1. 1.1 Variables
Objective: Students will be able to substitute numbers in for variables and evaluate problems. Students must demonstrate they know how to solve for any letter in the distance formula and also demonstrate they know how to find the perimeter of a shape.

2. Notes Variable: letter represents one or more numbers.
Values: number of the variable.
Variable expression: numbers, variables, and operations
Ex: 3x + 2
Ex: 2y
Ex: 3 + x
Evaluate expression: put number in for variable and solve.

3. Evaluate the expression
X + 5 when x = 4
g/8 when g = 72

4. Evaluate the expression
W + 8 when w = 3.6
7 – x when x = 2.9
2/3 + x when x = 2/3
24/y when y = 6

5. Distance problems d = rt
Solve for r.
Solve for t.
Find the average speed of a car that traveled 180 miles. It took 3hrs to get there.
A train travels 75 miles in 50 minutes. Find the average speed.

6. Real life problems simplify
Time = 20mi / 5mi per hr.
Distance = (60mi/hr)(2.5hr)
In 10 seconds, an athlete runs 40 ft.
A horse gallops 10km in 30 minutes.

7. Perimeter
If I have a triangle, whose sides are 3in, 5in, and 10in, what is the perimeter?
The perimeter of a square is equal to 4s, where s is the side length. If a side of the square is 10ft., what’s the perimeter?

8. Find the Perimeter 8 8 10 6 15 10

9. Simple Interest
Simple interest formula:
I = Prt
I = interest
P = principal
R = rate
T = time
A $1000 bond earns 6% interest. What’s the interest after 4 years.
Find the interest of P = $500, r = 7%, t = 4months

10. Find the simple interest
You invest $80 at 5%. How much interest would you make after 1.5yrs?
Sam deposited $300 at 3% for 9 months. Below is her work. Describe what is wrong and what the correct answer would be.
Work: 300 (3) .75 = $675

11. Wrap up
Questions/Comments
Hw: Text pg. 6-8, #’s: 2, 3, 5-16, 19-37