Transcript Lesson 5 Problems - Adjective Noun Math

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Algebra
Problems…
Solutions
Set 5
© 2007 Herbert I. Gross
By Herbert I. Gross and Richard A. Medeiros
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Problem #1
What signed number is
2
named by 3( 4) ?
Answer: -48
© 2007 Herbert I. Gross
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Answer: -3(-4)2 = -48
Solution:
We do exponents before we multiply.
What's being raised to the second power
(squared) is -4.
Since the product of two negative numbers is
positive, we see that (-4)2 = (-4)(-4) = +16
Replacing (-4)2 by +16, we obtain…
-3(-4)2 = -3(+16)
And since the product of two numbers that
have opposite signs is negative,
-3(+16) = -48
© 2007 Herbert I. Gross
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Careful
Don't overlook the exponent. That is, if
we look at the expression…
-3(-4)2
too quickly, it might seem that we are
multiplying two negative numbers. We
have to observe that -3 is not multiplying
-4, but rather (-4)2, which is +16.
© 2007 Herbert I. Gross
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Careful
Since we raise to a power before we
multiply, the exponent refers only to -4.
If instead we had wanted to multiply -4 by -3
before we raised the expression to the
second power, we would have had to write
the expression as…
[-3(-4)]2
In this case, the answer would have been
given by…
[-3(-4)]2 = [+12]2 = +144
© 2007 Herbert I. Gross
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PEMDAS
(Review)
Once again, we have to stress that there is
ambiguity when grouping symbols are
omitted; for that reason, we need some
general agreement concerning the
order of operations.
And the agreement used most generally,
and which we will use, goes under the
acronym PEMDAS.
© 2007 Herbert I. Gross
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PEMDAS Review
In PEMDAS the agreed Order of
Operations that we act on is…
Parentheses
Exponents
Multiplication
Division
Addition and
Subtraction
© 2007 Herbert I. Gross
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Problem #2
What signed number is
2
named by 8 – ( 3) ?
Answer: -1
© 2007 Herbert I. Gross
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Answer: 8 – (-3)2 = -1
Solution:
By our order of operations (PEMDAS
agreement) we read
8 – (-3)2 as 8 – [(-3)2].
That is, we do what's inside the parentheses
first, then raise to powers before we
subtract.
We know that (-3)2 means (-3)(-3) and that
(-3)(-3) = +9.
© 2007 Herbert I. Gross
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Solution (cont.):
Therefore we may rewrite the expression
8 – [(-3)2] in the equivalent form…
+8
– +9
By the add-the-opposite rule we may rewrite
the expression +8 – +9 in the equivalent form…
+8
+ -9 = -1
and by our rule for adding two numbers that
have different signs, we see that the answer to
our problem is -1.
© 2007 Herbert I. Gross
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Notes on #2
Again, notice the importance of the
grouping symbols. We multiply -3 by itself
before we subtract it from +8.
Yet if we had read the expression 8 – (-3)2
too quickly, we might have confused it with
the expression 8 – -(3)2.
© 2007 Herbert I. Gross
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Notes on #2
Then in calculating the expression
8 – -(3)2, we would have used the
add-the-opposite rule and our
answer would have been…
8 + +(3)2 = 17.
In short, 17 is the correct answer to
8 + (-3)2, BUT
is the incorrect answer to the
given problem, which was… 8 – (-3)2.
© 2007 Herbert I. Gross
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Problem #3
What signed number is
named by
(-15 + -3) × (-6 ÷ +2)?
Answer: +54
© 2007 Herbert I. Gross
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Answer: (-15 + -3) × (-6 ÷ +2) = +54
Solution:
Since there are no exponents, and we are
using the order of operations (PEMDAS), we
begin by working inside the parentheses
first. We know from the previous lessons
that…
-15
© 2007 Herbert I. Gross
+ -3 = -18
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Solution:
…and from the present lesson we know that
when we divide two signed numbers we
divide their magnitudes and choose the
positive sign for the quotient if the two
numbers have the same sign and the
negative sign otherwise. This means that…
-6
© 2007 Herbert I. Gross
÷ + 2 = -3
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Solution:
If we now substitute the results of our two
equations into the expression…
(-15-18
+ -3) × (-6 -÷3 +2)
= +54
we obtain the equivalent expression
which by our rule for multiplying two
numbers that have the same sign is equal
to +54
© 2007 Herbert I. Gross
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Solution:
In more compact form, what we did was…
( 15
+
-3)
-18
×
×
(6
÷
-3
=
+54
© 2007 Herbert I. Gross
+2)
=
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Careful
In this problem we had to perform three
different arithmetic operations with signed
numbers. When we have to perform several
different types of operations, we have to be
careful not to confuse one rule with another.
Unless we have internalized the rules, it is all
too easy for example to be confused about
why the sum of two negative numbers is
always negative, but the product of two
negative numbers is always positive.
© 2007 Herbert I. Gross
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Problem #4a
Evaluate (t2 – 5)t when
t = 5.
Answer: -100
© 2007 Herbert I. Gross
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Answer: -100
Solution:
(t2 – 5)t
Using PEMDAS we first work inside the
parentheses and replace by t by -5. Thus t2
becomes (-5)2; and (-5)2 means (-5) (-5). By
our rule for multiplying two numbers that
have the same sign (-5) (-5) = +25.
t2 = (-5)2(-5)
= +25
© 2007 Herbert I. Gross
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Solution:
Thus when t = -5.
2 – +5
= +t25
5
= +20
replacing t by -5,
(t2 – 5)t becomes...
which by our rule for
multiplying signed numbers
is equal to -100.
© 2007 Herbert I. Gross
-100
-5)
(20)(
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Note on #4a
Until you have internalized the arithmetic of
signed numbers and the language of
algebra, a good strategy is to translate the
expression into “English” first. More
specifically, given the algebraic expression
(t2 – 5)t
we work within the parentheses first and we
raise to the second power before we
subtract. After that, we do the indicated
multiplication.
© 2007 Herbert I. Gross
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(t2 – 5)t
Note on #4a
Thus, in the “plain English” format; the
expression (t2 – 5)t becomes…
Step 1
Step 2
Start with any number t
Multiply it by itself t2
Step 3
Step 4
Subtract 5 t2 – 5
Multiply this by t t2 – 5t
© 2007 Herbert I. Gross
-5
(-5) (- 5)
– +5
(20)(-5)
+ 25
=
=
=
+25
+20
-100
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Note on #4a
The table/recipe shows each step of the
calculation, and shows -5 replacing t.
Better yet, the order in which the steps
are written already indicates the
order of operations, in advance.
This lets us concentrate on the arithmetic,
without being distracted
by such questions as “Do I change
the sign first or do I multiply first”, etc.
© 2007 Herbert I. Gross
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Problem #4b
Evaluate t2 – 5t when
t = 5.
Answer: +50
© 2007 Herbert I. Gross
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Answer: +50
Solution:
Since no grouping symbols appear we use
our PEMDAS agreement to rewrite t2 – 5t…
in the form…
(t )2 – (5t )
In words the expression would read…
Start with any number (t)
Square it (t2)
Subtract 5 times the number you chose
(t2 – 5t)
© 2007 Herbert I. Gross
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Solution:
(t)2 – (5t)
So we start with t = -5…
We multiply it by itself; (t)2
We multiply t by 5…
We subtract 5t from t2 to obtain…
Using the add the opposite rule we obtain…
Start with any number t
Multiply it by itself
t2
Multiply t by 5
5t
Subtract 5t from t2 t2 – 5t
(Or by adding the opposite)
© 2007 Herbert I. Gross
-5
(-5) (- 5)
-25
+25 – -25
+25 + +25
=
+25
=
+50
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Note on #4b
Once you are comfortable working with the
algebraic notation, simply replace t by
(-5) wherever it appears in the expression.
(-5)t22 –– 55 t(-5)
+25 – -25
+25 +
+25
= +50
© 2007 Herbert I. Gross
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Note on #4b
However, until you feel comfortable with
manipulating the algebraic symbols,
practice translating the algebraic
expressions into “plain English”.
Otherwise, for example, it might be easy
to confuse (t2 – 5)t with t2 – 5t.
© 2007 Herbert I. Gross
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Problem #4c
Evaluate t2 – 5t when
t = 6?
Answer: +66
© 2007 Herbert I. Gross
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Answer: +66
Solution:
This is word-for-word the same as Exercise
4b except that we replace -5 by -6. More
specifically, in verbal form evaluating the
expression t2 – 5t when t = -6 means that…
Start with any number (t)
Square it (t2)
Subtract 5 times the number you chose
(t2 – 5t)
© 2007 Herbert I. Gross
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Solution:
(t)2 – (5t)
So we start with t = -6…
We multiply it by itself; (t)2
We multiply t by 5…
We subtract 5t from t2 to obtain…
Using the add the opposite rule we obtain…
Start with any number t
Multiply it by itself
t2
Multiply t by 5
5t
Subtract 5t from t2 t2 – 5t
(Or by adding the opposite)
© 2007 Herbert I. Gross
-6
(-6) (- 6)
-30
+36 – -30
+36 + +30
=
+36
=
+66
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Looking Ahead
The reason for giving two such closely
connected problems such as #4b and #4c
will become more transparent when we
present the solution for problem #4d.
The key point in #4d is that if t2 – 5t = +50
when t = -5, and it equals 66 when t = -6 then
if n represents any number that is between
50 and 66, there is at least one value
for t between -5 and -6 for which t2 – 5t = n.
In #4d, we chose 55 as the value of n.
© 2007 Herbert I. Gross
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Problem #4d
Use the answers to parts
(b) and (c) to find a value
of t for which
2
t – 5t = 55
(Round off your answer to
the nearest integer).
Answer: t = -5
© 2007 Herbert I. Gross
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Answer: t = -5
Solution:
In #4b, we saw that if we chose t to be -5 the
value of t2 – 5t would be 50; while in #4c, we
saw that if we chose t to be -6, the value of
t2 – 5t would have been 66.
Since 55 is less than 66 but greater than
50, we may conclude that there must be at
least one value of t between -5 and -6 such
that t2 – 5t = 55.
© 2007 Herbert I. Gross
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Solution:
Moreover because 50 is closer in value
to 55 than 66 is, we assume that the
required value of t is closer to -5 than to -6
which leads us to believe that the value of
t, rounded off to the nearest integer, is -5
© 2007 Herbert I. Gross
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Note on #4d
To verify our assumption (and we might
use a calculator to make the necessary
computations less tedious to obtain), we
could make a table such as…
t
t2
5t
t2 – 5t
-5.0
+25
-25
+25 – -25
-5.1
+26.01 -25.5
-5.2
+27.04
-5.3
+28.09 -26.5
-5.4
+29.16
© 2007 Herbert I. Gross
-26
-27
+26.01 – -25.5
= +50
= +51.5
+27.04 – -26
= +53.04
+28.09 – -26.5
= +54.59
+29.16 – -27
= +56.16
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Note on #4d
Since 55 is between 54.59 and 56.16, we
know that the required value of t is, in fact,
between -5.3 and -5.4…
t
t2
5t
t2 – 5t
-5.0
+25
-25
+25 – -25
-5.1
+26.01 -25.5
-5.2
+27.04
-5.3
+28.09 -26.5
-26
?
?
?
-5.4
+29.16
-27
© 2007 Herbert I. Gross
+26.01 – -25.5
= +50
= +51.5
+27.04 – -26
= +53.04
+28.09 – -26.5
= +54.59
+55
+29.16 – -27
= +56.16
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Note on #4d
If we wanted an even better approximation,
we could evaluate the expression t2 – 5t,
say when t = -5.31, -5.32, -5.33, -5.34, etc.
For example, if we let t = -5.33, we obtain…
t
t2
5t
-5.33 +28.4089 -26.65
t2 – 5t
+28.4089 – -26.65
= +55.0589
which indicates that the required value of t
is very nearly equal to -5.33.
© 2007 Herbert I. Gross
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This problem illustrates an important
technique for solving equations that may be
beyond the scope of our algebra knowledge.
Later in the course, we will learn how to obtain
exact algebraic solutions to equations such as
t2 – 5t = 55.
Meantime, there is nothing wrong with making
guesses (sometimes referred to as
trial-and-error or guess-and-check).
By repetitive guessing and checking, we can
correctly find the value of t accurate to any
required number of decimal places.
© 2007 Herbert I. Gross
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Careful
Be careful not to assume that there is only
one value of t for which t2 – 5t.
For example, when t = 10, t2 – 5t = 50.
And when t = 11, t2 – 5t = 66.
Hence there is another value of t that is
between 10 and 11 for which t2 – 5t = 55.
How can this be, and are there still other
values of t for which t2 – 5t = 55?
Questions such as these will be answered
as our course continues.
© 2007 Herbert I. Gross
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Problem #5a
The formula that relates the
Fahrenheit temperature (F) to the
Celsius temperature (C) is…
C = 5/9 (F – 32).
What is the Celsius temperature
when the Fahrenheit temperature is
14°?
© 2007 Herbert I. Gross
Answer: -10°C
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Answer: -10°C
Solution:
Starting with the formula C = 5/9 (F – 32), we
replace F by 14 and then obtain the
following sequence of steps…
C = 5/9( F – +32)
= 5/9(+14 – +32)
= 5/9(+14 + -32)
= 5/9(-18)
= -10
© 2007 Herbert I. Gross
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To review the logic behind each step, we
start with… We first replaced F by +14…
We then used the “add the opposite rule”
to obtain… And finally, we used our rules
of arithmetic to obtain…
F – +32)
C = 5/9(+14
C = 5/9(+14 + -32)
C = 5/9 ( -18)
© 2007 Herbert I. Gross
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Answer: -10°C
Solution:
Since we are multiplying a positive number
(5/9) by a negative number (-18) the product
will be negative and since 5/9 × 18 = 10, we
see that the equation can be written in the
form…
-10
5
C
=
C = /9(-18)
© 2007 Herbert I. Gross
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Problem #5b
The formula that relates the
Fahrenheit temperature (F) to the
Celsius temperature (C) is…
C = 5/9 (F – 32).
What is the Fahrenheit
temperature when the Celsius
temperature is -20°?
© 2007 Herbert I. Gross
Answer: -4°F
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Answer -4°F
Solution:
In part 5a we were faced with a direct
computation (arithmetic). In this part we are
faced with an indirect computation (algebra).
That is, we are still working
with the formula… but now we replace C by
-20 to obtain the equation.
-20
C
© 2007 Herbert I. Gross
= 5/9( F – +32)
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Solution:
Since 5/9 is multiplying what's inside the
parentheses, we “unblock” the parentheses
by dividing both sides of the equation
by 5/9. Dividing by 5/9 is equivalent to
multiplying by 9/5. Hence, we may rewrite
the equation as…
9/
© 2007 Herbert I. Gross
-20 = 5
9/ (× F
5/ –
+F
+32)
×
(
32)
–
5
5
9
9
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Solution:
Since 9/5 × -20 = -36 and 9/5 × 5/9 = 1, we may
rewrite the equation as...
9/
-20
-36==
91
5/– )+(32)
+32)
×
(
/
(F
×
F
–
5
5
9
-36
© 2007 Herbert I. Gross
= F – +32
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Solution:
By the “add the opposite” rule, we may
rewrite… and we now add +32 to both sides
of the equation to obtain… Since
-36 + +32 = -4 and -32 + +32 = 0, we may
rewrite the equation as…
= F –++-32
--36
36 ++ ++32
32 == FF ++ --32
32 ++ ++32
32
-4 =(F + 0
)
-36
© 2007 Herbert I. Gross
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Note on #5
Using the recipe model that we
introduced in Lesson 1 we can illustrate
the connection between #5a and #5b.
Namely, in words, the formula
C = 5/9(F – 32) tells us how to find the
Celsius temperature once we know the
Fahrenheit temperature.
© 2007 Herbert I. Gross
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“Recipe”
In terms of our “program”…
problem 5a becomes…
Converting Fahrenheit to Celsius
Start with Fahrenheit Temp.
F
14
Subtract 32
F – 32 14 – 32 = -18
5/ (F – 32)
5/ (-18)
Multiply by 5/9
9
9
-10
Answer is Celsius Temp. C = 5/9(F – 32)
© 2007 Herbert I. Gross
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Calculator
Notice that if your calculator doesn't have
a fraction mode, multiplying by 5/9 is the
same as dividing by 9 and then
multiplying by 5 (or first multiplying by 5
and then dividing by 9). Therefore, the
sequence of key strokes might be…
14
© 2007 Herbert I. Gross
–
32
÷
9
×
5
=
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Recipe
• For problem #5b we would want to use
the “undoing” process. That is, if the last
step was multiplying by 5, the first step in
the “undoing” program would be to start
with the Celsius temperature and then
divide by 5.
We would next “undo” dividing by 9 which
means multiplying by 9. Finally, we would
“undo” subtracting 32 by adding 32.
© 2007 Herbert I. Gross
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Doing &
Undoing
• In terms of a “plain English” chart…
Doing Recipe
Undoing Recipe
Start with F
Answer is F
Subtract 32
Add 32
Multiply by 5/9
Divide by 5/9
i.e. Multiply by 9/5
Answer is C
Start with C
© 2007 Herbert I. Gross
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Note on #5
Once we have used a “plain English” chart
long enough, it becomes more natural for
us to work with the algebraic format.
For example, starting with
C = 5/9( F – +32)
© 2007 Herbert I. Gross
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Note on #5
To “unblock” the parentheses (that is, to
“undo” the last step in the formula) we
multiply both sides by 9/5 to obtain…
9/
9/
59/ (F
+32)
5/ –
+32)]
C
=
[
(F
–
95
5
9
9/ • 5/ )(F – +32)
C
=
(
5
5
9
9/
© 2007 Herbert I. Gross
+32)
C
=
1(F
–
5
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Note on #5
In other words, the formula now becomes…
and if we now add 32 to both sides of the
above formula we see that…
9/ C
5
+
= 32
F –=32
F
F = 9/5C + 32
or in more traditional format…
© 2007 Herbert I. Gross
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Note on #5
The two formulas look quite different, but
they are equivalent ways to express the
relationship between F and C. Most likely if
we wanted to do only direct computations,
we would use formula C = 5/9(F – 32) if we
were told the value of F, and we wanted to
find the corresponding value of C; and we
would use formula F = 9/5C + 32 if we were
given the value of C, and we wanted to find
the corresponding value of F.
© 2007 Herbert I. Gross
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Note on #5
Recall that 0°C represents the freezing point
of water and 100°C represents the boiling
point of water. Thus, while we
don't often think of it in those terms, the
fact is that the Celsius scale is analogous to
percents. For example, if the temperature is
20°C it means that it is 20% of the way
between the freezing point of water and the
boiling point of water. That's one nice
advantage that Celsius has over Fahrenheit.
© 2007 Herbert I. Gross
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Note on #5
On the other hand, a degree on the
Celsius scale represents a greater
change in temperature than a degree on
the Fahrenheit scale. More specifically,
the rate is 5°C per 9°F. Thus, a Celsius
degree is almost double a Fahrenheit
degree. Therefore, being off by 1 degree
on the Celsius scale represents a
greater error than by being off 1 degree
on the Fahrenheit scale.
© 2007 Herbert I. Gross