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# 8.3 Creating and Evaluation Expressions

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## Jenna Jessup

on 9 April 2013

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#### Transcript of 8.3 Creating and Evaluation Expressions

Goal: Translate statements into algebraic expressions, and evaluate the expression 8.3 Creating and Evaluating
Expression Colin is helping to organize a school trip to Ottawa. He needs to figure out the cost of transportation and choose the least expensive bus company. A total of 260 students are going on the trip. He has prices from three bus companies for the cost of seven buses:
- Carry all bus company charges \$350 per bus
- School bus transport charges \$7 per student
- Zim transport charges \$500 to cover the cost for the drivers plus \$6 per student

- 260 Students
- Carryall bus company charges \$350 per bus
-School bus transport charges \$7 per student
-Zim transport charges \$500 to cover the cost for the drivers plus \$6 per student

Lets make an algebraic expression to solve the problem! What bus company should colin choose?

What information do we have? -260 Students/ 7 buses of students
-CarryAll bus company charges \$350 per bus
-School bus transport charges \$7 per student.
-Zim transport charges \$500 to cover the cost for the drivers plus \$6 per student.

First lets calculate the cost using the CarryAll bus company.

\$350 X 7 = \$2450

The cost will be \$2450 to use the Carry all bus company. Solving the problem -260 students / 7 buses
-CarryAll bus company charges \$350 per bus
-School bus transport charges \$7 per student
- Zim transport charges \$500 to cover the cost for the divers plus \$6 per student.

Lets create an algebraic expression for School bus transport.

Lets use the variable s to represent the number of students going on the trip.

-\$7 times 260 students
-we substitute 260 for s because 260 is the number of students going on the trip.

7s
=7(260)
=1820
Therefore school bus transport would cost \$1820.
-260 student/ 7 buses
-CarryAll bus company charges \$350 per bus.
-School bus transport charges \$7 per student.
-Zim Transport charges \$500 to cover the cost for the drivers plus \$6 per student.

Lets make an expression for Zim Transport.

500+6s
=500+6(260)
=500 + 1560
=2060

-We used the variable s to represent the number of students going on the trip.
-we wrote an expression to represent multiplying s by \$6 (the charge per student) and then adding \$500 (The cost of the drivers)
- we substituted 260 for s because 260 is the number of students going on the trip.
-we used the order of operations to evaluate the expression.
Therefor Zim transport would charge \$2060 for the buses.
Example #1 Question- Do you think that it would be easier to buy an ipad by paying an equal amount every month? or by putting down a deposit and then paying a lesser amount each month? Conclusion CarryAll Bus company cost \$2450
School Bus transport cost \$1820
Zim transport cost \$ 2060
Colin should hire School Bus transport to go take the class to Ottawa!

1) why do you think colin did not use a table of values to find the cost for school Bus Transport or Zim Transport?
2) Why do you think that Colin did not use a scatterplot to find the cost for school Bus transport or Zim transport?
3) Explain how to create an algebraic expression. Homework!!! -Work with a partner and complete question 4-6.
-Individually complete 10,12 and 16.
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