Transcript Glencoe Algebra 1 - Hays High School

Five-Minute Check (over Lesson 2–8)
CCSS
Then/Now
New Vocabulary
Example 1: Real-World Example: Mixture Problem
Example 2: Real-World Example: Percent Mixture Problem
Example 3: Real-World Example: Speed of One Vehicle
Example 4: Real-World Example: Speed of Two Vehicles
Over Lesson 2–8
Solve 6r + t = r – 1 for r.
A.
B.
C.
D.
Over Lesson 2–8
Solve 4c – d = 4a – 2c + 1 for c.
A.
B.
C.
D.
Over Lesson 2–8
for h.
A.
B.
C.
D.
Over Lesson 2–8
A. 4.6 cm
B. 5.8 cm
C. 6.2 cm
D. 6.4 cm
Over Lesson 2–8
A. 236.25 miles
B. 472.5 miles
C. 945 miles
D. 1285.75 miles
Content Standards
A.REI.1 Explain each step in solving a simple equation as
following from the equality of numbers asserted at the
previous step, starting from the assumption that the original
equation has a solution. Construct a viable argument to
justify a solution method.
A.REI.3 Solve linear equations and inequalities in one
variable, including equations with coefficients represented
by letters.
Mathematical Practices
1 Make sense of problems and persevere in solving them.
4 Model with mathematics.
Common Core State Standards © Copyright 2010. National Governors Association Center for Best Practices and Council of Chief State
School Officers. All rights reserved.
You translated sentences into equations.
• Solve mixture problems.
• Solve uniform motion problems.
• weighted average
• mixture problem
• uniform motion problem
• rate problem
Mixture Problem
PETS Mandisha feeds her cat gourmet cat food that
costs $1.75 per pound. She combines it with cheaper
food that costs $0.50 per pound. How many pounds
of cheaper food should Mandisha buy to go with
5 pounds of gourmet food, if she wants the average
price to be $1.00 per pound?
Let w = the number of pounds of cheaper cat food.
Make a table.
Mixture Problem
Write and solve an equation using the information
in the table.
Price of
gourmet
cat food
8.75
plus
price of
cheaper
cat food
equals
+
0.5w
=
price of
mixed
cat food.
1.00(5 + w)
8.75 + 0.5w = 1.00(5 + w)
Original equation
8.75 + 0.5w = 5 + 1w
Distributive
Property
8.75 + 0.5w – 0.5w = 5 + 1w – 0.5w
8.75 = 5 + 0.5w
Subtract 0.5w
from each side.
Simplify.
Mixture Problem
8.75 – 5 = 5 + 0.5w – 5
3.75 = 0.5w
Subtract 5 from
each side.
Simplify.
Divide each side
by 0.5.
7.5 = w
Simplify.
Answer: Mandisha should buy 7.5 pounds of cheaper
cat food to be mixed with the 5 pounds of
gourmet cat food so that the average price is
$1.00 per pound of cat food.
Cheryl bought 3 ounces of glass beads that cost
$1.79 an ounce. The seed beads cost $0.99 an
ounce. How many ounces of seed beads can she
buy if she only wants the beads to be $1.29 an
ounce for her craft project?
A. 3.1 ounces
B. 5 ounces
C. 4.6 ounces
D. 2 ounces
Percent Mixture Problem
AUTO MAINTENANCE A car’s radiator should
contain a solution of 50% antifreeze. Bae has
2 gallons of a 35% antifreeze. How many gallons of
100% antifreeze should Bae add to his solution to
produce a solution of 50% antifreeze?
Let g = the number of gallons of 100% antifreeze to be
added. Make a table.
Percent Mixture Problem
Write and solve an equation using the information
in the table.
Amount of
antifreeze in
35% solution
0.35(2)
plus
+
amount of
antifreeze in
100% solution
1.0(g)
0.35(2) + 1.0(g) = 0.50(2 + g)
0.70 + 1g = 1 + 0.50g
equals
amount of
antifreeze in
50% solution.
=
0.50(2 + g)
Original equation
Distributive
Property
0.70 + 1g – 0.50g = 1 + 0.50g – 0.50g Subtract 0.50g
from each side.
Percent Mixture Problem
0.70 + 0.50g = 1
0.70 + 0.50g – 0.70 = 1 – 0.70
0.50g = 0.30
Simplify.
Subtract 0.70
from each side.
Simplify.
Divide each
side by 0.50.
g = 0.6
Simplify.
Answer: Bae should add 0.6 gallon of 100%
antifreeze to produce a 50% solution.
A recipe calls for mixed nuts with 50% peanuts.
pound of 15% peanuts has already been used.
How many pounds of 75% peanuts needs to be
added to obtain the required 50% mix?
A.
lb
B.
lb
C.
lb
D.
lb
Speed of One Vehicle
AIR TRAVEL Nita took a non-stop flight to visit her
grandmother. The 750-mile trip took three hours and
45 minutes. Because of bad weather, the return trip
took four hours and 45 minutes. What was her
average speed for the round trip?
Understand
We know that Nita did not travel the
same amount of time on each portion of
her trip. So, we will need to find the
weighted average of the plane’s speed.
We are asked to find the average
speed for both portions of the trip.
Speed of One Vehicle
Plan
First, find the rate of the going portion,
and then the return portion of the trip.
Because the rate is in miles per hour,
convert 3 hours and 45 minutes to
3.75 hours and 4 hours 45 minutes to
4.75 hours.
Going
Formula for rate
Speed of One Vehicle
Return
Formula for rate
Because we are looking for a weighted average, we
cannot just average the speeds. We need to find the
weighted average for the round trip.
Speed of One Vehicle
Solve
Substitution
Simplify.
Speed of One Vehicle
Answer: The average speed was about 176 miles per
hour.
Check
The solution of 176 miles per hour is between
the going portion rate 200 miles per hour, and
the return rate, 157.9 miles per hour. So, the
answer is reasonable.
In the morning, when traffic is light, it takes
30 minutes to get to work. The trip is 15 miles
through towns. In the afternoon, when traffic is
a little heavier, it takes 45 minutes. What is the
average speed for the round trip?
A. 24 miles per hour
B. 30 miles per hour
C. 15 miles per hour
D. 45 miles per hour
Speeds of Two Vehicles
RESCUE A railroad switching operator has discovered
that two trains are heading toward each other on the
same track. Currently, the trains are 53 miles apart. One
train is traveling at 75 miles per hour and the other
40 miles per hour. The faster train will require 5 miles to
stop safely, and the slower train will require 3 miles to
stop safely. About how many minutes does the operator
have to warn the train engineers to stop their trains?
Step 1
Draw a diagram.
53 miles apart
Takes 5 miles to stop
53 – (5 + 3) = 45 miles
Takes 3 miles to stop
Speeds of Two Vehicles
Step 2 Let m = the number of hours that the operator
has to warn the train engineers to stop their trains safely.
Make a table.
Step 3 Write and solve an equation using the
information in the table.
Distance
traveled by
fast train
75m
plus
distance
traveled by
other train
equals
45 miles.
+
40m
=
45
Speeds of Two Vehicles
75m + 40m = 45
115m = 45
Original equation
Simplify.
Divide each side by 115.
m ≈ 0.39
0.39 × 60 = 23.4
Round to the nearest
hundredth.
Convert to minutes by
multiplying by 60.
Answer: The operator has about 23 minutes to warn
the engineers.
Two students left the school on their bicycles at the
same time, one heading north and the other
heading south. The student heading north travels
15 miles per hour, and the one heading south
travels at 17 miles per hour. After about how many
minutes will they be 7.5 miles apart?
A. 17 minutes
B. 15 minutes
C. 14 minutes
D. 30 minutes