Chapter+Three


 * Chapter 3**

**Sec 3.1**
 * Formulas are equations with variables. People use a formula by “plugging in” the values for the different variables to get the result. The answer depends on the values used, but the formula, or equation, stays the same no matter which values you use.
 * Perimeter – The distance around the outside of something.
 * Square = 4s or four times the length of one side.
 * Rectangle = 2L + 2W or 2 times the length plus 2 times the width
 * Find the missing side(s) of a square or rectangle by plugging in what you do know in the formula and solving for the variable that you don’t know.
 * Perimeter of odd shapes is easy as long as the sides of the shape are straight lines. You just add up the lengths of all of the sides.
 * Picture Frame problem – When you add a border to a rectangle, draw a picture of what is going on and re-measure the length and width. You will find that it increases by double what you added (the width of the frame).
 * Sec 3.2 **
 * Area of a rectangle A=L(W)
 * The measurements of both sides must be the same. You can’t multiply inches by feet, for example. If they give you different units of measure, convert one or the other first before doing the multiplication.
 * The answer to an area problem will be expressed in square units (like ft2) because both of the sides being multiplied have a unit, so when they get multiplied, the numbers are multiplied and the units get multiplied as well. Ft times ft is ft2.)
 * If you have the area of a rectangle and need to find the length of one side, just divide the area by the measure of the other side.
 * Area of a square is s2, because all of the sides are the same so it is just side times side.
 * If you have the area of a square and need to find the length of a side, you need to figure out what number, times itself, equals the area. (This is the same thing as taking the square root of the area, but we won’t cover that until a later chapter.)
 * Area of a parallelogram A=Base times Height
 * The height is not the same as the width. It will usually be indicated by a dotted line.
 * The base is the same thing as the length.
 * The tablecloth problem is like the picture frame problem from the previous section. If you want the tablecloth to hang down over the tabletop, you will need to add twice the amount that it hangs down to both the length and the width. Draw a picture and re-measure the length and width before doing the math.
 * Sec 3.3**
 * Translating word phrases into algebraic expressions (six more than a number = X + 6)
 * Is means equals
 * Sum is addition
 * “More than” is addition as well
 * Difference means subtraction
 * “Less than” means subtraction as well
 * “a number” is written as X or another variable
 * Solving application problems (Procedure on page 174)
 * Application problem is another term for word problems.
 * The steps are:
 * Read the problem to figure out what is being asked
 * Write down any details or facts you know
 * Write the problem as a sentence
 * Translate it into a math problem
 * Solve the problem
 * Check your answer
 * Solving application problems with one variable means that there will be only one unknown thing to figure out. They will give you some numbers and you will need to translate your own problem and then solve for X the same way you did in chapter 2.
 * Sec 3.4**
 * Solving an application problem with two unknown, related quantities
 * This means that you have two unknown things, but they are related, such as X = 2y or X = y + 2
 * Sometimes you will see these as words, such as Mary scores 10 more than John. You still don’t know how much each person scores, but when they give you more information, you will be able to figure both of them out.
 * To solve this type of problem, you are trying to get rid of one of the variables. If you know that X = 2y, use 2y if the rest of the problem has Y’s in it. Solve for whichever variable is left, then you might have to do one more math step to figure out what the other variable is. In this example, if you find out that Y = 5, you can figure out X because you know that x = 2y, or twice as much as y.
 * Geometry with two variables (The length is 5 in. more than the width. The perimeter is 46 inches. Find the length and width.) This is exactly the same as the last method, but they like to put this on tests because it requires you to know more than one thing (how to solve for two variables and also the formula for the area of a rectangle.)