing exponents, subtract the indecies. Ex: x
36
/x
6
=
x
30
3. For powers of a power, multiply the exponents. Ex: (x
2
)
4
= x
8
4. When the exponent is zero, the answer is 1. Ex: 1,000,000
0
= 1
Then we learned how to write scientific notation from common numbers.
Scientific Notation
To write scientific notation, the integer must be displayed as a # between 1 and 10. Then we multiply this number to 10 to a power.
Ex: 1,580,000
1.58*106
Exponents are yellow.
Afterward, we were first introduced to radicals and how are they connected to square #.
A radical is also called a square root. Radicals are completely opposite to a square #.
Example: 5*5=25, 5
2
=25
Therefore, the square root of 25 is
√25= 5
When the # of a radical aren't square numbers, we can symplify them into the simplest radical form
Ex: √50= √2 *√25= 5√2
Exponents and radicals are connected to the world in many ways. One of them is that if we wanted to calculate the interest rate in a home using exponents. The formula would be F = P (1+i)^n , where F is the future value and P is the present value, i is the interest rate and n is the number of years.
Unit # 5: Pythagoras
On this unit, the first thing we learned was what was Pythagoras theorem.
Pythagoras Theorem
In a right angled triangle, the hypotenuse is the longest length and the other lengths are a,b
a
2
+b
2
=c
2
Then we learned how to manipulate the equation to solve different parts of a right angled triangle.
Finding Hypotenuse and
arm length
To find the hypotenuse, we add a and b and square root the result into the simplest radical form.
Ex:
5
2
+3
2
= C
2
25+9=34
To find the length of any of the two arms we subtract c
2
-b
2
and we get a
2.
Then we square the result and simplify it.
Example
c
2
-b
2
=√a
2
5
2
-1
2
=√a
2
25-1= √24
√24= 2√6
This describes the relationship between the sides of a right angled triangle.
Then we also learned how to manipulate it to solve many different problems.
Solving problems with Pythagoras
Using Pythagoras, we can solve many different problems that at the end derive from Pythagoras.
Example:
3
4
If we wanted to find the diagonal of a rectangle we can use Pythagoras to find it.
3
2
+4
2
= C
2
9+16=25
√25=5
We can use Pythagoras Theorem in many ways in real life. If we wanted to calculate the distance from one point to another directly instead of going straight, we could use Pythagoras theorem to solve it.
On this unit the first thing we reviewed was common factors.
Common factors
Common factors are factors that integers have in common.
Ex:
15, 36
The common factors of 15 and 36 are:
1 and 3
Then we learned how to do factorization with variables, using expansion and simplifying.
Distributive property
On the distributive property, one expands the equation and then solves it. a(b+c)=ab+ac
Ex:

5(x+4)
5*X+5*4
=5x+20
Then we learned about trinomials and binomials.
Binomials and Trinomials.
Binomials are products that are in the form of (a+b) (c+d)
Ex:
(x+2) (x-1)
These can be transformed into trinomials (x
2
+bx+c)
using F.O.I.L
First: x*x= x
2
Outer:x*-1= -1x
Inner: 2*x= 2x
Last: 2*-1= -2
x
2
+x-2