The document discusses how many physics and mathematics formulas lead to second-degree equations. It provides an example of calculating the height, time, and maximum height of a stone thrown straight up in the air. Specifically, it shows:
1) If a stone is thrown up at 64 feet/second, it will be 48 feet high after 1 second.
2) It will take the stone 4 seconds to fall back to the ground.
3) The stone will reach its maximum height of 64 feet after 2 seconds.
Solving problems involving related factors that are multiplied leads to second-degree equations.
Mathematics 9 Lesson 1-D: System of Equations Involving Quadratic EquationsJuan Miguel Palero
This powerpoint presentation discusses or talks about the topic or lesson System of Equations involving Quadratic Equations. It also discusses and explains the rules, steps and examples of System of Equations involving Quadratic Equations
Mathematics 9 Lesson 1-D: System of Equations Involving Quadratic EquationsJuan Miguel Palero
This powerpoint presentation discusses or talks about the topic or lesson System of Equations involving Quadratic Equations. It also discusses and explains the rules, steps and examples of System of Equations involving Quadratic Equations
THE BINOMIAL THEOREM shows how to calculate a power of a binomial –
(x+ y)n -- without actually multiplying out.
For example, if we actually multiplied out the 4th power of (x + y) --
(x + y)4 = (x + y) (x + y) (x + y) (x + y)
-- then on collecting like terms we would find:
(x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4 . . . . . (1)
Model Attribute Check Company Auto PropertyCeline George
In Odoo, the multi-company feature allows you to manage multiple companies within a single Odoo database instance. Each company can have its own configurations while still sharing common resources such as products, customers, and suppliers.
Macroeconomics- Movie Location
This will be used as part of your Personal Professional Portfolio once graded.
Objective:
Prepare a presentation or a paper using research, basic comparative analysis, data organization and application of economic information. You will make an informed assessment of an economic climate outside of the United States to accomplish an entertainment industry objective.
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
For more information, visit-www.vavaclasses.com
Acetabularia Information For Class 9 .docxvaibhavrinwa19
Acetabularia acetabulum is a single-celled green alga that in its vegetative state is morphologically differentiated into a basal rhizoid and an axially elongated stalk, which bears whorls of branching hairs. The single diploid nucleus resides in the rhizoid.
Honest Reviews of Tim Han LMA Course Program.pptxtimhan337
Personal development courses are widely available today, with each one promising life-changing outcomes. Tim Han’s Life Mastery Achievers (LMA) Course has drawn a lot of interest. In addition to offering my frank assessment of Success Insider’s LMA Course, this piece examines the course’s effects via a variety of Tim Han LMA course reviews and Success Insider comments.
Introduction to AI for Nonprofits with Tapp NetworkTechSoup
Dive into the world of AI! Experts Jon Hill and Tareq Monaur will guide you through AI's role in enhancing nonprofit websites and basic marketing strategies, making it easy to understand and apply.
Biological screening of herbal drugs: Introduction and Need for
Phyto-Pharmacological Screening, New Strategies for evaluating
Natural Products, In vitro evaluation techniques for Antioxidants, Antimicrobial and Anticancer drugs. In vivo evaluation techniques
for Anti-inflammatory, Antiulcer, Anticancer, Wound healing, Antidiabetic, Hepatoprotective, Cardio protective, Diuretics and
Antifertility, Toxicity studies as per OECD guidelines
Palestine last event orientationfvgnh .pptxRaedMohamed3
An EFL lesson about the current events in Palestine. It is intended to be for intermediate students who wish to increase their listening skills through a short lesson in power point.
Instructions for Submissions thorugh G- Classroom.pptxJheel Barad
This presentation provides a briefing on how to upload submissions and documents in Google Classroom. It was prepared as part of an orientation for new Sainik School in-service teacher trainees. As a training officer, my goal is to ensure that you are comfortable and proficient with this essential tool for managing assignments and fostering student engagement.
Synthetic Fiber Construction in lab .pptxPavel ( NSTU)
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
2. The simplest type of formulas leads to 2nd degree equations
are the ones of the form AB = C.
2nd-Degree-Equation Word Problems
3. The simplest type of formulas leads to 2nd degree equations
are the ones of the form AB = C.
Specifically, if the two factors A and B are related linearly,
then their product is a 2nd degree expression.
2nd-Degree-Equation Word Problems
4. The simplest type of formulas leads to 2nd degree equations
are the ones of the form AB = C.
Specifically, if the two factors A and B are related linearly,
then their product is a 2nd degree expression.
2nd-Degree-Equation Word Problems
5. The simplest type of formulas leads to 2nd degree equations
are the ones of the form AB = C.
Specifically, if the two factors A and B are related linearly,
then their product is a 2nd degree expression.
Hence the problem of finding A and B when their product
is known, is a 2nd degree equation.
2nd-Degree-Equation Word Problems
Example A. We have two positive numbers.
The larger number is 2 less than three times of the smaller one.
The product of two numbers is 96. Find the two numbers.
6. The simplest type of formulas leads to 2nd degree equations
are the ones of the form AB = C.
Specifically, if the two factors A and B are related linearly,
then their product is a 2nd degree expression.
Hence the problem of finding A and B when their product
is known, is a 2nd degree equation.
2nd-Degree-Equation Word Problems
Example A. We have two positive numbers.
The larger number is 2 less than three times of the smaller one.
The product of two numbers is 96. Find the two numbers.
Let the smaller number be x.
7. The simplest type of formulas leads to 2nd degree equations
are the ones of the form AB = C.
Specifically, if the two factors A and B are related linearly,
then their product is a 2nd degree expression.
Hence the problem of finding A and B when their product
is known, is a 2nd degree equation.
2nd-Degree-Equation Word Problems
Example A. We have two positive numbers.
The larger number is 2 less than three times of the smaller one.
The product of two numbers is 96. Find the two numbers.
Let the smaller number be x. Hence the larger one is (3x – 2).
Their product of them is 96 so
x(3x – 2) = 96
8. The simplest type of formulas leads to 2nd degree equations
are the ones of the form AB = C.
Specifically, if the two factors A and B are related linearly,
then their product is a 2nd degree expression.
Hence the problem of finding A and B when their product
is known, is a 2nd degree equation.
2nd-Degree-Equation Word Problems
Example A. We have two positive numbers.
The larger number is 2 less than three times of the smaller one.
The product of two numbers is 96. Find the two numbers.
Let the smaller number be x. Hence the larger one is (3x – 2).
Their product of them is 96 so
x(3x – 2) = 96 expand
3x2 – 2x = 96
9. The simplest type of formulas leads to 2nd degree equations
are the ones of the form AB = C.
Specifically, if the two factors A and B are related linearly,
then their product is a 2nd degree expression.
Hence the problem of finding A and B when their product
is known, is a 2nd degree equation.
2nd-Degree-Equation Word Problems
Example A. We have two positive numbers.
The larger number is 2 less than three times of the smaller one.
The product of two numbers is 96. Find the two numbers.
Let the smaller number be x. Hence the larger one is (3x – 2).
Their product of them is 96 so
x(3x – 2) = 96 expand
3x2 – 2x = 96
3x2 – 2x – 96 = 0
10. The simplest type of formulas leads to 2nd degree equations
are the ones of the form AB = C.
Specifically, if the two factors A and B are related linearly,
then their product is a 2nd degree expression.
Hence the problem of finding A and B when their product
is known, is a 2nd degree equation.
2nd-Degree-Equation Word Problems
Example A. We have two positive numbers.
The larger number is 2 less than three times of the smaller one.
The product of two numbers is 96. Find the two numbers.
Let the smaller number be x. Hence the larger one is (3x – 2).
Their product of them is 96 so
x(3x – 2) = 96 expand
3x2 – 2x = 96
3x2 – 2x – 96 = 0
(3x + 16)(x – 6) = 0
So x = –16/3 or x = 6.
11. The simplest type of formulas leads to 2nd degree equations
are the ones of the form AB = C.
Specifically, if the two factors A and B are related linearly,
then their product is a 2nd degree expression.
Hence the problem of finding A and B when their product
is known, is a 2nd degree equation.
2nd-Degree-Equation Word Problems
Example A. We have two positive numbers.
The larger number is 2 less than three times of the smaller one.
The product of two numbers is 96. Find the two numbers.
Let the smaller number be x. Hence the larger one is (3x – 2).
Their product of them is 96 so
x(3x – 2) = 96 expand
3x2 – 2x = 96
3x2 – 2x – 96 = 0.
(3x + 16)(x – 6) = 0
So x = –16/3 or x = 6. Hence the two numbers are 6 and 16.
13. Many physics formulas are 2nd degree.
If a stone is thrown straight up on Earth then
h = -16t2 + vt
2nd-Degree-Equation Word Problems
14. Many physics formulas are 2nd degree.
If a stone is thrown straight up on Earth then
h = -16t2 + vt
2nd-Degree-Equation Word Problems
where
h = height in feet
t = time in second
v = upward speed in feet per second
15. Many physics formulas are 2nd degree.
If a stone is thrown straight up on Earth then
h = -16t2 + vt
2nd-Degree-Equation Word Problems
height = -16t2 + vt
after t secondswhere
h = height in feet
t = time in second
v = upward speed in feet per second
16. Many physics formulas are 2nd degree.
If a stone is thrown straight up on Earth then
h = -16t2 + vt
2nd-Degree-Equation Word Problems
height = -16t2 + vt
after t secondswhere
h = height in feet
t = time in second
v = upward speed in feet per second
Example B. If a stone is thrown straight
up at a speed of 64 ft per second,
a. how high is it after 1 second?
17. Many physics formulas are 2nd degree.
If a stone is thrown straight up on Earth then
h = -16t2 + vt
2nd-Degree-Equation Word Problems
height = -16t2 + vt
after t secondswhere
h = height in feet
t = time in second
v = upward speed in feet per second
Example B. If a stone is thrown straight
up at a speed of 64 ft per second,
a. how high is it after 1 second?
t = 1, v = 64,
18. Many physics formulas are 2nd degree.
If a stone is thrown straight up on Earth then
h = -16t2 + vt
2nd-Degree-Equation Word Problems
height = -16t2 + vt
after t secondswhere
h = height in feet
t = time in second
v = upward speed in feet per second
Example B. If a stone is thrown straight
up at a speed of 64 ft per second,
a. how high is it after 1 second?
t = 1, v = 64, so
h = -16(1)2 + 64(1)
19. Many physics formulas are 2nd degree.
If a stone is thrown straight up on Earth then
h = -16t2 + vt
2nd-Degree-Equation Word Problems
height = -16t2 + vt
after t secondswhere
h = height in feet
t = time in second
v = upward speed in feet per second
Example B. If a stone is thrown straight
up at a speed of 64 ft per second,
a. how high is it after 1 second?
t = 1, v = 64, so
h = -16(1)2 + 64(1)
h = -16 + 64 = 48 ft
20. b. How long will it take for it to fall back to the ground?
2nd-Degree-Equation Word Problems
21. b. How long will it take for it to fall back to the ground?
To fall back to the ground means the height h is 0.
2nd-Degree-Equation Word Problems
22. b. How long will it take for it to fall back to the ground?
To fall back to the ground means the height h is 0.
Hence
h = 0, v = 64, need to find t.
2nd-Degree-Equation Word Problems
23. b. How long will it take for it to fall back to the ground?
To fall back to the ground means the height h is 0.
Hence
h = 0, v = 64, need to find t.
0 = -16 t2 + 64t
2nd-Degree-Equation Word Problems
24. b. How long will it take for it to fall back to the ground?
To fall back to the ground means the height h is 0.
Hence
h = 0, v = 64, need to find t.
0 = -16 t2 + 64t
0 = -16t(t – 4)
2nd-Degree-Equation Word Problems
25. b. How long will it take for it to fall back to the ground?
To fall back to the ground means the height h is 0.
Hence
h = 0, v = 64, need to find t.
0 = -16 t2 + 64t
0 = -16t(t – 4)
t = 0
2nd-Degree-Equation Word Problems
26. b. How long will it take for it to fall back to the ground?
To fall back to the ground means the height h is 0.
Hence
h = 0, v = 64, need to find t.
0 = -16 t2 + 64t
0 = -16t(t – 4)
t = 0 or t – 4 = 0 or t = 4
2nd-Degree-Equation Word Problems
27. b. How long will it take for it to fall back to the ground?
To fall back to the ground means the height h is 0.
Hence
h = 0, v = 64, need to find t.
0 = -16 t2 + 64t
0 = -16t(t – 4)
t = 0 or t – 4 = 0 or t = 4
Therefore it takes 4 seconds.
2nd-Degree-Equation Word Problems
28. 2nd-Degree-Equation Word Problems
c. What is the maximum height obtained?
b. How long will it take for it to fall back to the ground?
To fall back to the ground means the height h is 0.
Hence
h = 0, v = 64, need to find t.
0 = -16 t2 + 64t
0 = -16t(t – 4)
t = 0 or t – 4 = 0 or t = 4
Therefore it takes 4 seconds.
29. 2nd-Degree-Equation Word Problems
c. What is the maximum height obtained?
Since it takes 4 seconds for the stone to fall back to the
ground, at 2 seconds it must reach the highest point.
b. How long will it take for it to fall back to the ground?
To fall back to the ground means the height h is 0.
Hence
h = 0, v = 64, need to find t.
0 = -16 t2 + 64t
0 = -16t(t – 4)
t = 0 or t – 4 = 0 or t = 4
Therefore it takes 4 seconds.
30. 2nd-Degree-Equation Word Problems
c. What is the maximum height obtained?
Since it takes 4 seconds for the stone to fall back to the
ground, at 2 seconds it must reach the highest point.
Hence t = 2, v = 64, need to find h.
b. How long will it take for it to fall back to the ground?
To fall back to the ground means the height h is 0.
Hence
h = 0, v = 64, need to find t.
0 = -16 t2 + 64t
0 = -16t(t – 4)
t = 0 or t – 4 = 0 or t = 4
Therefore it takes 4 seconds.
31. 2nd-Degree-Equation Word Problems
c. What is the maximum height obtained?
Since it takes 4 seconds for the stone to fall back to the
ground, at 2 seconds it must reach the highest point.
Hence t = 2, v = 64, need to find h.
h = -16(2)2 + 64(2)
b. How long will it take for it to fall back to the ground?
To fall back to the ground means the height h is 0.
Hence
h = 0, v = 64, need to find t.
0 = -16 t2 + 64t
0 = -16t(t – 4)
t = 0 or t – 4 = 0 or t = 4
Therefore it takes 4 seconds.
32. 2nd-Degree-Equation Word Problems
c. What is the maximum height obtained?
Since it takes 4 seconds for the stone to fall back to the
ground, at 2 seconds it must reach the highest point.
Hence t = 2, v = 64, need to find h.
h = -16(2)2 + 64(2)
= - 64 + 128
b. How long will it take for it to fall back to the ground?
To fall back to the ground means the height h is 0.
Hence
h = 0, v = 64, need to find t.
0 = -16 t2 + 64t
0 = -16t(t – 4)
t = 0 or t – 4 = 0 or t = 4
Therefore it takes 4 seconds.
33. 2nd-Degree-Equation Word Problems
c. What is the maximum height obtained?
Since it takes 4 seconds for the stone to fall back to the
ground, at 2 seconds it must reach the highest point.
Hence t = 2, v = 64, need to find h.
h = -16(2)2 + 64(2)
= - 64 + 128
= 64 (ft)
b. How long will it take for it to fall back to the ground?
To fall back to the ground means the height h is 0.
Hence
h = 0, v = 64, need to find t.
0 = -16 t2 + 64t
0 = -16t(t – 4)
t = 0 or t – 4 = 0 or t = 4
Therefore it takes 4 seconds.
34. 2nd-Degree-Equation Word Problems
c. What is the maximum height obtained?
Since it takes 4 seconds for the stone to fall back to the
ground, at 2 seconds it must reach the highest point.
Hence t = 2, v = 64, need to find h.
h = -16(2)2 + 64(2)
= - 64 + 128
= 64 (ft)
Therefore, the maximum height is 64 feet.
b. How long will it take for it to fall back to the ground?
To fall back to the ground means the height h is 0.
Hence
h = 0, v = 64, need to find t.
0 = -16 t2 + 64t
0 = -16t(t – 4)
t = 0 or t – 4 = 0 or t = 4
Therefore it takes 4 seconds.
37. 2nd-Degree-Equation Word Problems
Area of a Rectangle
Given a rectangle, let
L = length of a rectangle
W = width of the rectangle,
Formulas of area in mathematics also lead to 2nd degree
equations.
38. 2nd-Degree-Equation Word Problems
Area of a Rectangle
Given a rectangle, let
L = length of a rectangle
W = width of the rectangle,
the area A of the rectangle is
A = LW.
Formulas of area in mathematics also lead to 2nd degree
equations.
39. 2nd-Degree-Equation Word Problems
Area of a Rectangle
Given a rectangle, let
L = length of a rectangle
W = width of the rectangle,
the area A of the rectangle is
A = LW.
L
w A = LW
Formulas of area in mathematics also lead to 2nd degree
equations.
40. If L and W are in a given unit, then A is in unit2.
2nd-Degree-Equation Word Problems
Area of a Rectangle
Given a rectangle, let
L = length of a rectangle
W = width of the rectangle,
the area A of the rectangle is
A = LW.
L
w A = LW
Formulas of area in mathematics also lead to 2nd degree
equations.
41. If L and W are in a given unit, then A is in unit2.
For example: If L = 2 in, W = 3 in. then A = 2*3 = 6 in2
2nd-Degree-Equation Word Problems
Area of a Rectangle
Given a rectangle, let
L = length of a rectangle
W = width of the rectangle,
the area A of the rectangle is
A = LW.
L
w A = LW
Formulas of area in mathematics also lead to 2nd degree
equations.
42. If L and W are in a given unit, then A is in unit2.
For example: If L = 2 in, W = 3 in. then A = 2*3 = 6 in2
1 in
1 in
1 in2
2nd-Degree-Equation Word Problems
Area of a Rectangle
Given a rectangle, let
L = length of a rectangle
W = width of the rectangle,
the area A of the rectangle is
A = LW.
L
w A = LW
Formulas of area in mathematics also lead to 2nd degree
equations.
43. If L and W are in a given unit, then A is in unit2.
For example: If L = 2 in, W = 3 in. then A = 2*3 = 6 in2
1 in
1 in
1 in2
2 in
3 in
6 in2
2nd-Degree-Equation Word Problems
Area of a Rectangle
Given a rectangle, let
L = length of a rectangle
W = width of the rectangle,
the area A of the rectangle is
A = LW.
L
w A = LW
Formulas of area in mathematics also lead to 2nd degree
equations.
44. Example C. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
2nd-Degree-Equation Word Problems
L
w A = LW
45. Example C. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
2nd-Degree-Equation Word Problems
L
w A = LW
46. Example C. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
2nd-Degree-Equation Word Problems
L
w A = LW
47. Example C. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
x2 + 4x = 21
2nd-Degree-Equation Word Problems
L
w A = LW
48. Example C. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
x2 + 4x = 21
x2 + 4x – 21 = 0
2nd-Degree-Equation Word Problems
L
w A = LW
49. Example C. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
x2 + 4x = 21
x2 + 4x – 21 = 0
(x + 7)(x – 3) = 0
2nd-Degree-Equation Word Problems
L
w A = LW
50. Example C. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
x2 + 4x = 21
x2 + 4x – 21 = 0
(x + 7)(x – 3) = 0
x = - 7 or x = 3
2nd-Degree-Equation Word Problems
L
w A = LW
51. Example C. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
x2 + 4x = 21
x2 + 4x – 21 = 0
(x + 7)(x – 3) = 0
x = - 7 or x = 3
2nd-Degree-Equation Word Problems
L
w A = LW
52. Example C. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
x2 + 4x = 21
x2 + 4x – 21 = 0
(x + 7)(x – 3) = 0
x = - 7 or x = 3
Therefore, the width is 3 and the length is 7.
2nd-Degree-Equation Word Problems
L
w A = LW
53. Example C. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
x2 + 4x = 21
x2 + 4x – 21 = 0
(x + 7)(x – 3) = 0
x = - 7 or x = 3
Therefore, the width is 3 and the length is 7.
2nd-Degree-Equation Word Problems
Area of a Parallelogram
B=base
H=height
L
w A = LW
54. Example C. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
x2 + 4x = 21
x2 + 4x – 21 = 0
(x + 7)(x – 3) = 0
x = - 7 or x = 3
Therefore, the width is 3 and the length is 7.
2nd-Degree-Equation Word Problems
Area of a Parallelogram
A parallelogram is the area enclosed
by two sets of parallel lines.
B=base
H=height
L
w A = LW
55. Example C. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
x2 + 4x = 21
x2 + 4x – 21 = 0
(x + 7)(x – 3) = 0
x = - 7 or x = 3
Therefore, the width is 3 and the length is 7.
2nd-Degree-Equation Word Problems
Area of a Parallelogram
A parallelogram is the area enclosed
by two sets of parallel lines. If we
move the shaded part as shown,
H=height
B=base
L
w A = LW
56. Example C. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
x2 + 4x = 21
x2 + 4x – 21 = 0
(x + 7)(x – 3) = 0
x = - 7 or x = 3
Therefore, the width is 3 and the length is 7.
2nd-Degree-Equation Word Problems
Area of a Parallelogram
A parallelogram is the area enclosed
by two sets of parallel lines. If we
move the shaded part as shown,
H=height
B=base
L
w A = LW
57. Example C. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
x2 + 4x = 21
x2 + 4x – 21 = 0
(x + 7)(x – 3) = 0
x = - 7 or x = 3
Therefore, the width is 3 and the length is 7.
2nd-Degree-Equation Word Problems
Area of a Parallelogram
A parallelogram is the area enclosed
by two sets of parallel lines. If we
move the shaded part as shown, we
get a rectangle.
B=base
H=height
L
w A = LW
58. Example C. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
x2 + 4x = 21
x2 + 4x – 21 = 0
(x + 7)(x – 3) = 0
x = - 7 or x = 3
Therefore, the width is 3 and the length is 7.
2nd-Degree-Equation Word Problems
Area of a Parallelogram
A parallelogram is the area enclosed
by two sets of parallel lines. If we
move the shaded part as shown, we
get a rectangle. Hence the area A
of the parallelogram is A = BH.
H=height
B=base
L
w A = LW
59. Example D. The area of the parallelogram shown is 27 ft2.
Find x.
2nd-Degree-Equation Word Problems
2x + 3
x
60. Example D. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
2nd-Degree-Equation Word Problems
2x + 3
x
2x2 + 3x = 27
61. Example D. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
2nd-Degree-Equation Word Problems
2x + 3
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0
62. Example D. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
2nd-Degree-Equation Word Problems
2x + 3
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
(2x + 9)(x – 3) = 0
63. Example D. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
2nd-Degree-Equation Word Problems
2x + 3
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
(2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft
64. Example D. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
2nd-Degree-Equation Word Problems
2x + 3
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
(2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft
65. Example D. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
2nd-Degree-Equation Word Problems
Area of a Triangle
2x + 3
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
(2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft
66. Example D. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
2nd-Degree-Equation Word Problems
Area of a Triangle
Given the base (B) and the height (H) of a triangle as shown.
2x + 3
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
(2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft
B
H
67. Example D. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
2nd-Degree-Equation Word Problems
Area of a Triangle
Given the base (B) and the height (H) of a triangle as shown.
B
H
2x + 3
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
(2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft
Take another copy and place it
above the original one as shown .
68. Example D. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
2nd-Degree-Equation Word Problems
Area of a Triangle
Given the base (B) and the height (H) of a triangle as shown.
B
H
2x + 3
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
(2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft
Take another copy and place it
above the original one as shown.
69. Example D. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
2nd-Degree-Equation Word Problems
Area of a Triangle
Given the base (B) and the height (H) of a triangle as shown.
B
H
2x + 3
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
(2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft
Take another copy and place it
above the original one as shown.
We obtain a parallelogram.
70. Example D. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
2nd-Degree-Equation Word Problems
Area of a Triangle
Given the base (B) and the height (H) of a triangle as shown.
B
H
2x + 3
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
(2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft
Take another copy and place it
above the original one as shown.
We obtain a parallelogram.
If A is the area of the triangle,
71. Example D. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
2nd-Degree-Equation Word Problems
Area of a Triangle
Given the base (B) and the height (H) of a triangle as shown.
B
H
2x + 3
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
(2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft
Take another copy and place it
above the original one as shown.
We obtain a parallelogram.
If A is the area of the triangle,
then 2A = HB or .A = BH
2
72. Example E. The base of a triangle is 3 inches shorter than
the twice of the height. The area is 10 in2. Find the base
and height.
2nd-Degree-Equation Word Problems
73. Example E. The base of a triangle is 3 inches shorter than
the twice of the height. The area is 10 in2. Find the base
and height.
2nd-Degree-Equation Word Problems
74. Example E. The base of a triangle is 3 inches shorter than
the twice of the height. The area is 10 in2. Find the base
and height.
Let x = height, then the base = (2x – 3)
2nd-Degree-Equation Word Problems
75. Example E. The base of a triangle is 3 inches shorter than
the twice of the height. The area is 10 in2. Find the base
and height.
Let x = height, then the base = (2x – 3)
2nd-Degree-Equation Word Problems
2x– 3
x
76. Example E. The base of a triangle is 3 inches shorter than
the twice of the height. The area is 10 in2. Find the base
and height.
Let x = height, then the base = (2x – 3)
Hence, use the formula 2A = BH
2*10 = (2x – 3) x
2nd-Degree-Equation Word Problems
2x– 3
x
77. Example E. The base of a triangle is 3 inches shorter than
the twice of the height. The area is 10 in2. Find the base
and height.
Let x = height, then the base = (2x – 3)
Hence, use the formula 2A = BH
2*10 = (2x – 3) x
20 = 2x2 – 3x
2nd-Degree-Equation Word Problems
2x– 3
x
78. Example E. The base of a triangle is 3 inches shorter than
the twice of the height. The area is 10 in2. Find the base
and height.
Let x = height, then the base = (2x – 3)
Hence, use the formula 2A = BH
2*10 = (2x – 3) x
20 = 2x2 – 3x
0 = 2x2 – 3x – 20
2nd-Degree-Equation Word Problems
2x– 3
x
79. Example E. The base of a triangle is 3 inches shorter than
the twice of the height. The area is 10 in2. Find the base
and height.
Let x = height, then the base = (2x – 3)
Hence, use the formula 2A = BH
2*10 = (2x – 3) x
20 = 2x2 – 3x
0 = 2x2 – 3x – 20
0 = (x – 4)(2x + 5)
2nd-Degree-Equation Word Problems
2x– 3
x
80. Example E. The base of a triangle is 3 inches shorter than
the twice of the height. The area is 10 in2. Find the base
and height.
Let x = height, then the base = (2x – 3)
Hence, use the formula 2A = BH
2*10 = (2x – 3) x
20 = 2x2 – 3x
0 = 2x2 – 3x – 20
0 = (x – 4)(2x + 5)
x = 4 or x = -5/2
2nd-Degree-Equation Word Problems
2x– 3
x
81. Example E. The base of a triangle is 3 inches shorter than
the twice of the height. The area is 10 in2. Find the base
and height.
Let x = height, then the base = (2x – 3)
Hence, use the formula 2A = BH
2*10 = (2x – 3) x
20 = 2x2 – 3x
0 = 2x2 – 3x – 20
0 = (x – 4)(2x + 5)
x = 4 or x = -5/2
2nd-Degree-Equation Word Problems
2x– 3
x
82. Example E. The base of a triangle is 3 inches shorter than
the twice of the height. The area is 10 in2. Find the base
and height.
Let x = height, then the base = (2x – 3)
Hence, use the formula 2A = BH
2*10 = (2x – 3) x
20 = 2x2 – 3x
0 = 2x2 – 3x – 20
0 = (x – 4)(2x + 5)
x = 4 or x = -5/2
Therefore the height is 4 in. and the base is 5 in.
2nd-Degree-Equation Word Problems
2x– 3
x
83. Exercise A. Use the formula h = –16t2 + vt for the following
problems.
2nd-Degree-Equation Word Problems
1. A stone is thrown upward at a speed of v = 64 ft/sec,
how long does it take for it’s height to reach 48 ft?
2. A stone is thrown upward at a speed of v = 64 ft/sec,
how long does it take for it’s height to reach 28 ft?
3. A stone is thrown upward at a speed of v = 96 ft/sec,
a. how long does it take for its height to reach 80 ft? Draw a
picture.
b. how long does it take for its height to reach the highest
point?
c. What is the maximum height it reached?
4. A stone is thrown upward at a speed of v = 128 ft/sec,
a. how long does it take for its height to reach 256 ft?
Draw a picture. How long does it take for its height to reach
the highest point and what is the maximum height it reached?
84. B. Given the following area measurements, find x.
2nd-Degree-Equation Word Problems
5.
8 ft2
x + 2
x
6.
12 ft2
x
(x – 1)
7.
x + 2
8.
12 ft2 x
(x + 4)
9.
24 ft2
(3x – 1)
x
10.
15 ft2x
18 ft2 x
(4x + 1)
85. B. Given the following area measurements, find x.
2nd-Degree-Equation Word Problems
2x + 1
11.
x5cm2
12.
x
2x – 3
9cm2
2x + 1
13.
x
18km2
14.
(x + 3)
(5x + 3)
24km2
15. 16.
16km2
2
x
x + 1
35km2
2
x
2x – 1