Math 6H ( Periods 3, 6, & 7)

Multiplication and Division of Mixed Numbers 7-5

One method of finding the product of two mixed numbers is to first change the mixed numbers into improper fractions and then multiply

6 X 3 1/12 becomes
6/1 X 37/12
Simplify using the methods taught and reviewed this week
6/1 X 37/ 12 = 37/6 = 18 1/2

5 3/4 X 4 2/3
23/4 X 14/3 becomes
23/2 X 7/ 3 = 161/6 = 26 5/6

To divide one mixed number by another, we change the mixed numbers into improper fractions and use the methods from the previous lessons

2 2/3 ÷ 10 2/3 first change to improper fractions
8/3 ÷ 32/3 Remember. that dividing by a fraction is the same a multiplying by its reciprocal
8/3 X 3/32
Using the GCF to simplify first
1/1 X 1/4 = 1/4

What about
1 7/15 ÷ 5 1/2
22/15 ÷ 11/2
22/15 X 2/11 = 4/15

What if we have equations such as the book shows
n X 2 1/3 = 6 5/12
First rewrite it as (2 1/3)n = 6 5/12
Then change the mixed numbers to improper fractions
(7/3)n = 77/12
Now, we know to isolate the variable, we must use the inverse operation and in this case we would multiply 7/3 by its reciprocal 3/7 to both sides
(3/7)(7/3) n = (77/12)(3/7)
n = 11/4
n = 2 3/4

Math 6H ( Periods 3, 6, & 7)

Division of Fractions 7-4

Certain numbers when multiplied together have the product 1
5 X 1/5 = 1
3/4 X 4/3 = 1

Two numbers whose product is 1 are called reciprocals of each other.
Thus 3/4 is the reciprocal of 4/3.
Zero does not have a reciprocal


Look at the following:
We know 18 = 3 X 6 and we know 18 ÷ 6 = 3 as well as 18 X 1/6 = 3
Dividing a number by a fraction is the same as multiplying the number by the RECIPROCAL of the fraction
a/b ÷ c/d = a/b ÷ d/c
Remember- you are using the reciprocal of the divisor... that is , as students want to say "You FLIP the 2nd number!!"

42/ 55 ÷ 36/11
you must rewrite the problem using the reciprocal of the 2nd number
42/55 X 11/36
Now using your skills of observing GCF simplify before you multiply ( MUCH EASIER and FASTER)
42/ 5 X 1/36 which becomes 7/5 X 1/ 6 = 7/30

USING THE PYTHAGOREAN THEOREM - WORD PROBLEMS: 11-8
There are many real life examples where you can use the Pythagorean Theorem to find a length.

EXAMPLE: HOW HIGH A 10 FOOT LADDER REACHES ON A HOUSE
A 10 ft ladder is placed on a house 5 ft away from the base of the house.
Find how high up the house the ladder reaches.
The ladder makes a right triangle with the ground being one leg, the house the other, and the ladder is the hypotenuse ( see drawing in #1 on p. 515)
You need to find the distance on the house, so you're finding one leg.

ANOTHER EXAMPLE:
You're flying your kite for the kite project and you want to know how long the kite string must be so that it can reach a height of 13 ft in the air if you're standing 9 feet away from where the kite is in the air.
The string represents the hypotenuse.
You know one leg is the height in the air (13 ft) and the other leg is how far on the ground you are standing away from where the kite is flying (9 ft)
You need to find the hypotenuse.

EQUATIONS WITH RADICALS: 11-9
When you have an equation where the variable is under the √ sign,
simply square both sides to solve for the variable.

This is actually similar to regular equation balancing!
The goal is still to find out what the variable is.
But to find it, you need to ISOLATE the √ with the variable first
Then you square


Example #1 on p. 521
√x = 5
Square both sides and you'll get x = 25

Example # 9 is much harder
3 + √(x - 1) = 5
move the 3 to other side first √(x - 1) = 5 - 3
Now square both sides x - 1 = 22
Now add 1 to both sides: x = 4 + 1 x = 5

Look at Examples #15 & 16 ---- In both cases, there is no possible value for x because the square root of a number CAN NEVER BE NEGATIVE IN THE REAL NUMBER SYSTEM!

Try Example #17 yourself, and see what happens (you should also end up with no value, but why?)

Need to review how to estimate the value of square roots--Check out the blog post on Finding Square Roots

Math 6H ( Periods 3, 6, & 7)

Multiplication of Fractions 7-3

If a rectangle is divided into 4 equal parts, each part is ¼ of the whole. If each of these parts is then divided into 3 parts, that is into thirds, then there are 12 equal parts and each is 1/(3 ∙4) or 1/12 of the whole.

That is 1/3 of 1/4 is 1/(3 ∙4) or 1/12 and 1/3 ∙ 1/4 = 1/12 is

so another example 2/3 of 4/5 is 2∙4 /(3∙8) or 2/3 ∙4/5 = 8/15


Notice, that the numerator of the product, 8, is the product of the numerators 2 and 4. The denominator of the product, 15, is the product of the denominators 3 and 5

Rule
If a, b, c, and d are whole numbers with , then

a/b(c/d) = a∙c/(b∙d)


When multiplying two fractions, you can simplify the multiplication by dividing either of the numerators and either of the denominators by common factors

6/35 ( 7/3) we can simplify first because both 6 and 3 are divisible by 3
2/35 (7/1) and then both 35 and 7 are divisible by 7 so 2/5 (1(1) = 2/5

Try the following

25/6 ( 42/5) What can we do there?

7/8(20/21) How about with these two sets of fractions?

19/20 ( 25/38) … and these fractions?

Algebra Period 4

PYTHAGOREAN THEOREM 11-7
(an old friend) -
FOR RIGHT TRIANGLES ONLY!
2 legs - make the right angle - called ‘a’ and ‘b’
(doesn't matter which is which because you will add them and adding is COMMUTATIVE!)
hypotenuse - longest side across from the right angle - called ‘c’
You can find the third side of a right triangle as long as you know the other two sides:
a² + b² = c²
After squaring the two sides that you know, you'll need to find the square root of that number to find the length of the missing side (that's why it's in this chapter!)

EASIEST - FIND THE HYPOTENUSE (c)
Example #1 from p. 510
8² + 15² = c²
64 + 225 = c²
289 = c²
c = 17

A LITTLE HARDER - FIND A MISSING LEG (Either a or b)
Example #5 from p. 510
5² + b² = 13²
25 + b² = 169
b² = 169 - 25
b² = 144
b = 12

ONE THAT YOU WOULDN'T HAVE HAD IN PRE-ALGEBRA:
One of the legs = √5
√5² + b² = 13²
5 + b² = 169
b² = 169 - 5
b² = 164
√ b2 = √164
b = √4•41
b = 2√41

DISTANCE FORMULA
(based on the Pythagorean Theorem):
see p. 513 in book
The distance between any two points on the coordinate plane (x y plane)
The distance is the hypotenuse of a right triangle that you can draw using any two points on the coordinate plane (I'll show you how to draw it in class).

The formula is:
distance = √[( difference of the two x's)² + (difference of the two y's)²]
distance = √(x₁-x₂)² + (y₁- y₂)²
The difference between the 2x’s is the length of the leg parallel to the y axis and
the difference between the 2y’s is the length of the leg parallel to the x axis
The difference of the two x's is length of one of the two legs
The difference of the two y's is the length of the other leg
The distance is the length of the hypotenuse
So you could actually rewrite the distance formula to look like the Pythagorean Theorem:
c = √[a² + b²]

EXAMPLE: What is the distance between (3, -10) and (-7, -2)?
d = √[(3 - -7)² + (-10 - -2)²]
d = √[10² +( -8²)]
d = √(164)
Simplifying:
2√41
Check out the following website- with its 81 different proofs
Pythagorean Theorem and its many proofs. See if you can find the proof that is attributed to one of our US Presidents
Practice your skills with the Pythagorean Theorem using this Shodor Interactive Site at
Pythagorean Explorer

Math 6H ( Periods 3, 6, & 7)

Word Problems 7-1 & 7-2