Download Lecture Notes for section 1.4

Announcements
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The homework for section 1.3 is due by the start of class on
Monday, January 27.
Your rst quiz will be available on MyMathLab after class on
Wednesday. It will cover chapter 1 of the textbook.
I You have 15 minutes to complete the quiz.
I The quiz must be taken in one session.
I You need to complete the quiz before the start of class on
Wednesday, January 29.
1.4 Formulas and Problem Solving
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Objective 1: Solve a formula for a given variable
Objective 2: Use formulas to solve application problems
Solving a Formula for a Given Variable
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A formula is an equation that describes the relationship
between two or more variables.
Solving a Formula for a Given Variable
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A formula is an equation that describes the relationship
between two or more variables.
To solve a formula for a given variable means to isolate the
variable on one side of the equation.
Example 1
1.4.5 Solve
E
= I (r + R ) for r .
Example 1
1.4.5 Solve
E
= I (r + R ) for r .
E = I (r + R )
E
= (r + R )
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E
−R =r
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Example 2
1.4.10 Solve
A = 2lw + 2lh + 2wh for h.
Example 2
1.4.10 Solve
A = 2lw + 2lh + 2wh for h.
A = 2lw + 2lh + 2wh
A − 2wh = 2lw + 2lh
A − 2wh = l (2w + 2h)
A − 2wh
=l
2w + 2h
Use Formulas to Solve Application Problems
1.
2.
3.
4.
Dene the Problem.
Assign the Variables.
Translate into an Equation (or system of equations).
Solve the equation(s).
Formulas from Geometry You Should Know
Some Facts You Should Know
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The sum of the measures of the three interior angles of a
triangle is 180◦ .
(total value) = (value per item) · (number of items)
Uniform motion:(distance) = (rate) · (time)
(amount of componet) = (concentration)·(amount of mixture)
Simple Interest: (interest) = (principal) (interest rate) (time)
Example 3
1.4.12 The area of a triangular sail for a boat is 90 square feet. If
the base of the sail is 10 feet long, nd it's height.
Example 3
1.4.12 The area of a triangular sail for a boat is 90 square feet. If
the base of the sail is 10 feet long, nd it's height.
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Let
Let
Let
A be the area of the triangular sail.
b be the length of the base of the sail.
h be the height of the sail.
Example 3
1.4.12 The area of a triangular sail for a boat is 90 square feet. If
the base of the sail is 10 feet long, nd it's height.
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Let
Let
Let
A be the area of the triangular sail.
b be the length of the base of the sail.
h be the height of the sail.
From geometry, we know the area of a triangle is base times
height. Symbolically,
A = bh
Example 3
1.4.12 The area of a triangular sail for a boat is 90 square feet. If
the base of the sail is 10 feet long, nd it's height.
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Let
Let
Let
A be the area of the triangular sail.
b be the length of the base of the sail.
h be the height of the sail.
From geometry, we know the area of a triangle is base times
height. Symbolically,
A = bh
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In our case,
90 = 10h
9=h
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The height of the sale is 9 feet.
Example 4
1.4.16 An isosceles triangle has two equal angles. If the measure of
the third angle is 20◦ less than the sum of the two equal angles,
nd the measure of the three angles.
Example 4
1.4.16 An isosceles triangle has two equal angles. If the measure of
the third angle is 20◦ less than the sum of the two equal angles,
nd the measure of the three angles.
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Let θ be the measure of the two equal angles
Let φ be the measure of the remaining equals
Example 4
1.4.16 An isosceles triangle has two equal angles. If the measure of
the third angle is 20◦ less than the sum of the two equal angles,
nd the measure of the three angles.
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Let θ be the measure of the two equal angles
Let φ be the measure of the remaining equals
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As an equation, the second sentence states: φ = 2θ − 20.
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Example 4
1.4.16 An isosceles triangle has two equal angles. If the measure of
the third angle is 20◦ less than the sum of the two equal angles,
nd the measure of the three angles.
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Let θ be the measure of the two equal angles
Let φ be the measure of the remaining equals
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As an equation, the second sentence states: φ = 2θ − 20.
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This equation alone is not enough to solve this problem. We
also need a formula from geometry.
Example 4
1.4.16 An isosceles triangle has two equal angles. If the measure of
the third angle is 20◦ less than the sum of the two equal angles,
nd the measure of the three angles.
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Let θ be the measure of the two equal angles
Let φ be the measure of the remaining equals
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As an equation, the second sentence states: φ = 2θ − 20.
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This equation alone is not enough to solve this problem. We
also need a formula from geometry.
The sum of the interior angles of a triangle is 180◦
Example 4
1.4.16 An isosceles triangle has two equal angles. If the measure of
the third angle is 20◦ less than the sum of the two equal angles,
nd the measure of the three angles.
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Let θ be the measure of the two equal angles
Let φ be the measure of the remaining equals
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As an equation, the second sentence states: φ = 2θ − 20.
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This equation alone is not enough to solve this problem. We
also need a formula from geometry.
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The sum of the interior angles of a triangle is 180◦
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In our case, 2θ + φ = 180.
Example 4 (continued)
1.4.16 An isosceles triangle has two equal angles. If the measure of
the third angle is 20◦ less than the sum of the two equal angles,
nd the measure of the three angles.
I Now we have a system of linear equations to solve:
φ = 2θ − 20
2θ + φ = 180
Example 4 (continued)
1.4.16 An isosceles triangle has two equal angles. If the measure of
the third angle is 20◦ less than the sum of the two equal angles,
nd the measure of the three angles.
I Now we have a system of linear equations to solve:
φ = 2θ − 20
2θ + φ = 180
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The second equation is equivalent to 2θ = 180 − φ
Example 4 (continued)
1.4.16 An isosceles triangle has two equal angles. If the measure of
the third angle is 20◦ less than the sum of the two equal angles,
nd the measure of the three angles.
I Now we have a system of linear equations to solve:
φ = 2θ − 20
2θ + φ = 180
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The second equation is equivalent to 2θ = 180 − φ
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Substituting this into the rst equation produces:
φ = (180 − φ) − 20
2φ = 160
φ = 80◦
Example 4 (continued)
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We now know φ = 80◦ . But we also need to nd θ.
Example 4 (continued)
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We now know φ = 80◦ . But we also need to nd θ.
2θ = 180 − φ
= 180 − 80
= 100
Example 4 (continued)
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We now know φ = 80◦ . But we also need to nd θ.
2θ = 180 − φ
= 180 − 80
= 100
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Dividing by 2 shows us:
θ = 50◦
Example 5
1.4.21 A car leaves a house. Two hours later a second car leaves
the same house and travels the same path. If the second car drives
an average of 20 mph faster than the rst, what is the average
speed of each car if it takes the second car three hours to catch up
to the rst?
Example 5
1.4.21 A car leaves a house. Two hours later a second car leaves
the same house and travels the same path. If the second car drives
an average of 20 mph faster than the rst, what is the average
speed of each car if it takes the second car three hours to catch up
to the rst?
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Let
Let
Let
s1 be the average speed of the rst car.
s2 be the average speed of the second car.
t be the time after the second car leaves the house.
Example 5
1.4.21 A car leaves a house. Two hours later a second car leaves
the same house and travels the same path. If the second car drives
an average of 20 mph faster than the rst, what is the average
speed of each car if it takes the second car three hours to catch up
to the rst?
s1 be the average speed of the rst car.
s2 be the average speed of the second car.
t be the time after the second car leaves the house.
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Let
Let
Let
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s2 = s1 + 20
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Example 5
1.4.21 A car leaves a house. Two hours later a second car leaves
the same house and travels the same path. If the second car drives
an average of 20 mph faster than the rst, what is the average
speed of each car if it takes the second car three hours to catch up
to the rst?
s1 be the average speed of the rst car.
s2 be the average speed of the second car.
t be the time after the second car leaves the house.
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Let
Let
Let
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s2 = s1 + 20
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Recall: (distance) = (rate) · (time)
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Example 5
1.4.21 A car leaves a house. Two hours later a second car leaves
the same house and travels the same path. If the second car drives
an average of 20 mph faster than the rst, what is the average
speed of each car if it takes the second car three hours to catch up
to the rst?
s1 be the average speed of the rst car.
s2 be the average speed of the second car.
t be the time after the second car leaves the house.
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Let
Let
Let
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s2 = s1 + 20
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Recall: (distance) = (rate) · (time)
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The last part of the problem when combined with the second
sentence tells us:
3s2 = 5s1 .
Example 5 (continued)
1.4.21 A car leaves a house. Two hours later a second car leaves
the same house and travels the same path. If the second car drives
an average of 20 mph faster than the rst, what is the average
speed of each car if it takes the second car three hours to catch up
to the rst?
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equations:
s2 = s1 + 20
3s2 = 5s1
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Substituting the rst equation into the second gives us:
3 (s1 + 20) = 5s1
60 = 2s1
30 = s1
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We can then substitute this result to see that
s2 = s1 + 20 = 30 + 20 = 50.
Example 6
1.4.24 A barge travels on river from point A to point B . The barge
travels at a rate of 5 mph relative to the water. The river ows
downstream at a rate of 1 mph. If the trip upstream takes two
hours longer than the trip downstream, how far is it from point A
to point B ?
Example 6
1.4.24 A barge travels on river from point A to point B . The barge
travels at a rate of 5 mph relative to the water. The river ows
downstream at a rate of 1 mph. If the trip upstream takes two
hours longer than the trip downstream, how far is it from point A
to point B ?
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Let
Let
Let
tu be the time it takes the barge to travel upstream.
td be the time it takes the barge to travel downstream.
d be the distance from point A to point B .
Example 6
1.4.24 A barge travels on river from point A to point B . The barge
travels at a rate of 5 mph relative to the water. The river ows
downstream at a rate of 1 mph. If the trip upstream takes two
hours longer than the trip downstream, how far is it from point A
to point B ?
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Let
Let
Let
tu be the time it takes the barge to travel upstream.
td be the time it takes the barge to travel downstream.
d be the distance from point A to point B .
When the barge is traveling upstream, it will travel at a speed
of 4 (5-1) mph relative to the river bank.
When the barge is traveling downstream, it will travel at a
speed of 6 (5+1) mph relative to the river bank.
Example 6 (continued)
1.4.24 A barge travels on river from point A to point B . The barge
travels at a rate of 5 mph relative to the water. The river ows
downstream at a rate of 1 mph. If the trip upstream takes two
hours longer than the trip downstream, how far is it from point A
to point B ?
I tu = td + 2
Example 6 (continued)
1.4.24 A barge travels on river from point A to point B . The barge
travels at a rate of 5 mph relative to the water. The river ows
downstream at a rate of 1 mph. If the trip upstream takes two
hours longer than the trip downstream, how far is it from point A
to point B ?
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Since (distance) = (rate) · (time),
d = 4tu = 5td .
Example 6 (continued)
1.4.24 A barge travels on river from point A to point B . The barge
travels at a rate of 5 mph relative to the water. The river ows
downstream at a rate of 1 mph. If the trip upstream takes two
hours longer than the trip downstream, how far is it from point A
to point B ?
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Since (distance) = (rate) · (time),
d = 4tu = 5td .
In order to nd d , we rst have to solve the following system
of equations:
tu = td + 2
4tu = 5td
Example 6 (continued)
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In order to nd d , we rst have to solve the following system
of equations:
tu = td + 2
4tu = 5td
Example 6 (continued)
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In order to nd d , we rst have to solve the following system
of equations:
tu = td + 2
4tu = 5td
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After substituting the rst equation into the second equation
we have:
4 (td + 2) = 5td
8 = td
Example 6 (continued)
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In order to nd d , we rst have to solve the following system
of equations:
tu = td + 2
4tu = 5td
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After substituting the rst equation into the second equation
we have:
4 (td + 2) = 5td
8 = td
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Hence
tu = td + 2 = 10 and d = 5td = 5 · 8 = 40 miles.
Example 7
1.4.25 Suppose 8 pints of a 12% alcohol solution is mixed with 2
pints of a 60% alcohol solution. What is the concentration of
alcohol in the new 10-pint mixture?
Example 7
1.4.25 Suppose 8 pints of a 12% alcohol solution is mixed with 2
pints of a 60% alcohol solution. What is the concentration of
alcohol in the new 10-pint mixture?
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(amount of componet) = (concentration)·(amount of mixture)
Example 7
1.4.25 Suppose 8 pints of a 12% alcohol solution is mixed with 2
pints of a 60% alcohol solution. What is the concentration of
alcohol in the new 10-pint mixture?
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(amount of componet) = (concentration)·(amount of mixture)
The amount of alcohol in the 8 pint solution is 8 · 0.12 = 0.96
pints.
Example 7
1.4.25 Suppose 8 pints of a 12% alcohol solution is mixed with 2
pints of a 60% alcohol solution. What is the concentration of
alcohol in the new 10-pint mixture?
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(amount of componet) = (concentration)·(amount of mixture)
The amount of alcohol in the 8 pint solution is 8 · 0.12 = 0.96
pints.
The amount of alcohol in the second solution is 2 · 0.6 = 1.2
pints
Example 7
1.4.25 Suppose 8 pints of a 12% alcohol solution is mixed with 2
pints of a 60% alcohol solution. What is the concentration of
alcohol in the new 10-pint mixture?
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(amount of componet) = (concentration)·(amount of mixture)
The amount of alcohol in the 8 pint solution is 8 · 0.12 = 0.96
pints.
The amount of alcohol in the second solution is 2 · 0.6 = 1.2
pints
So the amount of alcohol in the nal solution is
0.96 + 1.2 = 2.16 pints.
Example 7
1.4.25 Suppose 8 pints of a 12% alcohol solution is mixed with 2
pints of a 60% alcohol solution. What is the concentration of
alcohol in the new 10-pint mixture?
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(amount of componet) = (concentration)·(amount of mixture)
The amount of alcohol in the 8 pint solution is 8 · 0.12 = 0.96
pints.
The amount of alcohol in the second solution is 2 · 0.6 = 1.2
pints
So the amount of alcohol in the nal solution is
0.96 + 1.2 = 2.16 pints.
So the concentration of alcohol in the new mixture is
(amount of alcohol)
.16
= 210
= 0.216 or 21.6%
(amount of mixture)