Introduction to Quadratic Equations 13-1
We learned from Chapter 12 that a quadratic function is a function that can be defined by an equation of the form
ax2 + bx + c = y where a is not equal to 0. This is a parabola when the domain is the set of REAL numbers.
When y = 0 in the quadratic function ax2 + bx + c = y we have an equation of the form ax2 + bx + c = 0. An equation that can be written in this form is called a quadratic equation.
STANDARD FORM is ax2 + bx + c = 0
4x2 + 7x = 5 write in standard form and determine a, b, and c
4x2 + 7x – 5 = 0
a= 4
b = 7
c = -5
CHAPTER 13 gives you several different ways to SOLVE QUADRATICS
Solving a quadratic means to find the x intercepts of a parabola.
There are different ways of asking the exact same question:
Find the.....
x intercepts = the roots = the solutions = the zeros of a quadratic
We'll answer this question one of the following ways:
1) Read them from the graph (read the x intercepts) That’s were y = 0 or where the parabola crosses the x-axis!!
but...graphing takes time and sometimes the intercepts are not integers
2) Set y or f(x) = 0 and then factor (we did this in Chapter 6)
but...some quadratics are not factorable
3) Square root each side (+ or - square root on the answer side)
but...sometimes the variable side is not a perfect square (it's irrational)
4) If not a perfect square on the variable side, complete the square, then solve using #3 method
Now this method ALWAYS works, but...it takes a lot of time and can get complicated
5) Quadratic Formula (works for EVERY quadratic)
Really easy if you just memorize the formula and how to use it! :)
Section 13-1
Reading the x intercepts from a graph or factoring and solving using the zero products property.
METHOD 1:
Where the graph crosses the x axis is/are the x intercepts. (Remember, y = 0 here!)
The x intercepts are the two solutions or roots of the quadratic.
METHOD 2:
When we factored in Chapter 6 and set each piece equal to zero, we were finding the x value when y was zero.
That means we were finding these two roots!
y2 – 5y = 6 = 6y – 18
first put this in standard form
y2 – 11y + 24 = 0
(y -8) (y-3) = 0
y = 8 or y = 3
Substitute to verify that 8 and 3 are solutions!!
More Solving Quadratic Equations 13-2
You did this in Chapter 11 for Pythagorean Theorem!
If there is no x term, it's easiest to just square root both sides to solve!
DIFFERENT FROM PYTHAGOREAN: NOT LOOKING FOR JUST THE PRINCIPAL SQUARE ROOT ANYMORE. NEED THE + OR - SYMBOL!!
This is also different from what we did when we had radical equations. Before we squared both sides to solve. It looks like these, but only after we squared both sides. Before we had to carefully check each answer—we had changed the equations by squaring. However, always check your solutions for any mistakes!!
3x2 = 18
divide both sides by 3 and get: x2 = 6
square root each side and get x = + SQRT 6 or - SQRT 6
(x - 5)2 = 9
SQRT each side and get: x - 5 = + or - 3
+ 5 to both sides: x = 5 + 3 or x = 5 - 3
So, the 2 roots are x = 8 or x = 2
(x + 2)2 = 7
SQRT each side and get: x + 2 = + or - SQRT of 7
-2 to both sides: x = -2+ SQRT 7 or x = -2 - SQRT 7
FORMULAS THAT ARE QUADRATICS:
Many formulas have a variable that is squared: compounded interest, height of a projectile (ball)
The formula for a projectile is h = -5t2 + v0t
We can find when a projectile is a ground level ( h= 0) by solving for
0 =-5t2 + v0t \If the projectile begins its flight at height c, its approximate height at time t is h = -5t2+v0t + c We can find when it hits the ground by solving 0 = -5t2 + v0t + c
For example: a slow-pitch softball player hits a pitch when the ball is 2 m above the ground. The ball pops up with an initial velocity of 9m/s If the ball is allowed to drop to the ground, how long will it be in the air?
When the ball hits the ground h = 0 so
0 = -5t2 + 9t + 2 or
-5t2 + 9t + 2 = 0
( 5t + 1)(t -2) = 0
5t = 1 and t = 2
t = -1/5 can’t be a solution since the answer should be positive t must be 2 seconds
Check out Purple Math for great help on quadratics
Friday, March 19, 2010
Thursday, March 18, 2010
Pre Algebra (Period 1)
Scientific Notation 4-9
You've had this since 6th grade!
You restate very big or very small numbers using powers of 10 in exponential form
Move the decimal so the number fits in this range: less than 10 and greater than or equal to 1
Count the number of places you moved the decimal and make that your exponent
Very big numbers - exponent is positive
Very small numbers (decimals) - exponent is negative (just like a fraction!)
Remember that STANDARD notation is what you expect (the normal number)
When you multiply or divide scientific notations, use the power rules!
Just be careful that is your answer does not fit the scientific notation range, that you restate it.
ORDERING SCIENTIFIC NOTATION NUMBERS:
As long as numbers are in scientific notation, they are easy to put in order from least to greatest!
1) If they are all different powers, simply order them by powers
2) If they have the same power, simply order them using your decimal ordering skills.
EXAMPLE 1: Order 3.7 x 108, 4.3 x 10-2, 9.3 x 105, and 8.7 x 10-5
8.7 x 10-5, 4.3 x 10-2, 9.3 x 105, 3.7 x 108
EXAMPLE 2: 3.7 x 108, 4.3 x 108, 9.3 x 108, and 8.7 x 108
3.7 x 108, 4.3 x 108, 8.7 x 108, 9.3 x 108
TRY THESE LINKS:
TRY THIS LINK THAT TAKES YOU FROM LARGE POWERS
TO
LITTLE POWERS (NEGATIVE POWERS OR DECIMALS)
Try this link to practice scientific notation!
You've had this since 6th grade!
You restate very big or very small numbers using powers of 10 in exponential form
Move the decimal so the number fits in this range: less than 10 and greater than or equal to 1
Count the number of places you moved the decimal and make that your exponent
Very big numbers - exponent is positive
Very small numbers (decimals) - exponent is negative (just like a fraction!)
Remember that STANDARD notation is what you expect (the normal number)
When you multiply or divide scientific notations, use the power rules!
Just be careful that is your answer does not fit the scientific notation range, that you restate it.
ORDERING SCIENTIFIC NOTATION NUMBERS:
As long as numbers are in scientific notation, they are easy to put in order from least to greatest!
1) If they are all different powers, simply order them by powers
2) If they have the same power, simply order them using your decimal ordering skills.
EXAMPLE 1: Order 3.7 x 108, 4.3 x 10-2, 9.3 x 105, and 8.7 x 10-5
8.7 x 10-5, 4.3 x 10-2, 9.3 x 105, 3.7 x 108
EXAMPLE 2: 3.7 x 108, 4.3 x 108, 9.3 x 108, and 8.7 x 108
3.7 x 108, 4.3 x 108, 8.7 x 108, 9.3 x 108
TRY THESE LINKS:
TRY THIS LINK THAT TAKES YOU FROM LARGE POWERS
TO
LITTLE POWERS (NEGATIVE POWERS OR DECIMALS)
Try this link to practice scientific notation!
Wednesday, March 17, 2010
Pre Algebra Period 1
Powers of Products & Quotients 5-9
(4∙2)3 = (4∙2)∙(4∙2)∙(4∙2)
= 4∙4∙4∙2∙2∙2
= (4∙4∙4)∙(2∙2∙2)
=43∙23
Raising a Product to a Power
(5∙3) 4 = 53∙33
or Algebraically:
(ab)m = ambm
Remember to simplify an expression, you write it with NO like terms or paranthese.
Simplify (4x2)3
Raise each factor to the power 3
= 43∙x2∙3
Use the rule for Raising a Power to a Power
=43∙x6
simplify
= 64x6
WE did several of these in class:
(2p)4 = (2∙2∙2∙2)∙(p∙p∙p∙p)
= 16p4
(xy2)5 = x5y10
(5x3)2 = 25x6
The location of a negative sign affects the value of an expression. Look at the differences between the following
(-5x2 = (-5)2x2 = (-5)(-5)x2 = 25x2
-(5x)2 = - (5)(5)x2 = -25x2
(-2y)4 = 16y4
-(2y)4 = -16y4
Do you see the subtle differences? Make sure you can determine why one is postive and the other is negative
Finding Powers of Quotients
(4/5)3 = (4/5)(4/5)(4/5)
or
4∙4∙4
5∙5∙5
=
43
53
= 64/125
To raise a quotient to a power, raise both the numerator and the denominator to the power
(2/3)4 =
24
34
=
16/81
(1/2)3 =
1/8
(-2/3)4
16/81
Why is it positive?
(2x2/3)3=
8x6/27
∙∙∙
(4∙2)3 = (4∙2)∙(4∙2)∙(4∙2)
= 4∙4∙4∙2∙2∙2
= (4∙4∙4)∙(2∙2∙2)
=43∙23
Raising a Product to a Power
(5∙3) 4 = 53∙33
or Algebraically:
(ab)m = ambm
Remember to simplify an expression, you write it with NO like terms or paranthese.
Simplify (4x2)3
Raise each factor to the power 3
= 43∙x2∙3
Use the rule for Raising a Power to a Power
=43∙x6
simplify
= 64x6
WE did several of these in class:
(2p)4 = (2∙2∙2∙2)∙(p∙p∙p∙p)
= 16p4
(xy2)5 = x5y10
(5x3)2 = 25x6
The location of a negative sign affects the value of an expression. Look at the differences between the following
(-5x2 = (-5)2x2 = (-5)(-5)x2 = 25x2
-(5x)2 = - (5)(5)x2 = -25x2
(-2y)4 = 16y4
-(2y)4 = -16y4
Do you see the subtle differences? Make sure you can determine why one is postive and the other is negative
Finding Powers of Quotients
(4/5)3 = (4/5)(4/5)(4/5)
or
4∙4∙4
5∙5∙5
=
43
53
= 64/125
To raise a quotient to a power, raise both the numerator and the denominator to the power
(2/3)4 =
24
34
=
16/81
(1/2)3 =
1/8
(-2/3)4
16/81
Why is it positive?
(2x2/3)3=
8x6/27
∙∙∙
Tuesday, March 16, 2010
Algebra Period 4
Quadratic Functions: 12- 4
A QUADRATIC FUNCTION is not y = mx + b
(which is a LINEAR function),
but instead is
y = ax2 + bx + c
OR
f(x) = ax2 + bx + c
where a, b, and c are all real numbers and
a cannot be equal to zero because
it must have a variable that is squared ( degree of 2)
[If a = 0, then we would end up with y = bx + c which is really y= mx + b]
Quadratics have a squared term, so they have TWO possible solutions also called roots. You already saw this in Chapter 6 when you factored the trinomial and used zero products property!! ( CHAPTER 6-- again)
If the domain is all real numbers, then you will have a PARABOLA which looks like
a smile when the a coefficient is positive or
looks like a frown when the a coefficient is negative.
Graphing quadratics:
You can graph quadratics exactly the way you graphed lines
by plugging in your choice of an x value and using the equation to find your y value.
Because it's a U shape, you should graph 5 points as follows:
STEP 1: determine Point 1: the vertex -
the minimum value of the smile or
the maximum value of the frown
The x value of the VERTEX = -b/2a
We get the values for a and b from the actual equation
f(x) = ax2 + bx + c
just plug in the b and the a value from your equation into -b/2a and you have the x-value of the vertex.
Now to find the y value -- take that x value and PLUG it into the equation
STEP 2: Next, draw the AXIS OF SYMMETRY : x = -b/2a
a line through the vertex parallel to the y axis . Draw this line as a dashed line. REMEMBER: It will be a dashed line parallel to the y-axis
STEP 3: Point 2- Pick an x value to the right or left of the axis and find its y by plugging into the equation.
STEP 4: Point 3- Graph its mirror image on the other side of the axis of symmetry by counting from axis of symmetry
STEP 5: Points 4 and 5- Repeat point 2 and 3 directions with another point even farther from the vertex
JOIN YOUR 5 POINTS IN A "U" SHAPE AND EXTEND LINES WITH ARROWS ON END
Parabolas that are functions have domains that are ALL REAL NUMBERS
Their ranges depend on where the vertex is and also if the a coefficient is positive or negative
EXAMPLE: f(x) = -3x2 (or y = -3x2)
the a coefficient is negative so it is a frown face
the x value of the vertex (maximum) is -b/2a or 0/2(-3) = 0
the y value of the vertex is 0
So the vertex is (0, 0)
To graph this function:
1) graph vertex (0, 0)
2) Draw dotted line x = 0 (actually this is the y axis!)
3) Pick x value to the right of axis of symmetry, say x = 1
Plug it in the equation: y = -3(1) = -3
Plot (1, -3)
4) count steps from axis of symmetry and place another point to the LEFT of axis in same place
5) pick another x value to the right of the axis of symmetry, say x = 2
plug it in the equation y = -3(2) = -6
plot ( 2, -6). Count the steps from the axis of symmetry and place another point to the LEFT of the axis in the same place.
The domain is all real numbers.
The range is y is less than or equal to zero
A QUADRATIC FUNCTION is not y = mx + b
(which is a LINEAR function),
but instead is
y = ax2 + bx + c
OR
f(x) = ax2 + bx + c
where a, b, and c are all real numbers and
a cannot be equal to zero because
it must have a variable that is squared ( degree of 2)
[If a = 0, then we would end up with y = bx + c which is really y= mx + b]
Quadratics have a squared term, so they have TWO possible solutions also called roots. You already saw this in Chapter 6 when you factored the trinomial and used zero products property!! ( CHAPTER 6-- again)
If the domain is all real numbers, then you will have a PARABOLA which looks like
a smile when the a coefficient is positive or
looks like a frown when the a coefficient is negative.
Graphing quadratics:
You can graph quadratics exactly the way you graphed lines
by plugging in your choice of an x value and using the equation to find your y value.
Because it's a U shape, you should graph 5 points as follows:
STEP 1: determine Point 1: the vertex -
the minimum value of the smile or
the maximum value of the frown
The x value of the VERTEX = -b/2a
We get the values for a and b from the actual equation
f(x) = ax2 + bx + c
just plug in the b and the a value from your equation into -b/2a and you have the x-value of the vertex.
Now to find the y value -- take that x value and PLUG it into the equation
STEP 2: Next, draw the AXIS OF SYMMETRY : x = -b/2a
a line through the vertex parallel to the y axis . Draw this line as a dashed line. REMEMBER: It will be a dashed line parallel to the y-axis
STEP 3: Point 2- Pick an x value to the right or left of the axis and find its y by plugging into the equation.
STEP 4: Point 3- Graph its mirror image on the other side of the axis of symmetry by counting from axis of symmetry
STEP 5: Points 4 and 5- Repeat point 2 and 3 directions with another point even farther from the vertex
JOIN YOUR 5 POINTS IN A "U" SHAPE AND EXTEND LINES WITH ARROWS ON END
Parabolas that are functions have domains that are ALL REAL NUMBERS
Their ranges depend on where the vertex is and also if the a coefficient is positive or negative
EXAMPLE: f(x) = -3x2 (or y = -3x2)
the a coefficient is negative so it is a frown face
the x value of the vertex (maximum) is -b/2a or 0/2(-3) = 0
the y value of the vertex is 0
So the vertex is (0, 0)
To graph this function:
1) graph vertex (0, 0)
2) Draw dotted line x = 0 (actually this is the y axis!)
3) Pick x value to the right of axis of symmetry, say x = 1
Plug it in the equation: y = -3(1) = -3
Plot (1, -3)
4) count steps from axis of symmetry and place another point to the LEFT of axis in same place
5) pick another x value to the right of the axis of symmetry, say x = 2
plug it in the equation y = -3(2) = -6
plot ( 2, -6). Count the steps from the axis of symmetry and place another point to the LEFT of the axis in the same place.
The domain is all real numbers.
The range is y is less than or equal to zero
Math 6H ( Periods 3, 6, & 7)
Scale Drawing 7-9
Opening your books to page 237, you will notice a drawing of a house. In this drawing of the house, the actual height of 9 meters is represented by a length of 3 centimeters, and the actual length of 21 meters is represented by a length of 7 centimeters.
This means that 1 cm in the drawing represents 3 m in the actual building. Such a drawing in which all lengths are in the same ratio to actual lengths is called a scale drawing.
The relationship of length in the drawing to actual length is called the scale. In the drawing of the house the scale is 1cm: 3m
We can express the scale as a ratio, called the scale ratio, if a common unit of measure is used. Since 3 m = 300 cm, the scale ratio is 1/300
Using the book’s drawing on page 237, find the length and width of the room shown, if the scale of the drawing is 1cm: 1.5 m
Measuring the drawing, we find that it has a length of 4 cm and a width of 3 cm
Method 1: write a proportion for the length
Let l = the actual length
1/1.5 = 4/ l
l= 4 (1.5)
l = 6
The room is 6 m long
Write a proportion for the width
Let w = the actual width
1/1.5 = 3/w
w = 3 (1.5)
w = 4.5 m
The room is 4.5 m wide
Method 2 : Use the scale ratio
1 cm/ 1.5 m = 1/cm/150 cm = 1/150
You need to change the units to the same and then set up a ratio
The actual length is 150 times the length in the drawing so
l =150 (4) = 600 cm = 6 m
w =150 (3) = 450 cm = 4.5 m
The scale on a map is 1 cm to 240 m
The distance from Ryan’s house to his school is 10 cm on the map. What is the actual distance?
let d = the distance from Ryan’s house to school
1/240 = 10/d so d = 240(10)
d = 2400m or
The distance from Ryan's house to school is 2.4 km
A picture of an insect has a scale 7 to 1. The length of the insect in the picture is 5.6 cm. What is the actual length of the insect?
Let l = the actual length of the insect
7/1 = 5.6/ l
7l = 5.6
l = .8 cm
The actual length of the insect is 0.8 cm which is 8 mm
Opening your books to page 237, you will notice a drawing of a house. In this drawing of the house, the actual height of 9 meters is represented by a length of 3 centimeters, and the actual length of 21 meters is represented by a length of 7 centimeters.
This means that 1 cm in the drawing represents 3 m in the actual building. Such a drawing in which all lengths are in the same ratio to actual lengths is called a scale drawing.
The relationship of length in the drawing to actual length is called the scale. In the drawing of the house the scale is 1cm: 3m
We can express the scale as a ratio, called the scale ratio, if a common unit of measure is used. Since 3 m = 300 cm, the scale ratio is 1/300
Using the book’s drawing on page 237, find the length and width of the room shown, if the scale of the drawing is 1cm: 1.5 m
Measuring the drawing, we find that it has a length of 4 cm and a width of 3 cm
Method 1: write a proportion for the length
Let l = the actual length
1/1.5 = 4/ l
l= 4 (1.5)
l = 6
The room is 6 m long
Write a proportion for the width
Let w = the actual width
1/1.5 = 3/w
w = 3 (1.5)
w = 4.5 m
The room is 4.5 m wide
Method 2 : Use the scale ratio
1 cm/ 1.5 m = 1/cm/150 cm = 1/150
You need to change the units to the same and then set up a ratio
The actual length is 150 times the length in the drawing so
l =150 (4) = 600 cm = 6 m
w =150 (3) = 450 cm = 4.5 m
The scale on a map is 1 cm to 240 m
The distance from Ryan’s house to his school is 10 cm on the map. What is the actual distance?
let d = the distance from Ryan’s house to school
1/240 = 10/d so d = 240(10)
d = 2400m or
The distance from Ryan's house to school is 2.4 km
A picture of an insect has a scale 7 to 1. The length of the insect in the picture is 5.6 cm. What is the actual length of the insect?
Let l = the actual length of the insect
7/1 = 5.6/ l
7l = 5.6
l = .8 cm
The actual length of the insect is 0.8 cm which is 8 mm
Monday, March 15, 2010
Algebra Period 4
Relations & Functions 12-1 and 12-2
RELATIONS: Set of ordered pairs where the x values are the DOMAIN and the y values are the RANGE.
FUNCTIONS: Relations where there is just one y value for each x value IN OTHER WORDS----YOU CAN'T HAVE TWO y VALUES for the SAME x value!!!
If you see x repeated twice, it's still a relation, but it's not a function.
In the real world, there are excellent examples....pizza prices.
A restaurant can't have two different prices for the same size cheese pizza.
If you charge $10 and $12 on the same day for the same pizza, you don't have a function.
But, you certainly can charge $10 for a cheese pizza and $12 for a pepperoni pizza.
VERTICAL LINE TEST: When you graph a function, if you draw a vertical line anywhere on the graph, that line will only intersect the function at one point!!!!
If it intersects at 2 or more, it's a relation, but not a function.
So a horizontal line function, y = 4, is a function, but a vertical line function, x = 4 is not.
Any line, y = mx + b, is a function.
INPUTS: x values
OUTPUTS: y values
f(x) means the value of the function at the given x value
You can think of f(x) as the y value
Finding the value of a function: Plug it in, plug it in!
f(x) = 2x + 7
Find f(3)
f(3) = 2(3) + 7 = 13
The function notation gives you more information than using y
If I tell you y = 13 you have no idea what the x value was at that point
But if I tell you f(3) = 13, you know the entire coordinate (3, 13)
Domain of a function = all possible x values (inputs) that keep the solution real
Range of a function = all possible y values (outputs) that result from the domain
EXAMPLE:
f(x) = x + 10 has the domain of all real numbers and the same range because every value will keep the answer f(x) a real number
EXAMPLE:
f(x) = x2 has the domain again of all real numbers, BUT the range is greater than or = to zero
because when a number is squared it will never be negative! So f(x) will always be 0 or positive
EXAMPLE:
f(x) = absolute value of x has the domain of all real numbers, but again the range will be greater than or equal to zero because absolute value will never be negative
EXAMPLE:
f(x) = 1/x has a domain of all real numbers EXCEPT FOR ZERO because it would be undefined if zero was in the denominator. The range is all real numbers except zero as well.
This function will approach both axes but never intersect with them.
The axes are called asymptotes which means that they will get very close but never reach them
EXAMPLE:
f(x) = (x - 10)/x + 3
Domain is all real numbers EXCEPT -3 because -3 will turn the denominator into zero (undefined)
What is the range?
RELATIONS: Set of ordered pairs where the x values are the DOMAIN and the y values are the RANGE.
FUNCTIONS: Relations where there is just one y value for each x value IN OTHER WORDS----YOU CAN'T HAVE TWO y VALUES for the SAME x value!!!
If you see x repeated twice, it's still a relation, but it's not a function.
In the real world, there are excellent examples....pizza prices.
A restaurant can't have two different prices for the same size cheese pizza.
If you charge $10 and $12 on the same day for the same pizza, you don't have a function.
But, you certainly can charge $10 for a cheese pizza and $12 for a pepperoni pizza.
VERTICAL LINE TEST: When you graph a function, if you draw a vertical line anywhere on the graph, that line will only intersect the function at one point!!!!
If it intersects at 2 or more, it's a relation, but not a function.
So a horizontal line function, y = 4, is a function, but a vertical line function, x = 4 is not.
Any line, y = mx + b, is a function.
INPUTS: x values
OUTPUTS: y values
f(x) means the value of the function at the given x value
You can think of f(x) as the y value
Finding the value of a function: Plug it in, plug it in!
f(x) = 2x + 7
Find f(3)
f(3) = 2(3) + 7 = 13
The function notation gives you more information than using y
If I tell you y = 13 you have no idea what the x value was at that point
But if I tell you f(3) = 13, you know the entire coordinate (3, 13)
Domain of a function = all possible x values (inputs) that keep the solution real
Range of a function = all possible y values (outputs) that result from the domain
EXAMPLE:
f(x) = x + 10 has the domain of all real numbers and the same range because every value will keep the answer f(x) a real number
EXAMPLE:
f(x) = x2 has the domain again of all real numbers, BUT the range is greater than or = to zero
because when a number is squared it will never be negative! So f(x) will always be 0 or positive
EXAMPLE:
f(x) = absolute value of x has the domain of all real numbers, but again the range will be greater than or equal to zero because absolute value will never be negative
EXAMPLE:
f(x) = 1/x has a domain of all real numbers EXCEPT FOR ZERO because it would be undefined if zero was in the denominator. The range is all real numbers except zero as well.
This function will approach both axes but never intersect with them.
The axes are called asymptotes which means that they will get very close but never reach them
EXAMPLE:
f(x) = (x - 10)/x + 3
Domain is all real numbers EXCEPT -3 because -3 will turn the denominator into zero (undefined)
What is the range?
Pre Algebra ( Period 1)
EXPONENTS 4-7 & 4-8
MULTIPLYING Powers with LIKE BASES:
Simply ADD THE POWERS
WITH VARIABLES:
m5m3 = m8
You can check this by EXPANDING:
(mmmmm)(mmm) = m8
WITH NUMBERS:
(25)(23) = 28
DIVIDING Powers with LIKE BASES:
Simply SUBTRACT the POWERS
m8 = m3
m5
Again, you can check this by EXPANDING:
mmmmmmmm
mmmmm
ZERO POWERS:
Anything to the zero power = 1
(except zero to the zero power is undefined or indeterminate)
Proof of this was given in class:
1 =
mmmmmmmm
mmmmmmmm
=
m8 = m0 (by power rules for division)
m8
: 1 = m0
By the transitive property of equality
MULTIPLYING Powers with LIKE BASES:
Simply ADD THE POWERS
WITH VARIABLES:
m5m3 = m8
You can check this by EXPANDING:
(mmmmm)(mmm) = m8
WITH NUMBERS:
(25)(23) = 28
DIVIDING Powers with LIKE BASES:
Simply SUBTRACT the POWERS
m8 = m3
m5
Again, you can check this by EXPANDING:
mmmmmmmm
mmmmm
ZERO POWERS:
Anything to the zero power = 1
(except zero to the zero power is undefined or indeterminate)
Proof of this was given in class:
1 =
mmmmmmmm
mmmmmmmm
=
m8 = m0 (by power rules for division)
m8
: 1 = m0
By the transitive property of equality
Math 6H ( Periods 3, 6, & 7)
Problem Solving: Using Proportion 7-8
Proportions can be used to solve word problems. Use the following steps to help you in solving problems using proportions
~ Decide which quantity is to be found and represent it by a variable
~ Determine whether the quantities involved can be compared using ratios (rates)
~ Equate the ratios in a proportion
~ Solve the proportion
Taylor’s mom bought 4 tires for her car at a cost of $264. How much would 5 tires cost at the same rate?
Let c = the cost of 5 tires. Set up a proportion
4/264 = 5/c
Solve the proportion
4c = 5(264)
Now divide both sides by 4
4c/4 = 5(264)/4
c = 330
Therefore, 5 tires would cost $330
Notice, the proportion in this example could also have been written as
2/264 = c/5
In fact--Any of the following proportions can be used to solve the problem
4/5 = 264/c 5/4 = c/264 4/264 = 5/c 264/4 = c/ 5
All of the above proportions result in the same equation
4c = 5(264)
But be careful you need to use a proportion that does relate.
4/5 DOES NOT EQUAL c/264. That is not an accurate proportion and would result in an inaccurate solution.
For every 5 sailboats in a harbor, there are 3 motorboats. If there are 30 sailboats in the harbor, how many motorboats are there?
Let m = the number of motorboats
5/3 = 30/m
5m = 3(30) divide both sides by 5 5m/5 = 3(30)/5
m=18
There are 18 motorboats in the harbor.
Some guidelines you can use to determine when it is appropriate to use a proportion to solve a word problem.
Ask the following questions
If one quantity increase does the other quantity also increase? (If one quantity decreases, does the other quantity decrease?) When the number of tires is increase, the cost is also increased.
Does the amount of change (increase or decrease) o one quantity depend upon the amount of change (increase or decrease) of the other quantity? The amount of increase in the cost depends upon the number of additional tires bought.
Does one quantity equal some constant times the other quantity? The total costs equals the cost of one tire times the number of tires. The cost of one tire is constant.
If the answers to all the questions above is YES, then it is appropriate to use a proportion.
Sometimes setting up a table can be useful
Although the problems in this lesson may be solved without using proportions, I must insist that you write a proportion for each problem and solve using this method. You may check your work using another other method you know.
Proportions can be used to solve word problems. Use the following steps to help you in solving problems using proportions
~ Decide which quantity is to be found and represent it by a variable
~ Determine whether the quantities involved can be compared using ratios (rates)
~ Equate the ratios in a proportion
~ Solve the proportion
Taylor’s mom bought 4 tires for her car at a cost of $264. How much would 5 tires cost at the same rate?
Let c = the cost of 5 tires. Set up a proportion
4/264 = 5/c
Solve the proportion
4c = 5(264)
Now divide both sides by 4
4c/4 = 5(264)/4
c = 330
Therefore, 5 tires would cost $330
Notice, the proportion in this example could also have been written as
2/264 = c/5
In fact--Any of the following proportions can be used to solve the problem
4/5 = 264/c 5/4 = c/264 4/264 = 5/c 264/4 = c/ 5
All of the above proportions result in the same equation
4c = 5(264)
But be careful you need to use a proportion that does relate.
4/5 DOES NOT EQUAL c/264. That is not an accurate proportion and would result in an inaccurate solution.
For every 5 sailboats in a harbor, there are 3 motorboats. If there are 30 sailboats in the harbor, how many motorboats are there?
Let m = the number of motorboats
5/3 = 30/m
5m = 3(30) divide both sides by 5 5m/5 = 3(30)/5
m=18
There are 18 motorboats in the harbor.
Some guidelines you can use to determine when it is appropriate to use a proportion to solve a word problem.
Ask the following questions
If one quantity increase does the other quantity also increase? (If one quantity decreases, does the other quantity decrease?) When the number of tires is increase, the cost is also increased.
Does the amount of change (increase or decrease) o one quantity depend upon the amount of change (increase or decrease) of the other quantity? The amount of increase in the cost depends upon the number of additional tires bought.
Does one quantity equal some constant times the other quantity? The total costs equals the cost of one tire times the number of tires. The cost of one tire is constant.
If the answers to all the questions above is YES, then it is appropriate to use a proportion.
Sometimes setting up a table can be useful
Although the problems in this lesson may be solved without using proportions, I must insist that you write a proportion for each problem and solve using this method. You may check your work using another other method you know.
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