The Quadratic Formula: 12-3
Method 5:
THE QUADRATIC FORMULA
Let's all sing it to "Pop Goes the Weasel!
"
x = -b plus or minus the square root of b squared minus 4ac all over 2a
x =
-b ± √ (b2 - 4ac)
2a
Notice how the first part is the x value of the vertex -b/2a
The plus or minus square root of b squared minus 4ac represents
how far away the two x intercepts (or roots) are from the vertex!!!!
Very few real world quadratics can be solved by factoring or square rooting each side.
And completing the square always works, but it long and cumbersome!
All quadratics can be solved by using the QUADRATIC FORMULA.
You will find out that some quadratics have NO REAL solutions, which means that there are no x intercepts - the parabola does not cross the x axis!
Think about what kinds of parabolas would do this....ones that are smiles that have a vertex above the x or ones that are frowns that have a vertex below the x axis.
You will find out in Algebra II that these parabolas have IMAGINARY roots
So now you know 5 ways that you know to find the roots:
1. graph
2. factor if possible
3. square root each side
4. complete the square - that's what the quadratic formula is based on!
5. plug and chug in the Quadratic Formula -
This method always works if there's a REAL solution!
DON'T FORGET TO PUT THE QUADRATIC IN STANDARD FORM BEFORE PLUGGING THE VALUES INTO THE QUADRATIC FORMULA!
You should know that for the quadratic formula, you don't need the "a" coefficient to be positive!
That's important if you use factoring, SQRTing each side, and completing the square.
But for the quadratic formula, either way, you'll get the same roots!
EXAMPLE: x2 + 8x = 48
You can move the 48 over or move the x2 + 8x over and you'll get the same answers!
Let's look at:
"a" coefficient positive *****vs***** "a" coefficient negative
x2 + 8x - 48 = 0 ****VS**** -x2 - 8x + 48 = 0
First x2 + 8x - 48 = 0
-8 ± √[64 - 4(1)(-48)]
2(1)
-8 ± √(64 + 192)
2
-8 ± √(256)
2
-8 ± 16
2
(-8 + 16)/2 and (-8 - 16)/2
8/2 and -24/2
4 and -12
Now let's compare -x2 - 8x + 48 = 0
8 ± √[64 - 4(-1)(48)]
-2(1)
8 ± √(64 + 192)
-2
8 ± √(256)
-2
8 ± 16
-2
(8 + 16)/-2 and (8 - 16)/-2
24/-2 and -8/-2
-12 and 4
So all the signs are simply the opposite of each other and therefore the answers are the same
Monday, March 24, 2014
Math 7 ( Period 4)
Scientific Notation 6.9
You have had this since 5th or sixth grade... but check out the links at the bottom of these notes
Powers of ten provide a convenient way to write very large numbers. You restate very big numbers or very small numbers using powers of ten in exponential form. Numbers that are expressed as products of two factors
(1) a number greater than or equal to 1, but less than 10,
AND
(2) a power of ten
are said to be written in scientific notation.
We can write 'a number greater than or equal to 1, but less than 10' as an mathematical inequality 1 ≤ n < 10 To write a number in scientific notation we move the decimal point to the left until the resulting number is between 1 and 10. We then multiply this number by the power of 10, whose exponent is equal to the number of places we moved the decimal point. 4,592,000,000 in scientific notation First move the decimal point to the left to get a number between 1 and 10 4,592,000,000 the first factor in scientific notation becomes 4.592 Since the decimal point was moved 9 places, we multiply 4.592 by 109 to express the number in scientific notation
4.592 x 109 (Yes, you get to use the × symbol for multiplication .. but only for this!!
5,900,000,000 was given as the population of the world in 1998. I wonder what it is now? Anyone find out?
Anyway to write that large number in scientific notation it would be the product of two factors 5.9 and a power of ten
5.9 X 109
It is a way to write very large numbers AND very small numbers
Numbers expressed as products of a number greater than or equal to 1 BUT less than 10, AND a power of ten are called Scientific Notation.
Two Factors
91) 1≤ n < 10 (2) Power of 10 4,592,000,000 becomes 4.592 X 109
moved the decimal 9 places so we must multiply our number by a power of 109
98,000,000 = 9.8 X 107
320,000 = 3.2 X 105
What if I give you 7.04 X 108 and ask you to put it back into STANDARD NOTATION:
704,000,000.
0.0000207
0.0031 = 3.1 X 10-3
It isn't a negative number its just a very tiny number
1≤ n < 10 0.16 becomes 1.6 x 10 -1
0.0000207 becomes 2.07 X 10 -5
0.000345 becomes 3.45 X 10 -4
We also changed scientific notation back into standard notation ( the regular number)
9.32 X 10 -2 becomes 0.0932
we practiced some more
14,500 first we changed the number into something that fit the first factor 1 ≤ n < 10
so that becomes 1.45
Then we need to figure out how many places we moved the decimal
1.45 X 104
that works!
73,000,000 = 7.3 X 107
63,500 = 6.35 X 104
Write 3.45 X 104 into standard notation
34, 500
What happens if you are given 1/8000 this is a fraction and we need to change it to a decimal BEFORE we can write it in scientific notation
1/8000 = 0.000125
so it is
1.25 X 10-4
suppose you are multiplying two scientific notation numbers-- you do NOT need to change them into standard notation
(2.5 X 103 )(4.4 X10-5).
We found that it was
11 X 10-2
But 11 isn't 1 ≤ n < 10 so we need to change 11 into scientific notation
that alone is 1.1 X 101
Now we have 1.1 X 101 X 10-2
so 1.1 X 10-1 interesting
(6.2 X102 )( 8 X103) = (6.2)(8) X 102 X103 and
49.6 X 105 but 49.6 isn't in scientific notation
so
49.6 becomes 4.96 X101
4.96 X 101 X105 is
4.96 X 106
So
Try
(9.1 X 10-1 ) ( 5.4 X107)
(9.1)(5.4) X 10-1 X 107
So
49.14 X 106 and again 49.14 isn't in scientific notation
changing just 49.14 into scientific notation we get 4.914 X 101
so we have
4.914 X101 X 106
which
is
4.914 X 107
List from least to greatest
3.4 X 10-4 3.4 X 10-5 and 3.4 X 10-6
We noticed that the -6 exponent would be a smaller number-- when we wrote it in standard form so
3.4 X 10-6 < 3.4 X 10-5 < 3.4 X 10-4 would be the listing
Check out the following links:
this one takes you from huge numbers to teeny tiny ones
this one lets you practice scientific notation
this one lets you practice multiplying scientific notation
this one lets you practice dividing sn
this one lets you practice even harder ones!
You have had this since 5th or sixth grade... but check out the links at the bottom of these notes
Powers of ten provide a convenient way to write very large numbers. You restate very big numbers or very small numbers using powers of ten in exponential form. Numbers that are expressed as products of two factors
(1) a number greater than or equal to 1, but less than 10,
AND
(2) a power of ten
are said to be written in scientific notation.
We can write 'a number greater than or equal to 1, but less than 10' as an mathematical inequality 1 ≤ n < 10 To write a number in scientific notation we move the decimal point to the left until the resulting number is between 1 and 10. We then multiply this number by the power of 10, whose exponent is equal to the number of places we moved the decimal point. 4,592,000,000 in scientific notation First move the decimal point to the left to get a number between 1 and 10 4,592,000,000 the first factor in scientific notation becomes 4.592 Since the decimal point was moved 9 places, we multiply 4.592 by 109 to express the number in scientific notation
4.592 x 109 (Yes, you get to use the × symbol for multiplication .. but only for this!!
5,900,000,000 was given as the population of the world in 1998. I wonder what it is now? Anyone find out?
Anyway to write that large number in scientific notation it would be the product of two factors 5.9 and a power of ten
5.9 X 109
It is a way to write very large numbers AND very small numbers
Numbers expressed as products of a number greater than or equal to 1 BUT less than 10, AND a power of ten are called Scientific Notation.
Two Factors
91) 1≤ n < 10 (2) Power of 10 4,592,000,000 becomes 4.592 X 109
moved the decimal 9 places so we must multiply our number by a power of 109
98,000,000 = 9.8 X 107
320,000 = 3.2 X 105
What if I give you 7.04 X 108 and ask you to put it back into STANDARD NOTATION:
704,000,000.
0.0000207
0.0031 = 3.1 X 10-3
It isn't a negative number its just a very tiny number
1≤ n < 10 0.16 becomes 1.6 x 10 -1
0.0000207 becomes 2.07 X 10 -5
0.000345 becomes 3.45 X 10 -4
We also changed scientific notation back into standard notation ( the regular number)
9.32 X 10 -2 becomes 0.0932
we practiced some more
14,500 first we changed the number into something that fit the first factor 1 ≤ n < 10
so that becomes 1.45
Then we need to figure out how many places we moved the decimal
1.45 X 104
that works!
73,000,000 = 7.3 X 107
63,500 = 6.35 X 104
Write 3.45 X 104 into standard notation
34, 500
What happens if you are given 1/8000 this is a fraction and we need to change it to a decimal BEFORE we can write it in scientific notation
1/8000 = 0.000125
so it is
1.25 X 10-4
suppose you are multiplying two scientific notation numbers-- you do NOT need to change them into standard notation
(2.5 X 103 )(4.4 X10-5).
We found that it was
11 X 10-2
But 11 isn't 1 ≤ n < 10 so we need to change 11 into scientific notation
that alone is 1.1 X 101
Now we have 1.1 X 101 X 10-2
so 1.1 X 10-1 interesting
(6.2 X102 )( 8 X103) = (6.2)(8) X 102 X103 and
49.6 X 105 but 49.6 isn't in scientific notation
so
49.6 becomes 4.96 X101
4.96 X 101 X105 is
4.96 X 106
So
Try
(9.1 X 10-1 ) ( 5.4 X107)
(9.1)(5.4) X 10-1 X 107
So
49.14 X 106 and again 49.14 isn't in scientific notation
changing just 49.14 into scientific notation we get 4.914 X 101
so we have
4.914 X101 X 106
which
is
4.914 X 107
List from least to greatest
3.4 X 10-4 3.4 X 10-5 and 3.4 X 10-6
We noticed that the -6 exponent would be a smaller number-- when we wrote it in standard form so
3.4 X 10-6 < 3.4 X 10-5 < 3.4 X 10-4 would be the listing
Check out the following links:
this one takes you from huge numbers to teeny tiny ones
this one lets you practice scientific notation
this one lets you practice multiplying scientific notation
this one lets you practice dividing sn
this one lets you practice even harder ones!
Math 6A (Periods 1 & 2)
Subtracting Integers 11-3
The life story about someone who was so negative-- you wanted to take a little negativity away but since you can't do that you add a little positiveness-- works in math as well!!
Instead of subtracting ... "ADD THE OPPOSITE!"
We will prove it in class with our little red and yellow tiles... If you need to review, make your own out of red and yellow paper-- or whatever colors you want!!
In life-- to take away a little negative-- add some positive
To take away an integer... add its opposite.
Rule from our textbook
for all integers a and b
a - b = a + (the opposite of b) or
a - b = a + (-b)
Instead of subtracting.. "ADD THE OPPOSITE"
make sure you do the check, check.. you need to have two check marks.. one changing the subtraction to addition and the other changing the sign of the 2nd number to its opposite.
5 - - 2=
5 + + 2= 7
-2 - 5 =
- 2 + - 5 =
before I give you the answer look... we are looking at
-2 + -5
We are adding two negatives.. so we are back to the rules from Section 11-2...
when adding the same sign just add the number and use their sign so
- 2 + - 5 = -7
But what about -2 - -5 ?
adding the opposite, we get
-2 + + 5 .
Now, the signs are different so the rule from Section 11-2 is
ask yourself... "Who wins?" and "By how much?"
Okay, here the positive wins so I know the answer will be +
and by how much means.. to take the difference
5-2 = 3
so -2 + + 5 = +3
Do you need to put the + sign? No, but I like to in the beginning to show that I checked WHO WON!!
What about
-120 - -48?
add the opposite
-120 + + 48 follows the
Different signs rule... so
ask yourself
Who wins? answer: the negative.. so I know the answer will be negative..
And "By How Much?" take the difference 120-48 = 72
so
-120 + + 48 = -72
or
-120 - 48
-120 + (-48) = -168
NEVER EVER CHANGE THE FIRST NUMBER'S SIGN!!
WALK THE LINE, the number line-- that is!!
Remember... Attitude is such a little thing... but it makes a BIG difference!!
Always start with a positive attitude!!
When you walk the line, Which way are you always facing when you start???
Always attempt to get everything into addition so we can follow the rules of Section 11-2 Adding Integers.
1) SAME SIGN rule---> just add the numbers and use their sign
-4 + -5 = -9
2) DIFFERENT SIGNS rule
ask yourself those 2 important questions
a) Who wins? (answer is either negative or positive)
b) By how much? (Take the difference)
Page 376 answers to # 2-18 (evens)
2. -9
4. -5
6. 22
8. -9
10. -74
12. -32
14. -160
16. 498
18. -284
The life story about someone who was so negative-- you wanted to take a little negativity away but since you can't do that you add a little positiveness-- works in math as well!!
Instead of subtracting ... "ADD THE OPPOSITE!"
We will prove it in class with our little red and yellow tiles... If you need to review, make your own out of red and yellow paper-- or whatever colors you want!!
In life-- to take away a little negative-- add some positive
To take away an integer... add its opposite.
Rule from our textbook
for all integers a and b
a - b = a + (the opposite of b) or
a - b = a + (-b)
Instead of subtracting.. "ADD THE OPPOSITE"
make sure you do the check, check.. you need to have two check marks.. one changing the subtraction to addition and the other changing the sign of the 2nd number to its opposite.
5 - - 2=
5 + + 2= 7
-2 - 5 =
- 2 + - 5 =
before I give you the answer look... we are looking at
-2 + -5
We are adding two negatives.. so we are back to the rules from Section 11-2...
when adding the same sign just add the number and use their sign so
- 2 + - 5 = -7
But what about -2 - -5 ?
adding the opposite, we get
-2 + + 5 .
Now, the signs are different so the rule from Section 11-2 is
ask yourself... "Who wins?" and "By how much?"
Okay, here the positive wins so I know the answer will be +
and by how much means.. to take the difference
5-2 = 3
so -2 + + 5 = +3
Do you need to put the + sign? No, but I like to in the beginning to show that I checked WHO WON!!
What about
-120 - -48?
add the opposite
-120 + + 48 follows the
Different signs rule... so
ask yourself
Who wins? answer: the negative.. so I know the answer will be negative..
And "By How Much?" take the difference 120-48 = 72
so
-120 + + 48 = -72
or
-120 - 48
-120 + (-48) = -168
NEVER EVER CHANGE THE FIRST NUMBER'S SIGN!!
WALK THE LINE, the number line-- that is!!
Remember... Attitude is such a little thing... but it makes a BIG difference!!
Always start with a positive attitude!!
When you walk the line, Which way are you always facing when you start???
Always attempt to get everything into addition so we can follow the rules of Section 11-2 Adding Integers.
1) SAME SIGN rule---> just add the numbers and use their sign
-4 + -5 = -9
2) DIFFERENT SIGNS rule
ask yourself those 2 important questions
a) Who wins? (answer is either negative or positive)
b) By how much? (Take the difference)
Page 376 answers to # 2-18 (evens)
2. -9
4. -5
6. 22
8. -9
10. -74
12. -32
14. -160
16. 498
18. -284
Thursday, March 20, 2014
Algebra Honors ( Periods 6 & 7)
Completing the Square: 12-2
METHOD 4:
Now this is completely new to you!!!
When does the square root = ± square root method work well?
When the side with the variable is a PERFECT SQUARE! We saw that in the previous section.
perfect square = k ( when k ≥ 0)
So what if that side is not a perfect BINOMIAL SQUARED?
It may be possible to trasform it into one... by COMPLETING THE SQUARE
You can follow steps to make it into one!
Why is this good?
Because then you can just square root each side to find the roots!
THIS METHOD ALWAYS WORKS!
EXAMPLE:
x2 - 3x -18 = 0
(Head's up-- I wouldn't use this method here because I can see that it factors easily... into (x-6)(x+3) = 0 so the solution set is {-3,6}
However, knowing what I need to get for my solutions might be a good way to practice Completing the square..
so
x2 - 3x = 18
Step 1) b/2
That is, take half of the b coefficient ,
or in this case (- 3/2)
Step 2) Square b/2
(b/2)2
(-3/2 ⋅ -3/2 = 9/4)
Step 3) Add (b/2)2 to both sides of the equation
x2 - 3x + 9/4 = 18 +9/4
Step 4) Factor to a binomial square
(x - 3/2)2 = 18 + 9/4 = 81/4
Step 5) Square root each side and solve
√(x - 3/2)2 = √ 81/4
x - 3/2 = ± 9/2
x = 3/2 ± 9/2
x = 3/2 + 9/2 AND x = 3/2 - 9/2
x = 6 and x = -3
{-3, 6}
We got the same solutions !! Yay!!
x2 - 10x = 0
Not a TRINOMIAL SQUARE so it would not factor to a BINOMIAL SQUARED.
But here's how you can make it one!
Step 1) b/2
That is, take half of the b coefficient ,
or in this case (- 10/2 = -5)
Step 2) Square b/2
(b/2)2
(-5 x -5 = 25)
Step 3) Add (b/2)2 to both sides of the equation
x2 - 10x + 25 = +25
Step 4) Factor to a binomial square
(x - 5)2 = 25
Step 5) Square root each side and solve
√(x - 5)2 = √ 25
x - 5 = ± 5
Step 6) ADD 5 TO BOTH SIDES
x - 5 = ± 5
+ 5 = +5
x = 5 ± 5
Step 7) Simplify if possible
x = 5 + 5 and x = 5 - 5
x = 10 and x = 0
So the 2 roots (solutions/zeros/x intercepts) are 0 and 10.
YOU DON'T NEED TO GRAPH THE PARABOLA, BUT IF YOU DID, IT WOULD CROSS THE X AXIS AT 0 AND 10.
I don't know where the vertex is, but I don't need to because it's not the solution to the quadratic (although I certainly could find the vertex by using x = -b/2a)
Notice that I could also factor x2 - 10x = 0 to get the solution more easily.
So don't complete the square if the quadratic factors easily!
IF THERE IS A "c", first move the c constant to the other side of the equation before completing the square:
x2 - 10x - 11 = 0
x2 - 10x - 11 + 11 = 0 + 11
x2 - 10x = 11
NOW COMPLETE THE SQUARE AS ABOVE:
x2 - 10x + 25 = +25 + 11
(x - 5)2 = 36
√ (x - 5)2 = √ 36
x - 5 = ± 6
x = 5 ± 6
x = 11 and x = -1
Again, this one factored easily so I wouldn't have even used completing the square. ALWAYS CHECK IF IT FACTORS FIRST!
Now an example that DOES NOT FACTOR: x2 - 10x - 18 = 0
x2 - 10x - 18 = 0
x2 - 10x - 11 + 18 = 0 + 18
x2 - 10x = 18
NOW COMPLETE THE SQUARE AS ABOVE:
x2 - 10x + 25 = +25 + 18
(x - 5)2 = 43
√ (x - 5)2 = √ 43
x - 5 = ± √ 43
x = 5 ± √ 43
x = 5 + √ 43 and x = 5 - √ 43
When there is an IRRATIONAL square root, always SIMPLIFY if possible!
IF THERE IS AN "a" COEFFICIENT, YOU MUST DIVIDE EACH TERM BY IT BEFORE YOU CAN COMPLETE THE SQUARE:
Example: 2x2 - 3x - 1 = 0
Move the 1 to the other side of the equation: 2x2 - 3x = 1
Divide each term by the "a" coefficient:
x2 - 3/2 x = 1/2
Now find the completing the square term and add it to both sides:
[(-3/2)(-3/2)]2 = 9/16
x2 - 3/2 x + 9/16 = 1/2 + 9/16
(x - 3/4)2 =8/16 + 9/16
(x - 3/4)2 = 17/16
√[(x - 3/4)2 ] = ±√ [17/16]
x - 3/4 = ±[√17] /4
x = 3/4 ±[√ 17] /4
x = (3 ± [√17]) /4
or written better x =
3 ± √ 17
4
METHOD 4:
Now this is completely new to you!!!
When does the square root = ± square root method work well?
When the side with the variable is a PERFECT SQUARE! We saw that in the previous section.
perfect square = k ( when k ≥ 0)
So what if that side is not a perfect BINOMIAL SQUARED?
It may be possible to trasform it into one... by COMPLETING THE SQUARE
You can follow steps to make it into one!
Why is this good?
Because then you can just square root each side to find the roots!
THIS METHOD ALWAYS WORKS!
EXAMPLE:
x2 - 3x -18 = 0
(Head's up-- I wouldn't use this method here because I can see that it factors easily... into (x-6)(x+3) = 0 so the solution set is {-3,6}
However, knowing what I need to get for my solutions might be a good way to practice Completing the square..
so
x2 - 3x = 18
Step 1) b/2
That is, take half of the b coefficient ,
or in this case (- 3/2)
Step 2) Square b/2
(b/2)2
(-3/2 ⋅ -3/2 = 9/4)
Step 3) Add (b/2)2 to both sides of the equation
x2 - 3x + 9/4 = 18 +9/4
Step 4) Factor to a binomial square
(x - 3/2)2 = 18 + 9/4 = 81/4
Step 5) Square root each side and solve
√(x - 3/2)2 = √ 81/4
x - 3/2 = ± 9/2
x = 3/2 ± 9/2
x = 3/2 + 9/2 AND x = 3/2 - 9/2
x = 6 and x = -3
{-3, 6}
We got the same solutions !! Yay!!
x2 - 10x = 0
Not a TRINOMIAL SQUARE so it would not factor to a BINOMIAL SQUARED.
But here's how you can make it one!
Step 1) b/2
That is, take half of the b coefficient ,
or in this case (- 10/2 = -5)
Step 2) Square b/2
(b/2)2
(-5 x -5 = 25)
Step 3) Add (b/2)2 to both sides of the equation
x2 - 10x + 25 = +25
Step 4) Factor to a binomial square
(x - 5)2 = 25
Step 5) Square root each side and solve
√(x - 5)2 = √ 25
x - 5 = ± 5
Step 6) ADD 5 TO BOTH SIDES
x - 5 = ± 5
+ 5 = +5
x = 5 ± 5
Step 7) Simplify if possible
x = 5 + 5 and x = 5 - 5
x = 10 and x = 0
So the 2 roots (solutions/zeros/x intercepts) are 0 and 10.
YOU DON'T NEED TO GRAPH THE PARABOLA, BUT IF YOU DID, IT WOULD CROSS THE X AXIS AT 0 AND 10.
I don't know where the vertex is, but I don't need to because it's not the solution to the quadratic (although I certainly could find the vertex by using x = -b/2a)
Notice that I could also factor x2 - 10x = 0 to get the solution more easily.
So don't complete the square if the quadratic factors easily!
IF THERE IS A "c", first move the c constant to the other side of the equation before completing the square:
x2 - 10x - 11 = 0
x2 - 10x - 11 + 11 = 0 + 11
x2 - 10x = 11
NOW COMPLETE THE SQUARE AS ABOVE:
x2 - 10x + 25 = +25 + 11
(x - 5)2 = 36
√ (x - 5)2 = √ 36
x - 5 = ± 6
x = 5 ± 6
x = 11 and x = -1
Again, this one factored easily so I wouldn't have even used completing the square. ALWAYS CHECK IF IT FACTORS FIRST!
Now an example that DOES NOT FACTOR: x2 - 10x - 18 = 0
x2 - 10x - 18 = 0
x2 - 10x - 11 + 18 = 0 + 18
x2 - 10x = 18
NOW COMPLETE THE SQUARE AS ABOVE:
x2 - 10x + 25 = +25 + 18
(x - 5)2 = 43
√ (x - 5)2 = √ 43
x - 5 = ± √ 43
x = 5 ± √ 43
x = 5 + √ 43 and x = 5 - √ 43
When there is an IRRATIONAL square root, always SIMPLIFY if possible!
IF THERE IS AN "a" COEFFICIENT, YOU MUST DIVIDE EACH TERM BY IT BEFORE YOU CAN COMPLETE THE SQUARE:
Example: 2x2 - 3x - 1 = 0
Move the 1 to the other side of the equation: 2x2 - 3x = 1
Divide each term by the "a" coefficient:
x2 - 3/2 x = 1/2
Now find the completing the square term and add it to both sides:
[(-3/2)(-3/2)]2 = 9/16
x2 - 3/2 x + 9/16 = 1/2 + 9/16
(x - 3/4)2 =8/16 + 9/16
(x - 3/4)2 = 17/16
√[(x - 3/4)2 ] = ±√ [17/16]
x - 3/4 = ±[√17] /4
x = 3/4 ±[√ 17] /4
x = (3 ± [√17]) /4
or written better x =
3 ± √ 17
4
Wednesday, March 19, 2014
Math 6A ( Periods 1 & 2)
Adding Integers 11-2
Rules: The sum of two positive integers is a positive integer.
The sum of two negative integers is a negative integer.
So- if the two numbers have the same sign, use their sign and just add the numbers.
-15 + -13 = - 28
-10 + -4 = -14
Rules: The sum of a positive integer and a negative integer is :
POSITIVE… IF the positive number has a greater absolute value
NEGATIVE… IF the negative number has a greater absolute value
ZERO… IF both numbers have the same absolute value
Think of a game between two teams- The POSITIVE TEAM and The NEGATIVE TEAM.
30 + -16 … ask yourself the all important question…
“WHO WINS?
in this case the positive and then ask
“BY HOW MUCH?”
take the difference 14
14 + - 52…
“WHO WINS?”
the negative… “BY HOW MUCH?”
38
so the answer is -38
(-2 + 3) + - 6 you can work this 2 ways
(-2 + 3) + - 6 = 1 + -6 = -5 or
using all the properties that work for whole numbers
Commutative and Associative properties of addition
can change expression to (-2 + -6) + 3 or -8 + 3 = -5 you still arrive at the same solution.
You want to use these properties when you are adding more than 2 integers.
First look for zero pairs—you can cross them out right away!!
3 + (-3) = 0
-9 + 9 = 0
Then you can use C(+) to move the integers around to make it easier to add them together rather than adding them in the original order. In addition, you can use A(+) to group your positive and negative numbers in ways that make it easier to add as well.
One surefire way is to add all the positives up… and then add all the negatives up.
At this point ask yourself that all important question… WHO WINS? …
use the winner’s sign..
and then ask yourself..
BY HOW MUCH?
example:
-4 + 27 +(-6) + 5 + (-4) + (6) + (-27) + 13
Taking a good scan of the numbers, do you see any zero pairs?
YES—so cross them out and you are left with
-4 + 5 + (-4) + 13
add your positives 5 + 13 = 18
add your negatives and use their sign – 4 + -4 = -8
Okay, Who wins? the positive
By how much? 10
so
-4 + 27 +(-6) + 5 + (-4) + (6) + (-27) + 13 = 10
Rules: The sum of two positive integers is a positive integer.
The sum of two negative integers is a negative integer.
So- if the two numbers have the same sign, use their sign and just add the numbers.
-15 + -13 = - 28
-10 + -4 = -14
Rules: The sum of a positive integer and a negative integer is :
POSITIVE… IF the positive number has a greater absolute value
NEGATIVE… IF the negative number has a greater absolute value
ZERO… IF both numbers have the same absolute value
Think of a game between two teams- The POSITIVE TEAM and The NEGATIVE TEAM.
30 + -16 … ask yourself the all important question…
“WHO WINS?
in this case the positive and then ask
“BY HOW MUCH?”
take the difference 14
14 + - 52…
“WHO WINS?”
the negative… “BY HOW MUCH?”
38
so the answer is -38
(-2 + 3) + - 6 you can work this 2 ways
(-2 + 3) + - 6 = 1 + -6 = -5 or
using all the properties that work for whole numbers
Commutative and Associative properties of addition
can change expression to (-2 + -6) + 3 or -8 + 3 = -5 you still arrive at the same solution.
You want to use these properties when you are adding more than 2 integers.
First look for zero pairs—you can cross them out right away!!
3 + (-3) = 0
-9 + 9 = 0
Then you can use C(+) to move the integers around to make it easier to add them together rather than adding them in the original order. In addition, you can use A(+) to group your positive and negative numbers in ways that make it easier to add as well.
One surefire way is to add all the positives up… and then add all the negatives up.
At this point ask yourself that all important question… WHO WINS? …
use the winner’s sign..
and then ask yourself..
BY HOW MUCH?
example:
-4 + 27 +(-6) + 5 + (-4) + (6) + (-27) + 13
Taking a good scan of the numbers, do you see any zero pairs?
YES—so cross them out and you are left with
-4 + 5 + (-4) + 13
add your positives 5 + 13 = 18
add your negatives and use their sign – 4 + -4 = -8
Okay, Who wins? the positive
By how much? 10
so
-4 + 27 +(-6) + 5 + (-4) + (6) + (-27) + 13 = 10
Tuesday, March 18, 2014
Math 6A ( Periods 1 & 2(
Negative Numbers 11-1
On a horizontal number line we use negative numbers for the coordinates of points to the left of zero. We denote the number called ‘negative four’ by the symbol -4. The symbol -4 is normally read ‘ negative 4’ but we can also say ‘ the opposite of 4.’
The graphs of 4 and -4 are the same distance from 0—>but in opposite directions. Thus they are opposites. -4 is the opposite of 4.
The opposite of 0 is 0
Absolute Value is a distance concept. Absolute value is the graph of the distance of a number from 0 on a number line. The absolute value of a number can NEVER be negative!!
Counting (also known as Natural) numbers: 1, 2, 3, 4, ….
Whole numbers 0, 1, 2, 3, 4….
Integers are natural numbers and their opposites AND zero
…-4, -3, -2, -1, 0, 1, 2, 3, 4….
The opposite of 0 is 0.
The integer 0 is neither positive nor negative.
The farther we go to the right on a number line--- the bigger the number. We can compare two integers by looking at their position on a number line.
if x < 0 what do we know? x is negative number if x > 0, what do we know? x is a positive number
We have been practicing representing integers by their graphs, that is, by points on a number line.
Make sure that your number line includes arrows at both ends and a line indicating where zero falls on your number line.
The graph of a number MUST have a closed dot right on the number line at that specific number.
Please see our textbook page 366 for an accurate example.
On a horizontal number line we use negative numbers for the coordinates of points to the left of zero. We denote the number called ‘negative four’ by the symbol -4. The symbol -4 is normally read ‘ negative 4’ but we can also say ‘ the opposite of 4.’
The graphs of 4 and -4 are the same distance from 0—>but in opposite directions. Thus they are opposites. -4 is the opposite of 4.
The opposite of 0 is 0
Absolute Value is a distance concept. Absolute value is the graph of the distance of a number from 0 on a number line. The absolute value of a number can NEVER be negative!!
Counting (also known as Natural) numbers: 1, 2, 3, 4, ….
Whole numbers 0, 1, 2, 3, 4….
Integers are natural numbers and their opposites AND zero
…-4, -3, -2, -1, 0, 1, 2, 3, 4….
The opposite of 0 is 0.
The integer 0 is neither positive nor negative.
The farther we go to the right on a number line--- the bigger the number. We can compare two integers by looking at their position on a number line.
if x < 0 what do we know? x is negative number if x > 0, what do we know? x is a positive number
We have been practicing representing integers by their graphs, that is, by points on a number line.
Make sure that your number line includes arrows at both ends and a line indicating where zero falls on your number line.
The graph of a number MUST have a closed dot right on the number line at that specific number.
Please see our textbook page 366 for an accurate example.
Monday, March 17, 2014
Algebra Honors ( Period 6 & 7)
Quadratic Equations with Perfect Squares 12-1
In Chapter 5 you learned how to solve certain quadratic equations by factoring and in Chapter 11 you learned how to solve quadratics in the for
x2 = k
as in x2 = 49
x = ±7
This lesson extend to any quadratic equation involving a perfect square
if we have x2 = k
if:
k > 0 then x2 = k has two real-numbered roots with x = ±√k
if k = 0 then x2 = k has one real numbered root x = 0
if k<0 nbsp="" sup="" then="" x="">2
= k has no real numbered roots
m2 = 49
√(m2) = ±√49
m = ±7
{-7, 7}
5r2= 45
5r2 /5= 45/5
r2= 9
√r2= ±√9
r = ±3
{-3,3}
(x + 6)2 = 64
√(x+6)2 = ±√64
x+6 = ±8
be careful here
you now have
x = -6±8 which means
x = -6-8 = -14 AND x = -6+8 = 2
{-14, 2}
9r2 = 121
9r2/9 = 121/9
r2 = 121/9
√r2= ±√121/9
r = ±11/3
Make sure to check your solutions to insure that they both work
(x-3)2 = 100
√ (x-3)2 = ±√100
(x-3) = ±10
x = 3 ±10
x = 3 +10 = 13 and x = 3-10 = -7
{-7, 13}
5(x-4)2 = 40
5(x-4)2/5 = 40/5
(x-4)2 = 8
√ (x-4)2 = ±√8
Now you need to simplify the Radical
√ (x-4)2 = ±2√2
(x-4) = ±2√2
x = 4 ±2√2
{4 - 2√2, 4 + 2√2}
7(x - 8)2 = -28
7(x-8)2/7 = -28/7
(x-8)2 = -4
WAIT!!! that can't happen no real numbered solutions...
An equation that has a negative on one side and a perfect square as the other has NO REAL NUMBER Solutions
y2 + 6y + 9 = 49
(y+3)2 = 49
√(y+3)2 = ±√49
(y+3) = ±7
y = -3 ±7
y = -3-7 = -10 and y = -3 +7 = 4
{-10. 4}
Perfect Squares like (x -4) 2 and ( y + 3) 2
The square of any real number is always a non negative real number.
2(3x-5)2 + 15 = 7 becomes
2(3x -5)2 = -8
or (3x -5)2 = -4... NO REAL number solution!!
In Chapter 5 you learned how to solve certain quadratic equations by factoring and in Chapter 11 you learned how to solve quadratics in the for
x2 = k
as in x2 = 49
x = ±7
This lesson extend to any quadratic equation involving a perfect square
if we have x2 = k
if:
k > 0 then x2 = k has two real-numbered roots with x = ±√k
if k = 0 then x2 = k has one real numbered root x = 0
if k<0 nbsp="" sup="" then="" x="">2
= k has no real numbered roots
m2 = 49
√(m2) = ±√49
m = ±7
{-7, 7}
5r2= 45
5r2 /5= 45/5
r2= 9
√r2= ±√9
r = ±3
{-3,3}
(x + 6)2 = 64
√(x+6)2 = ±√64
x+6 = ±8
be careful here
you now have
x = -6±8 which means
x = -6-8 = -14 AND x = -6+8 = 2
{-14, 2}
9r2 = 121
9r2/9 = 121/9
r2 = 121/9
√r2= ±√121/9
r = ±11/3
Make sure to check your solutions to insure that they both work
(x-3)2 = 100
√ (x-3)2 = ±√100
(x-3) = ±10
x = 3 ±10
x = 3 +10 = 13 and x = 3-10 = -7
{-7, 13}
5(x-4)2 = 40
5(x-4)2/5 = 40/5
(x-4)2 = 8
√ (x-4)2 = ±√8
Now you need to simplify the Radical
√ (x-4)2 = ±2√2
(x-4) = ±2√2
x = 4 ±2√2
{4 - 2√2, 4 + 2√2}
7(x - 8)2 = -28
7(x-8)2/7 = -28/7
(x-8)2 = -4
WAIT!!! that can't happen no real numbered solutions...
An equation that has a negative on one side and a perfect square as the other has NO REAL NUMBER Solutions
y2 + 6y + 9 = 49
(y+3)2 = 49
√(y+3)2 = ±√49
(y+3) = ±7
y = -3 ±7
y = -3-7 = -10 and y = -3 +7 = 4
{-10. 4}
Perfect Squares like (x -4) 2 and ( y + 3) 2
The square of any real number is always a non negative real number.
2(3x-5)2 + 15 = 7 becomes
2(3x -5)2 = -8
or (3x -5)2 = -4... NO REAL number solution!!
Algebra Honors ( Periods 6 & 7)
Introduction to Quadratic Equations
We learned from Chapter 8 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 12 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 where 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 5)
but...some quadratics are not factorable
3) Square root each side (+ or - square root on the answer side) We did this in Chapter 11
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! :)
REVIEW:
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 5 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!!
METHOD 3:
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
METHOD 4:
The Quadratic Formula!!
We will get to this one...
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
We learned from Chapter 8 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 12 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 where 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 5)
but...some quadratics are not factorable
3) Square root each side (+ or - square root on the answer side) We did this in Chapter 11
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! :)
REVIEW:
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 5 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!!
METHOD 3:
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
METHOD 4:
The Quadratic Formula!!
We will get to this one...
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
Wednesday, March 12, 2014
Algebra Honors ( Periods 6 & 7)
Direct & Inverse Variations 8-9 and 8-10
Direct Variations 8-9
f(x) = mx + b
It is a linear function. the f(x) is dependent on the x value.
A direct variation is a function defined by an equation in the form
y = kx, where k is a non zero constant. When graphing, k is the slope
You can say that y varies directly as x
that is, as x goes up in value so does y or
as x decreases à so does y
When the domain is the set of all real numbers, the graph of a direct variation is a straight line with slope k that passes through the origin.
Given that m varies directly as n, and that m= 42 when n = 2, the constant of variation can be found by writing m = kn and substituting in those values. You find that
42 = 2k , so k = 21.
You can write the function as m = 21n.
Now when asked to find the value of m when n = 3 you just plug and chug…
m = 21(3) so m = 63
You can write the function as m = 21n.
Now when asked to find the value of m when n = 3 you just plug and chug…
m = 21(3) so m = 63
Could we have found this a different way? YES..
Suppose ( x1, y1) and ( x2, y2) are two ordered pairs of a direct variation defined by
y = kx and you know that neither x1 nor x2 are zero.
You know that
y1 = kx1 and that y2= kx2
Solving both for k you discover that
y1/x1 = k and so does y2/x2 = k
Since each ratio equals k, the ratios are equal and you can set them in an equation
y1/x1 = y2/x2 and you read this… “y1 is to x1 as y2 is to x2.”
This is a proportion!!
For this reason k is sometimes called the constant of proportionality and
y is said to be directly proportional to x
y is said to be directly proportional to x
When you use a proportion to solve a problem, you will want to recall that the product of the extremes equals the product of the means.
We reviewed a few equations and found the following to be direct variations:
y = 3x
p = 9s
d = 3.3t
even y/x = -5
But the following were determined NOT to be direct variations:
y = 3x2
xy = 4
Eample:
y varies directly as x
y = 6 and x = 72 Find the constant of variation
y = kx
6 = k(72)
6 = 72k
k = 1/12
Inverse Variations 8-10
An inverse variation is a function defined by an equation in the form:
xy = k where k is a NON ZERO constant
or y = k/x where x ≠ 0
You say that y varies inversely as x or that y is inversely proportional to x.
The constant k is the constant of variation.
The graph of an inverse function is NOT a straight line!
xy = k is NOT linear!
Your graph can be a hyperbola
When k is positive the branches of the graph are in Quadrants I and III
When k is negative the branches of the graph are in Quadrants II and IV
Similarly to direct variation, you can compare two ordered pairs of the same inverse variation. Since the coordinates must satisfy the equation xy = k you know that
x1y1=k and x2y2= k
or
x1y1= x2y2
Reviewing :
Direct Variation is y = kx or y1/x1 = y2/x2
Inverse Variation is xy = k or x1y1= x2y2
These equations show that for direct variation the quotients of the coordinates are constant and for inverse variation the products of the coordinates are constant.
Is it Direct or Inverse Variation?
y/x = k (careful this becomes y = kx so its direct)
y= k/x inverse
p = k/z inverse
xy = 25 inverse
d= 40t direct
m/n= 5/8 direct
x/y = 1/k direct
kxy = 5 inverse
WORD PROBLEMS FOR DIRECT & INVERSE VARIATIONS ( 2nd day of lesson)
Several examples were given in class:
Truck rental Company charges $35 a day plus 21 cents per mile. Normally your questions in the past were "What is the cost for a rental of a truck for ...:
1 day and 340 miles?" or
"2 days and 450 miles?"
You would just plug in and figure out the exact cost...
Then we discussed the rental of a chain saw (from the textbook)... remember it is for cutting down trees... like those which were knocked down by those tremendous winds we had last year… remember those winds…
We are given that the rental is $5.90 a hour and you must pay $6.50 for 1 can of gas. Again, in the past your questions would be something like...
"How much would it cost for 7.5 hours?"
But... you could write a linear function to represent the cost for all different rental hours.
Let h represent the hours
s(h) = 5.9h + 6.5
Now, no matter how many hours you rent the chain saw, you can figure out the cost.
Phone bills in the past charged 15 cents per message + a base charge. Let's say you were given the July bill of $18 which included 62 messages. What was the base charge?
First find out the cost of the messages (.15)(62) = $9.30 and subtract that from $18.
Or just realize it would be 18 - (.15)(62) = $ 8.70
Then you could write a linear function p(x) = .15x + 8.7
August had 76 messages... what was the bill becomes easy to solve-- just use the linear equation.
p(76) = .15(76) + 8.7 = 20.10. August bill was $20.10
Turn to Page 394
#20
distance on a map varies directly to actual distance
m= distance on the map
d= actual distance
m = kd
Given that 1 in on the map ---> 10 miles
1 = k(10)
1 = 10k
k = 0.1
so formula is m = 0.1d
writing as a proportion you would have
1/10 = m2 / d2
# 22 Volume directly proportional to temp T in Kelvin
5 Liters 300 degrees
V = kT
5= 300k
k = 1/60
so formula is V = (1/60)T
and as a formula before you simplify
5/300 = V2 /T2
or 1/60 = V2 /T2
Math 6A ( Periods 1 & 2)
Scale Drawing 7-9
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
A scale drawing is a diagram of an object where its length and width aare proportional to the actual length and width.
The scale for the drawing gives the relationship between the drawing's measurements and the actual measurements.
For example 1 inch represents 2 feet
The Scale Factor is the ratio of the length in the drawing to the corresponding actual lenth.
A Scale Factor on a matchbox car of 1/64 would mean that every part of the matchbox car was 1/64 of the actual car-- the windshield, the door, the wheels, the roof... would all have the same relationship.
Scale Factor MUST HAVE the same units of measurements.
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
Open 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
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 in millimeters ?
Let a= the actual length of the insect in cm
7/1 = 5.6/ a
7a = 5.6
a = .8 cm
or 7/a = 56/
The actual length of the insect is 0.8 cm which is 8 mm
OR change the given 5.6 cm into 56 mm to begin with!
this time
let a = the actual length of the insect in mm
7/1 = 56/a
Now we can simplify easily
a = 8
so
the actual length of the insect is 8 mm
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
A scale drawing is a diagram of an object where its length and width aare proportional to the actual length and width.
The scale for the drawing gives the relationship between the drawing's measurements and the actual measurements.
For example 1 inch represents 2 feet
The Scale Factor is the ratio of the length in the drawing to the corresponding actual lenth.
A Scale Factor on a matchbox car of 1/64 would mean that every part of the matchbox car was 1/64 of the actual car-- the windshield, the door, the wheels, the roof... would all have the same relationship.
Scale Factor MUST HAVE the same units of measurements.
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
Open 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
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 in millimeters ?
Let a= the actual length of the insect in cm
7/1 = 5.6/ a
7a = 5.6
a = .8 cm
or 7/a = 56/
The actual length of the insect is 0.8 cm which is 8 mm
OR change the given 5.6 cm into 56 mm to begin with!
this time
let a = the actual length of the insect in mm
7/1 = 56/a
Now we can simplify easily
a = 8
so
the actual length of the insect is 8 mm
Algebra Honors ( Periods 6 & 7)
Linear and Quadratic Functions 8-8
This lesson will be completed in two parts:
Linear Functions:
The function g defined by g(x) = 2x - 3 is called a linear function
Notice the word line in linear
If its domain is the set of all real numbers then the straight line that is the graph of
y = g(x) = 2x - 3 is the graph of g .
The slope of the graph is 2 and the y-intercept is -3.
A function f defined by f(x) = mx + b is a linear function
If the domain of f is the set of all real numbers, then its graph is the straight line with slope m and y-intercept b.
Relations:
A relation is any set of ordered pairs.
The set of the first coordinates of the ordered pairs is the domain of the relation
The set of the second coordinates of the ordered pair is the range.
A function is a relation in which different ordered pairs have different FIRST coordinates.
Tuesday, March 11, 2014
Math 7 (Period 4)
Multiplying and
Dividing Powers 6.7
We are discovering
the basic multiplication rules for powers having the same bases.
Multiplying Powers
a5 ∙a3=
a∙a∙a∙a∙a∙a∙a∙a = a8
RULE….to multiply two powers
with the SAME BASES just add their exponents
am ∙
an = am+n
32∙33
= 35
( -½)2
(-½ ) = (-½ )3
Multiplygin
Variable Expressions:
2x2∙3x4
= (2∙3)(x2∙x4) = 6x2+4 = 6x6
Dividing
Powers
c5/c3
c∙c∙c∙c∙c
c∙c∙c
c∙c∙c
= c∙c
= c2
c5/c3
= c5-3 = c2
To divide two powers with the SAME BASES just subtract the
exponents
am/an
= am-n
27/24=
27-4 = 23
(-5)3/-5
Careful
with this one… it is really
(-5)3/(-5)1
Now they
are the same base so it is just a subtraction problem with the exponents
(-5)3-1
=(-5)2 = (-5)(-5) = 25
So let’s
look at two examples
(-2)5
means (-2)(-2)(-2)(-2)(-2) = -32
But
(-2)4
means (-2)(-2)(-2)(-2) = + 16
(-2)11
à negative
(-2)10
à positive
***Odd/Even Rule with powers***
If there
is a negative inside the parentheses
odd # of
negative signs or odd power à it will be negative
even #
of negative signs or even power à it will be positive
(-4)
55 Ã negative
(-4)56
à positive
(-1)15
= -1
(-1)100
= +1
If there
is a negative but NO parentheses (NO HUGS!!)
such as
-25 or -24
….IT
will ALWAYS be negative
-25
means take the opposite of 25
and -24
means take the opposite of 24
-25
= -2∙2∙2∙2∙2 = -32
-24
= -2∙2∙2∙2 = -16
(-10)4/(-10)2
= (-10)4-2 = (-10)2
Be
careful
-102
≠ (-10)2
-100 ≠ 100
(2/5)6
(2/5)3
(2/5)3
= (2/5)6-3
=8/125
Math 6A ( Periods 1 & 2)
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
Mrs. Nelson bought 4 tires for her car ( an Infiniti- what else would a math teacher own!) 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.
tires ___________ = _____________tires
cost _____________=_____________cost
If 4 bars of soap cost $1.50, how much would 6 bars of soap cost?
let x = the cost of 6 bars of soap
bars ___________ = _____________bars
cost _____________=_____________cost
4/1.50 = 6/x
simplify first up and down ( BUT NEVER DIAGONALLY!!)
x = 3 (.75)
x = 2.25
$2.25
6 bars would cost $2.25
If 9 kg of fertilizer will feed 300 m2 of grass, how much fertilizer would be required to feed 500 m2 of grass?
kg ___________ = _____________kg
m2 _____________=_____________m2
9/300 = n/500
SIMPLIFY FIRST
9/3 = n/5
3/1 = n/5
n = 15
15kg of fertilzer
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
Mrs. Nelson bought 4 tires for her car ( an Infiniti- what else would a math teacher own!) 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.
tires ___________ = _____________tires
cost _____________=_____________cost
If 4 bars of soap cost $1.50, how much would 6 bars of soap cost?
let x = the cost of 6 bars of soap
bars ___________ = _____________bars
cost _____________=_____________cost
4/1.50 = 6/x
simplify first up and down ( BUT NEVER DIAGONALLY!!)
x = 3 (.75)
x = 2.25
$2.25
6 bars would cost $2.25
If 9 kg of fertilizer will feed 300 m2 of grass, how much fertilizer would be required to feed 500 m2 of grass?
kg ___________ = _____________kg
m2 _____________=_____________m2
9/300 = n/500
SIMPLIFY FIRST
9/3 = n/5
3/1 = n/5
n = 15
15kg of fertilzer
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.
Tuesday, March 4, 2014
Math 6A ( Periods 1 & 2)
Ratio 7-6 Word Problems
Since all of the word problems from Ratios 7-6 are excellent examples of using Ratios, I thought I would create a blog posting with the solutions. Although I will list the actually word problems, you might need to turn to the textbook and page 229 for any charts included.
Read each problem carefully and set up the solutions based on the question asked
Page 229
1. What is the cost of grapes in dollars per kilogram if 4.5 kg of grapes costs $ 7.56?
Since they want $/kg you need to have 7.56/ 4.5 = 1.68; so the grapes cost $ 1.68/kg.
2. This one looked confusing, but just read it a few times: The index of refraction of a transparent substance is the ratio of the speed of light in space to the speed of light in the substance. (Read it again—it is just a ratio). Using the table on Page 229, find the index of refraction of
A. Glass
B. Water
According to the table Speed of light in space is 300,000 km/sec
Speed of light in glass is 200,000 km/ sec.
So
300,000
200,000
or 3/2
Speed of Light in Water is 225,000
so that ratio is
300,000
225,000
This takes a little more time dividing but you can simplify it to 4/3 Since it is a ratio you want to LEAVE it as an improper fraction.
3. The mechanical advantage of a simple machine is the ratio of the weight lifted by the machine to the force necessary to lift it. Now—just read that again—you don’t need to truly understand physics to get this problem. It is a RATIO again!! What is the mechanical advantage of a jack that lifts a 3200-pound car with a force of 120 pounds?
3200 =
120
80/3; so the mechanical advantage is 80/3
4. The C-String of a cello vibrates 654 times in 5 seconds, How many vibrations per second is that? (We are finding unit rate)
It’s a rate of time/sec.
654/ 5 = 130 4/5 vibrations per second
5. A four cubic foot volume of water at sea level weighs 250 lb. What is the density of water in pounds per cubic foot? (Do you need to understand density to do this problem?) No—just unit rates!! We want lbs/cubic foot. Look at the information and place it with lbs/ft and then divide to get the unit rate.
250lb/ 4 ft3. That equals 62 1/2 lb/ft3 .
6. A share of stock that costs $88 earned $16 last year. What was the price-to-earnings ratio of this stock? (Wish that was happening now!!)
Price/earnings so look carefully 88/16 = 11/2 so the Ratio is 11/2
In exercises 7 and 8 you will need to look at the diagrams on page 230. You are finding the ratio in lowest terms.
Look at both the small and large triangles. For A you are finding the ratio between segment AB of the larger triangle and DE of the smaller triangle.
4.8
3.2
Simplify ( I like to clear decimals to make it easier but in this case you could easily divide
= 3/2
B.
Perimeter of Triangle ABC
Perimeter of Triangle DEF
First you need to find the perimeter of the larger triangle by adding up the three sides
6.3 + 4.8 + 3 = 14.1
and then finding the perimeter of the smaller triangle by adding up those three sides
4.2 + 3.2 + 2 = 9.4
14.1
9.4
I like to clear decimals by multiplying by 10
141
94
= 3/2
What did you notice?
8. Again look at Page 230 for the figure
We are finding the following Ratios:
PQ:TU 2.6:6.5 or 26:65 = 2:5
QR:UV 1.2:3 = 12:30 = 2:5
Perimeter of PQRS: Perimeter of TUVW
You need to add up all the sides of each of the figures
2.2 + 2.6 + 1.2 + 3 = 9
and 5.5 + 6.5 + 3 + 7.5 = 22.5
9:22.5 = 90:225 = 2:5
What did you notice here?
For exercised 9-12 you need to use the Table on Page 230 as well
9. The population of Centerville in 1980 to its population in 1970?
44/36 = 11/9
10. The growth in the population of Easton to its 1980 population.
You need to subtract the 1970 population from the 1980 population to get the growth so
28-16 = 12 12/18 = 3/7
11. The total population of both towns in 1970 to their total population in 1980
36 + 16 = 52 and 44 + 28 = 72 52/72 = 13/ 18
12. The total growth in the population of both towns to their total 1980 population
Like number 10 you need to subtract to find the growth
(44-36) + (28-16) = 8 + 12 = 20
44 + 28 = 72 (from before)
20/72 = 5/18
13. During a season a baseball player hit safely in 135times at bat; the player struck out or was fielded out in 340 times at bat. What is the player’s ratio of hits to times at bat?
You take the 135 + 340 = 475 (the total times at bat)
135/475 = 37/95
14. A fruit drink recipe requires fruit juice and milk in the ration 3:5. What fraction of the drink is milk? What fraction is juice? Remember we need to find the total first. 3 + 5 = 8
so the fraction that is milk is 5/8 and the fraction that is fruit juice is 3/8
Since all of the word problems from Ratios 7-6 are excellent examples of using Ratios, I thought I would create a blog posting with the solutions. Although I will list the actually word problems, you might need to turn to the textbook and page 229 for any charts included.
Read each problem carefully and set up the solutions based on the question asked
Page 229
1. What is the cost of grapes in dollars per kilogram if 4.5 kg of grapes costs $ 7.56?
Since they want $/kg you need to have 7.56/ 4.5 = 1.68; so the grapes cost $ 1.68/kg.
2. This one looked confusing, but just read it a few times: The index of refraction of a transparent substance is the ratio of the speed of light in space to the speed of light in the substance. (Read it again—it is just a ratio). Using the table on Page 229, find the index of refraction of
A. Glass
B. Water
According to the table Speed of light in space is 300,000 km/sec
Speed of light in glass is 200,000 km/ sec.
So
300,000
200,000
or 3/2
Speed of Light in Water is 225,000
so that ratio is
300,000
225,000
This takes a little more time dividing but you can simplify it to 4/3 Since it is a ratio you want to LEAVE it as an improper fraction.
3. The mechanical advantage of a simple machine is the ratio of the weight lifted by the machine to the force necessary to lift it. Now—just read that again—you don’t need to truly understand physics to get this problem. It is a RATIO again!! What is the mechanical advantage of a jack that lifts a 3200-pound car with a force of 120 pounds?
3200 =
120
80/3; so the mechanical advantage is 80/3
4. The C-String of a cello vibrates 654 times in 5 seconds, How many vibrations per second is that? (We are finding unit rate)
It’s a rate of time/sec.
654/ 5 = 130 4/5 vibrations per second
5. A four cubic foot volume of water at sea level weighs 250 lb. What is the density of water in pounds per cubic foot? (Do you need to understand density to do this problem?) No—just unit rates!! We want lbs/cubic foot. Look at the information and place it with lbs/ft and then divide to get the unit rate.
250lb/ 4 ft3. That equals 62 1/2 lb/ft3 .
6. A share of stock that costs $88 earned $16 last year. What was the price-to-earnings ratio of this stock? (Wish that was happening now!!)
Price/earnings so look carefully 88/16 = 11/2 so the Ratio is 11/2
In exercises 7 and 8 you will need to look at the diagrams on page 230. You are finding the ratio in lowest terms.
Look at both the small and large triangles. For A you are finding the ratio between segment AB of the larger triangle and DE of the smaller triangle.
4.8
3.2
Simplify ( I like to clear decimals to make it easier but in this case you could easily divide
= 3/2
B.
Perimeter of Triangle ABC
Perimeter of Triangle DEF
First you need to find the perimeter of the larger triangle by adding up the three sides
6.3 + 4.8 + 3 = 14.1
and then finding the perimeter of the smaller triangle by adding up those three sides
4.2 + 3.2 + 2 = 9.4
14.1
9.4
I like to clear decimals by multiplying by 10
141
94
= 3/2
What did you notice?
8. Again look at Page 230 for the figure
We are finding the following Ratios:
PQ:TU 2.6:6.5 or 26:65 = 2:5
QR:UV 1.2:3 = 12:30 = 2:5
Perimeter of PQRS: Perimeter of TUVW
You need to add up all the sides of each of the figures
2.2 + 2.6 + 1.2 + 3 = 9
and 5.5 + 6.5 + 3 + 7.5 = 22.5
9:22.5 = 90:225 = 2:5
What did you notice here?
For exercised 9-12 you need to use the Table on Page 230 as well
9. The population of Centerville in 1980 to its population in 1970?
44/36 = 11/9
10. The growth in the population of Easton to its 1980 population.
You need to subtract the 1970 population from the 1980 population to get the growth so
28-16 = 12 12/18 = 3/7
11. The total population of both towns in 1970 to their total population in 1980
36 + 16 = 52 and 44 + 28 = 72 52/72 = 13/ 18
12. The total growth in the population of both towns to their total 1980 population
Like number 10 you need to subtract to find the growth
(44-36) + (28-16) = 8 + 12 = 20
44 + 28 = 72 (from before)
20/72 = 5/18
13. During a season a baseball player hit safely in 135times at bat; the player struck out or was fielded out in 340 times at bat. What is the player’s ratio of hits to times at bat?
You take the 135 + 340 = 475 (the total times at bat)
135/475 = 37/95
14. A fruit drink recipe requires fruit juice and milk in the ration 3:5. What fraction of the drink is milk? What fraction is juice? Remember we need to find the total first. 3 + 5 = 8
so the fraction that is milk is 5/8 and the fraction that is fruit juice is 3/8
Subscribe to:
Posts (Atom)
