Licensing information• Users should treat this material as a working draft. This

material can be used in its current form, customized, redistributed and/or printed or displayed by the user.

• The author(s) request feedback on all materials so that they can be continually improved and updated.

• This material is licensed under the Creative Commons Attribution license: (http://creativecommons.org/licenses/by/2.5/).

• Author: Kevin Hall

Wording for the legal statement above is adapted from the legal statement for Trigonometry, published in 2009 by The CK-12 Foundation: http://about.ck12.org/

Today’s topic is rates. What are some examples of rates?

Warmup:

5 pencils cost $.60. How much would 13 pencils cost?

A cheeseburger with 10 fries has a total of 622.5 calories. The same cheeseburger with 20 fries has a total of 855 calories.

How many calories are there per fry?

Try on your own:

A bus with 4 people on it weighs a total of 29,440 lbs.The same bus with 11 people on it weighs a total 30,735 lbs.

On average, how much does each person weigh?

A bus with 4 people on it weighs a total of 29,440 lbs.

29,440 4

= 7360 Does each person weigh 7360 lbs?

Why does this method give us the wrong answer?When you divide up the picture into 4 pieces, what does each piece represent?

The bus with 11 people on it weighs a total of 30,735 lbs.

30,735 11

= 2,794.1 Does each person weigh 2,794.1 lbs?

Again, why does this method give us the wrong answer?Why is this answer different from our last answer?

The correct solution:What’s the difference between the 2 pictures?

A bus with 4 people on it weighs a total of 29,440 lbs.The same bus with 11 people on it weighs a total 30,735 lbs.

On average, how much does each person weigh?

The correct solution:

How does this relate to the formula for slope?:

A cheeseburger with 10 fries has a total of 622.5 calories. The same cheeseburger with 20 fries has a total of 855 calories.

How many calories are there per fry?

The correct answer on the cheeseburger problem:

The next goal for today is to use slope in real-life scenarios.

At 8:51, plane starts descending.

At 8:56, plane is at 28,130 ft.

At 9:05, plane is at 24,258 ft.

If the plane is descending at a constant rate, when will it land?

First we’ll do an example problem.

Based on the data below, when will you have only $200 left?

Days $

0 1,500

1,4942

1,4884

1,4826

- 6

- 6

- 6

Based on the data below, when will you have only $200 left?

Days $

0 1,500

1,4942

1,4884

1,4826

Actually, the rate is NOT -6.

The rate is ____.

Why?

-3

Based on the data below, when will you have only $200 left?

Days $

0 1,500

1,4942

1,4884

1,4826

- 3 - 3 - 3

1 1,497

3 1,491

5 1,495

y = mx + by = -3x + b

1,500

Based on the data below, when will you have only $200 left?

Days $

0 1,500

1,4942

1,4884

1,4826

- 3 - 3 - 3

1 1,497

3 1,491

5 1,495

y = mx + b y = -3x + 1500

1,500

200

Does the $200 represent x or y?

Now just solve the equation.

Today we’re leaning how to find the rate when the x-values are not increasing by 1.

x y

0 1,500

1,49421 1,497

3 1,491

1,500

We already know this

x y

0 1,500

1,4942412 1,497

36 1,491

1,500

We need to know this

At 8:51, plane starts descending.

At 8:56, plane is at 28,130 ft.

At 9:05, plane is at 24,258 ft.

If the plane is descending at a constant rate, when will it land?

At 8:51, plane starts descending.

At 8:56, plane is at 28,130 ft.

At 9:05, plane is at 24,258 ft.

If the plane is descending at a constant rate, when will it land?

Image credit of bus (from Wikimedia Commons): “This image has been released into the public domain by its author, Radagast. This applies worldwide.

http://commons.wikimedia.org/wiki/File:Grt_nova_bus.png