Transcript Lecture 1 Basic Concepts and Applications

Basic Math Conversions
Math for Water Technology
MTH 082
Fall 07
Chapters 1, 2, 4, and 7
Lecture 1
Today’s Outline
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How much math have you taken?
1. High School Math
2. College Math (MTH10MTH50)
3. College Algebra I (MTH 060)
4. Intermediate Algebra II (MTH
065)
5. Calculus or higher (MTH
251+)
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Have you taken the Math
Placement Exam for incoming
students in the testing center?
1. Yes
2. No
I have had MTH 065 (Intermediate
College Algebra II) and thus
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Although I have not completed the
prerequisite for this course, I am
willing to work _________% harder
than my teammates?
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2. 50%
3. 25%
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RULES TO SOLVING MATH
PROBLEMS
1. READ THE PROBLEM FIRST (AND PUT IT INTO
YOUR OWN WORDS)
2. LAY OUT THE PROBLEM=DRAW A DIAGRAM
3. DETERMINE WHAT YOU HAVE AND WHAT YOU
NEED (YOU MAY HAVE EXTRA)
4. PERFORM CONVERSIONS
5. ARTICULATE THE REASON FOR USING AN
EQUATION
6. DO DIMENSIONAL ANALYSIS FIRST
7. APPLY THE EQUATION---DO NOT PLUG AND
CHUG
8. SOLVE THE PROBLEM
9. CHECK YOUR WORK
Decimal Places
Greater than 1
Less than 1
100
10
1
1
10
http://www.gomath.com/htdocs/lesson/decimal_lesson1.htm
1
100
1
1000
Basic Math Conversions
Chapter 1
Power and Scientific Notation
Rules of Power and Scientific
Notation
Rule 1 = when a number is taken out of scientific
notation a positive exponent value indicates a move
of the decimal point to the right and a negative
exponent value indicates a decimal point move to the
left!
Rule 2 = when a number is PUT into scientific notation
a decimal point to the left indicates a positive exponent
and a decimal point move to the right indicates and
negative exponent values!
Rules of Scientific Notation
Rule 4 = when you multiply the numbers in scientific
notation, multiply the numbers but add the exponents.
Rule 5 = when you divide the numbers in scientific
notation, divide the numbers but subtract the
exponents.
POWER
Numeric
20=1
21 =2
22= 2 X 2 = 4
23= ( ) X ( ) X ( )= ________
English
ft2= ft X ft
m3= meter X meter X meter
POWER
Numeric Expanded and Exponential Form
2 2
2 2
( )  ( )( )
3
3 3
3 2  (
1
)
2
3
42
(4)( 4)
( 3)(
)
5
(5)(5)(5)
8 2  (
1
1
)

(
)
2
8
(8)(8)
English Expanded and Exponential Form
in 2
(in)(in)
in 2
( 2)(
)( )
ft
( ft)( ft)
ft
Your Turn
km2
( )( )
( 3 )(
)
m
( )( )( )
Scientific Notation
Scientific Notation = number multiplied by power of ten
5.4 101  (5.4 10)  54
1.2 103  (1.2  (10)(10)(10))  1,200
1 1
3.62
3.62 10  2  (3.62  ( )( ))  ( 2 )  0.0362
10 10
10
Your Turn (Write It All out!!!)
4.800 103 
350 10 3 
Scientific Notation
Scientific Notation = Taken out!
Rule 1 = when a number is taken out of scientific
notation a positive exponent value indicates a move of
the decimal point to the right and a negative exponent
value indicates a decimal point move to the left!
9.789 104  9.7890  97,890 17,890 105  17890  0.17890
Positive four places to right
Negative five places to left
Your Turn
0.5200 103 
0.039 10 2 
Scientific Notation
Scientific Notation = Put Into!
Rule 2 = when a number is PUT into scientific notation
a decimal point to the left indicates a positive
exponent and a decimal point move to the right
indicates a negative exponent values!
75  7.5  101
.0362  3.62  10-2
1200  1.2  103
Your Turn
0.0041 
420 
Multiplying in Scientific
Notation
Rule 4 = when you multiply the numbers in scientific
notation, multiply the numbers but add the exponents.
(3  104 )  (4 103 )  (3  4)  (10( 43) )  12 107
(5  10-4 )  (6 103 )  (5  6)  (10( 4 3) )  30 107
(.003)  (0.2)  (0.0006)  (3 103 )  (2 101 )  (6 104 )
(3  2  6)  (36)  (10( 3 1 4) )  36 108
Your Turn
(.02)  (0.2)  (0.002) 
Dividing in Scientific Notation
Rule 5 = when you divide the numbers in scientific
notation, divide the numbers but subtract the
exponents.
(4  104 )
4
.
 ( )  (10( 43) )  2 101
(2 103 )
2
(800)
8 102
8
2( 1)
3
(
)

(

10
)

4

10
(0.2)
2 101
2
Your Turn
(0.006)

(0.3)
Basic Math Conversions
Chapter 2
Dimensional Analysis
MATT’S RULE
ALWAYS USE
DIMENSIONAL ANALYSIS
BEFORE YOU PLUG AND
CHUG!
Dimensional Analysis
gal
gal
gal ft 
or 3
(ft) (ft) (ft)
ft
3
2
(km)(km)
km
km 2 mm 3 
or
(mm) (mm) (mm)
mm 3
gal
gal
gal/min
min
min


gal
gal
gal/ft 3
(ft)(ft)(f t)
ft 3
(km)(km)
km 2 mm 3 (mm) (mm) (mm)
(km)(km)
(sec)(sec)



m
m / sec 2
(mm) (mm) (mm)
m
(sec)(sec)
Your Turn
ft 3 / sec

ft / sec
Dividing is the same as multiplying by the INVERSE
Dimensional Analysis
Multiplication and Division
Need answer in gallons
(gal ft 3 )(ft 3 ) 
gal
(ft)(ft)(f t) gal


 gal
(ft) (ft) (ft)
1
1
Need answer in square feet
4
(ft)(ft)(f
t)
(ft)
ft
(80 ft 3 sec)  (3.5 ft/sec) 


 WRONG!
2
(sec)
(sec) (sec)
(ft)(ft)(f t)
(ft)(ft)(f t) (sec)
(sec)
3
(80 ft sec)/(3.5 ft/sec) 


 ft 2
ft
(sec)
(ft)
sec
Dimensional Analysis
Multiplication and Division
Need answer in cubic meters per second
(m)
(m)
1
m
(sec)
2
(100 m sec)/(5 m ) 

 2 
 UGLY!
2
(m)(m) (sec) m
(sec)m
1
2
3
(m)
(m
)
m
(100 m sec)  (5 m 2 ) 


 YEP!
(sec) (1)
(sec)
WORD PROBLEM
The flow rate in a water line is 2.3 ft3/sec. What is the
flow rate as gallons per minute?
Step 1: Use your own words. Got a pipe with a known flow rate, need to convert that
value from one unit to another. This is a simple conversion problem
Step 2: Draw a diagram
2.3 ft3/sec
Step 3: Conversions?
GIVEN: 2.3 ft3/sec
CONVERSIONS:
7.48 ft3/gal
60 sec/min
gal/min?
NEED: gal/min
WORD PROBLEM
The flow rate in a water line is 2.3 ft3/sec. What is the
flow rate as gallons per minute?
2.3 ft3/sec
Step 4: Convert ft3/sec to
divide?
(2.3 ft 3 sec)  (7.48
gal min.
gal/min?
Dimensional Analysis First. To multiply or
gal
(ft)(ft)(f t)
gal
sec
gal
)

60
sec/min




(Yes )
3
ft
sec
(ft)(ft)(f t) min min
Step 5: Solve the problem.
gal
gal
(2.3 ft sec)  (7.48 3 )  60 sec/min  1032
ft
min
3
WORD PROBLEM
A channel is 3 ft wide with water flowing to a depth of 2 ft. The velocity in the channel
is found to be 1.8 ft/sec. What is the flow rate in the channel in cubic feet per
second?
Step 1: Use your own words. Got a channel with known dimensions and a flow rate,
need to convert that value from one unit to another. This is a simple conversion
problem
Step 2: Draw a diagram
3 ft
1.8 ft/sec
ft3/sec?
2 ft
Step 3: Conversions?
GIVEN: 1.8 ft/sec , 3ft, 2 ft
NEED: ft3/sec
CONVERSIONS:
None necessary
WORD PROBLEM
A channel is 3 ft wide with water flowing to a depth of 2 ft. The velocity in the channel
is found to be 1.8 ft/sec. What is the flow rate in the channel in cubic feet per
second?
3 ft
1.8 ft/sec
ft3/sec?
2 ft
Step 3: Conversions?
GIVEN: f=1.8 ft/sec, w=3ft, d=2 ft
CONVERSIONS: None necessary
NEED: ft3/sec
Step 4 Equation : flow in channel (FC) = f X w X d
Step 5: Solve Dimensional Analysis First!
ft ft ft ft 3
(1.8 ft sec)  (3 ft)  (2 ft) 
  
 YES!
sec 1 1 sec
WORD PROBLEM
A channel is 3 ft wide with water flowing to a depth of 2 ft. The velocity in the channel
is found to be 1.8 ft/sec. What is the flow rate in the channel in cubic feet per
second?
3 ft
1.8 ft/sec
ft3/sec?
2 ft
Step 6: Solve Problem
Equation : flow in channel (FC) = f X w X d
where f = flow
w = width of channel
d = depth of channel
ft 3
(1.8 ft sec)  (3 ft)  (2 ft)  10.8
sec
Basic Math Conversions
Chapter 3
Rounding and Estimating
Decimal Places
Greater than 1
Less than 1
100
10
1
1
10
http://www.gomath.com/htdocs/lesson/decimal_lesson1.htm
1
100
1
1000
Basic Rules
of Rounding
A ≈ indicates a number or answer has been rounded
Rule 1: When rounding to any desired place if the digit directly to
the right of that place is less then 5 replace all digits to the right
with zeros.
Rule 2: When rounding to any desired place if the digit directly to
the right of that place is greater then 5, increase the digit in the
rounding place by 1 and replace all digits to the right of the
increase with zeros.
Rule 3: When rounding decimal numbers to the right of the
decimal point, drop the rounded digits
Rounding
Rule 1: When rounding to any desired place if the digit
directly to the right of that place is less then 5 replace
all digits to the right with zeros.
Round 342,427 to the nearest thousandths
342,427 ≈ 342,400
Rounding place (less then 5 everything to right =0)
hundredths place
Your Turn
Round 1,342,427 to the nearest hundredth thousandths place
1,342,427 ≈
Rounding
Rule 2: When rounding to any desired place if the digit
directly to the right of that place is greater then 5,
increase the digit in the rounding place by 1 and
replace all digits to the right of the increase with zeros.
Round 37,926 to the nearest tenth
37,926 ≈ 37,930
Rounding place (greater then 5 increase value by 1)
tenths place
Your Turn
Round 1,377,427 to the nearest hundredth thousandths place
1,377,427 ≈
Rounding
Rule 3: When rounding decimal numbers to the right of
the decimal point, drop the rounded digits.
Round 5.654 to the nearest tenth
5.654 ≈ 5.7
Rounding place (greater then 5 increase value by 1)
tenths place
Your Turn
Round 483.16 to the nearest unit
483.16 ≈
Estimating
Factoid: Estimating indicates the approximate size of a
calculated answer.
Estimate the value of 20 X 30 = (2 X 3) = 6 with two zeros at the end =600
Estimate the value of 40 X 600 = (4 X 6) = 24 with three zeros at the end =24,000
Estimate the value of 40 X 20 X 500 = (4 X 2 X 5) = 40 with four zeros at the end =400,000
Estimate the value of 9 X 700 X 60 X 70 = 2800 with four zeros at the end =28,000,000
9X7
63
≈
60 X 6
360
≈
400 X 7
Estimating
Factoid: Estimating indicates the approximate size of a
calculated answer.
Estimate the value of 40,000/200 = cancel zeros 400/2=200
Estimate the value of 700/6,000 = cancel zeros 7/60=
0 .1
60 7.0  0.1
Estimate the value of (20)(400)/(50)(80) = cancel zeros =(20)(4)/(5)(8)= 80/40= cancel zeros =8/4=2
Basic Math Conversions
Chapter 7
Percents
Basic Rules
of Percents
FACTOID. The term efficacy refers to a percent
Rule 1. In calculations greater than 100 percent, the
numerator of the percent equation must always be
larger than the denominator.
PART
PERCENT 
100
WHOLE
Percents
20
20% 
 0.20 or
100
95
95% 
 0.95
100
2
2% 
 0.02
100
.05
.05% 
 0.005
100
20
 0.20  20%
100
Percent Word Problems
A certain piece of equipment is having mechanical
difficulties. If the equipment fails 6 times out of 25
tests, what percent failure does this represent?
PART
PERCENT 
100
WHOLE
6
PERCENT  100  0.24 100  24% failure
25
Percent Word Problems
A certain piece of equipment is having mechanical
difficulties. If the equipment fails 6 times out of 25
tests, what percent failure does this represent?
PART
PERCENT 
100
WHOLE
6
PERCENT  100  0.24 100  24% failure
25
Percent Word Problems
The raw water entering a treatment plant has a turbidity
of 10 ntu. If the turbidity of the finished water is 0.5 ntu,
what is the turbidity removal efficacy of the treatment
plant.
PART
PERCENT 
 100
WHOLE
Percent is unknown and 10 ntu = whole. However,
0.5 ntu is not the part removed. It is the turbidity still in
the water. Thus, 10 ntu-0.5 ntu= 9.5 ntu
9.5
PERCENT 
100  95% turbidity removal efficacy
10
Percent Word Problems
Rule 1. In calculations greater than 100 percent, the
numerator of the percent equation must always be
larger than the denominator.
A treatment plant was designed to treat 60 Ml/d. One
day it treated 66 Ml. What % of the design capacity
does this represent.
PART
PERCENT 
 100
WHOLE
66
PERCENT  100  1.1100  110%
60
Basic Math Conversions
Chapter 5
Ratios and Proportions
Rules of Ratios and
Proportions
Rule 1 = If the unknown is expected to be smaller than the
known value, put an x in the numerator of the first fraction, and
put the known value of the same unit in the denominator.
Rule 2 = If the unknown is expected to be larger than the
known value, put an x in the denominator of the first fraction,
and put the known value of the same unit in the numerator.
Rule 3 = Make the two remaining values of the problem into
the second fraction. (smaller in numerator, larger in
denominator)
Ratios and Proportions
1
2
and proportional ?
2
4
(2)( 2)  4 and (4)(1)  4 so yes!
Are
6
x
and
solve for x.
13
40
(6)( 40)  (13)( x)
(6)(40)
x
13
18.46  x
3
42
and
proportional ?
7
98
(3)(98)  294 and (98)(3)  294 so yes!
Are
Ratios and Proportions
Rule 1 = If the unknown is expected to be smaller than the
known value, put an x in the numerator of the first fraction, and
put the known value of the same unit in the denominator.
.......lbs ....
.......mon ey....
smaller va lue smaller va lue

larger val ue
larger val ue
Problem = If three men can do a certain job in 10 hours, how
long would it take five men to do the same job?
What is the unknown? Time and it will be smaller…so
x
3

10 hr
5
5 x  30hrs
30
x
5
x  6 hrs
Ratios and Proportions
Rule 2 = If the unknown is expected to be larger than the known value, put an x in
the denominator of the first fraction, and put the known value of the same unit in the
numerator.
.......lbs ....
.......mon ey....
smaller va lue smaller va lue

larger val ue
larger val ue
Problem = If 5 lb of chemical are mixed with 2,000 gallons of water to obtain a desired
solution, how many pounds of chemical would be mixed with 10,000 gallons of water to obtain
a solution of the same concentration?
What is the unknown? lbs…so
5 lb
2,000 gal

x
10,000 gal
2,000 x  (5lb ) (10,000 gal )
(5lb ) (10,000 gal )
2,000 gal
(5lb )(10 gal )
x
 25lb
2 gal
x
Mixed numbers
Mixed Numbers as Fractions uses Circles to demonstrate how a fraction can be
renamed from mixed form to fraction form.
The circles below show the mixed number 2 2/5. You are to write 2 2/5 in fraction form
with only a numerator and denominator.
http://www.visualfractions.com/MixtoFrCircle.html
To write the example, you can think of each whole number as 5/5. So in the above
example you would have:
On the pretest, you can think of 13/8
.
13 8 5
5
5
   1  1
8 8 8
8
8