Math 6 Honors Periods 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

Ø Decide which quantity is to be found and represent it by a variable

Ø Determine whether the quantities involved can be compared using ratios (rates)

Ø Equate the ratios in a proportion

Ø Solve the proportion

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

Number of tires	Cost
4	$264
5	c

4c = 5(264)

Although the problems in this lesson may be solved with out 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.

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

Algebra Period 3 (Tuesday)

Simplify, multiply, and divide RATIONAL EXPRESSIONS: 10-1, 10-2, & 10-3
Rational Expressions = Expressions in fraction format (division) with a variable in the denominator
You have already been simplifying, multiplying and dividing these throughout this year!

10-1: SIMPLIFY
You will need to FACTOR (Chapter 6) both the numerator and denominator and "cross out" common factors in both (their quotient is 1!)

EXAMPLE:

Simplify

y² + 3y + 2the=the(y + 2)(y + 1)thie= th y + 2
Ss y² - 1 andtheisto(y - 1)(y + 1)thethietiy - 1

10-2: MULTIPLY
Factor if possible, cross cancel if possible, multiply numerators, then denominators, simplify

EXAMPLE:

[(y + 4)³][y² + 4y + 4] THEI = (y + 4)³ (y + 2)² R= R y + 4
[(y + 2)³][y² + 8y + 16] THISTH(y + 2)³ (y + 4)² THEy + 2

10-3: DIVIDE
Same as the previous example, only this time you will need to FLIP THE SECOND FRACTION
then MULTIPLY!!!!!

EXAMPLE:
x + 1 ÷ x + 1 T= ( x + 1) ( x² - 2x + 1) =E( x + 1)( x - 1)(x - 1) =
x² – 1Tx² - 2x + 1 T(x + 1)( x - 1) (x + 1)TH(x + 1)( x - 1)(x + 1) TH

x - 1
x + 1

Algebra Period 3 (Monday)

Radical Equations with Quadratics 13-6

THIS IS A REVIEW OF CHAPTER 11!!!

You square both sides to get rid of the radical sign
For these problems, you'll get a quadratic on one side after you square
BE SURE TO CHECK BOTH ANSWERS TO MAKE SURE THEY BOTH CHECK!

WORD PROBLEMS WITH QUADRATICS 13-7
There are 2 major types: Frame problems, triangle problems

FRAME PROBLEMS:
You know the frame's dimensions and the picture's area inside. You need to find the width of frame.
1)Set x = width of the frame
2) Write the area formula using the dimensions given in the problem minus 2x
as the base and height of the area of the picture.
Why are we subtracting 2x? Because there are two widths, one on the top and one on the bottom, one on the right side and one on the left side
3) Set this product equal to the area of the picture given in the problem
EXAMPLE:
A picture frame is 20 cm by 12 cm. The picture has an area of 84 cm²
(20 - 2x)(12 - 2x) = 84
FOIL
240 - 40x - 24x + 4x² = 84
Combine like terms and bring the 84 over
4x² -64x + 240 - 84 = 84 - 84
4x² - 64x + 156 = 0
Divide by 4 on each side
x² - 16x + 39 = 0
FACTOR or use the QUADRATIC FORMULA
(x - 3)(x - 13 ) = 0
x = 3 cm or x = 13 cm
The 13 cm does not make sense! How can the width of the frame be more than one of its dimensions (the height was only 12 cm!!!)???
Therefore, the only answer is that the frame is 3 cm wide.

The right triangle example in the book on p. 603 uses the Pythagorean Theorem

Pre Algebra Periods 1, 2, & 4

Space Figures 10-4 (solids)
SPACE FIGURES OR SOLIDS OR 3 DIMENSIONAL FIGURES

prisms = 2 congruent parallel bases - all other sides are rectangles
cylinder = 2 congruent circle bases
When you remove one base from a prism, it becomes a pyramid - all other sides are triangles
When you remove one base from a cylinder, it becomes a cone
When you have a set of points in all directions that are equal distance from a central point, you have a sphere
Vertices - the points where edges connect (the corners)
Edges - the line segments that connect the vertices

You should be able to visualize what a figure will look like if you could cut it apart and open it—that is called a net

SURFACE AREA OF CYLINDERS & PRISMS 10-5
Surface area is just the sum of the areas of all the sides

SHORTCUT TO ADDING UP ALL THE SIDES:
Just find the perimeter of ONE base and then multiply that by the height of the prism
Add to that the areas of the TWO bases, and you'll have the surface area
S.A. = area of 2 bases + (perimeter of one base)(height of prism)

According to our textbook:

the perimeter of one base X height = LATERAL AREA (L.A.)

the area of the base (base X height) = B

so

S.A. = L.A. + 2B
For a cylinder, do the same exact thing except now it's the CIRCUMFERENCE of the base
S.A. = area of the 2 bases + (circumference of one base)(height of cylinder)

S.A. OF PYRAMIDS, CONES, AND SPHERES
PYRAMIDS 10-6
For pyramids, you can simply add up the areas of all the sides
S.A. = area of 1 base + lateral areas (sides)
Lateral areas are all triangles in a pyramid and A = 1/2 bh
This time the height means the height of the triangle that is a side. It has a special name = slant height
The slant height is denoted as a cursive l

so S.A. = area of base + (1/2 bl)(# of sides)

CONES
For cones, you have one circle base plus a sort of triangular piece with the base being rounded
The cursive l now stands for the length of the slant side of this piece

S.A. = area of the circle + (pi)(radius of base)(slant height)
S.A. = (pi)r² + (pi)rl [pronounce this pie roll!)

SPHERES
S.A. + 4(pi)r²
(I always thought of this as you need 4 circles to wrap the basketball!)