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-1:t QUESTIONS Fltltt.Yi-ANSWERED
DUE: 21 ~PRIL 2D26
DISCLAIMER:
TRlnJDCUMENf,S DR-RE ERENq AND1iU1DANDCTURP SE
- DNL:Y, I DD NDTt TA E-RESPPNSIBll.;ln'~FDR-A~Y- Pl.AGIARISM, MISUS , DR
ACADEMIC MISCONDUCT RESUllTING FROM THEIR USE. IT IS YDURLJ_.., I-: : "_.,.,_
RESPONSIBILITY TQ ENSU E ORIGINAL TY lND COMPLIANCE WITL I
RELEVANT GUIDELINES DR STAND RD . t-~ ~ r
l
, PLEASE USE THIS DOCUMENT AS A GUIDE ONLY
Question 1
1.1.1. Conceptual Difference in the Role of k
The symbol k plays a fundamentally different role in each statement based on the mathematical
context:
In Statement A (5k−10=25): The symbol k acts as an unknown. In this context, the equation is
a "closed" sentence. There is a specific, fixed value that makes the statement true. The goal is
to "find" or solve for this missing value.
In Statement B (y=5k−10): The symbol k acts as a variable. In this functional relationship, k
does not have one single value; instead, it represents a range of possible values (the input). As
the value of k changes, the value of y changes accordingly, describing a continuous
relationship between two quantities.
1.1.2. 'Specific Value' vs. 'Domain of Input Values'
To determine which represents which, we can perform the calculations for both:
Statement A: 5k−10=25
To find the value of k:
Add 10 to both sides: 5k=35
Divide by 5: k=7
Classification: This represents a 'specific value'. The symbol k can only be 7 for the statement to be
mathematically valid.
Statement B: y=5k−10
If we attempt to solve for k:
1. Y + 10 -== 5k
2. k -:: 1t 1.lll
5
Classification: This represents a 'domain of input values'. Because k is dependent on y (and vice
versa), k can take on any value from a defined set (usually all real numbers), producing a
corresponding output for y. It defines a line on a coordinate plane rather than a single point on a
number line.
, Statement Role of k Representation
II_II
A: 5k - 10 = 25 Unknown Specific Value (k = 7)
======:II ~=1
B: y = 5k - 10 Variable Domain======:
of Input Values
1.2. Testing Sam's Logic with Numbers
To see why 6S=P is incorrect, let's use the hint and test the relationship with a concrete example.
1. The Goal:
The sentence states: "There are six times as many students as professors."
This means the group of students is much larger. If we have:
Professors (P) = 2
Students (S) should be 2×6=12
2. Testing Sam’s Equation (6S=P):
If we plug these numbers into Sam’s equation to see if it holds true:
6(12) == 2
72 2
The equation is mathematically false for the scenario. In fact, if we use Sam's equation to find the
number of professors (P) when there are 12 students (S), it would suggest there are 72 professors.
This makes the smaller group (professors) even larger, which is the opposite of the intended
meaning.
Why the Logic is Flawed
Sam’s error stems from Literal Translation. He likely read the sentence from left to right and
matched the symbols to the words:
"Six times..." → 6
"...students..." → S
"...as/is..." → =
"...professors." → P
This creates 6S=P. In algebra, however, the coefficient (6) must be placed next to the smaller
quantity to bring it up to the size of the larger quantity to create an equality (a balance).
C ->
I
t
l
-1:t QUESTIONS Fltltt.Yi-ANSWERED
DUE: 21 ~PRIL 2D26
DISCLAIMER:
TRlnJDCUMENf,S DR-RE ERENq AND1iU1DANDCTURP SE
- DNL:Y, I DD NDTt TA E-RESPPNSIBll.;ln'~FDR-A~Y- Pl.AGIARISM, MISUS , DR
ACADEMIC MISCONDUCT RESUllTING FROM THEIR USE. IT IS YDURLJ_.., I-: : "_.,.,_
RESPONSIBILITY TQ ENSU E ORIGINAL TY lND COMPLIANCE WITL I
RELEVANT GUIDELINES DR STAND RD . t-~ ~ r
l
, PLEASE USE THIS DOCUMENT AS A GUIDE ONLY
Question 1
1.1.1. Conceptual Difference in the Role of k
The symbol k plays a fundamentally different role in each statement based on the mathematical
context:
In Statement A (5k−10=25): The symbol k acts as an unknown. In this context, the equation is
a "closed" sentence. There is a specific, fixed value that makes the statement true. The goal is
to "find" or solve for this missing value.
In Statement B (y=5k−10): The symbol k acts as a variable. In this functional relationship, k
does not have one single value; instead, it represents a range of possible values (the input). As
the value of k changes, the value of y changes accordingly, describing a continuous
relationship between two quantities.
1.1.2. 'Specific Value' vs. 'Domain of Input Values'
To determine which represents which, we can perform the calculations for both:
Statement A: 5k−10=25
To find the value of k:
Add 10 to both sides: 5k=35
Divide by 5: k=7
Classification: This represents a 'specific value'. The symbol k can only be 7 for the statement to be
mathematically valid.
Statement B: y=5k−10
If we attempt to solve for k:
1. Y + 10 -== 5k
2. k -:: 1t 1.lll
5
Classification: This represents a 'domain of input values'. Because k is dependent on y (and vice
versa), k can take on any value from a defined set (usually all real numbers), producing a
corresponding output for y. It defines a line on a coordinate plane rather than a single point on a
number line.
, Statement Role of k Representation
II_II
A: 5k - 10 = 25 Unknown Specific Value (k = 7)
======:II ~=1
B: y = 5k - 10 Variable Domain======:
of Input Values
1.2. Testing Sam's Logic with Numbers
To see why 6S=P is incorrect, let's use the hint and test the relationship with a concrete example.
1. The Goal:
The sentence states: "There are six times as many students as professors."
This means the group of students is much larger. If we have:
Professors (P) = 2
Students (S) should be 2×6=12
2. Testing Sam’s Equation (6S=P):
If we plug these numbers into Sam’s equation to see if it holds true:
6(12) == 2
72 2
The equation is mathematically false for the scenario. In fact, if we use Sam's equation to find the
number of professors (P) when there are 12 students (S), it would suggest there are 72 professors.
This makes the smaller group (professors) even larger, which is the opposite of the intended
meaning.
Why the Logic is Flawed
Sam’s error stems from Literal Translation. He likely read the sentence from left to right and
matched the symbols to the words:
"Six times..." → 6
"...students..." → S
"...as/is..." → =
"...professors." → P
This creates 6S=P. In algebra, however, the coefficient (6) must be placed next to the smaller
quantity to bring it up to the size of the larger quantity to create an equality (a balance).
