Answer:
No, the reference point selected for the properties of a substance does't have any effect on thermodynamic analysis.
Step-by-step explanation:
Consider the provided information.
We need to determine that the reference point selected for the properties of a substance have any effect on thermodynamic analysis.
The answer for this question is No.
The reference point selected for the properties of a substance does't have any effect on thermodynamic analysis.
Because in thermodynamic analysis we are deal with the change in the properties.
If we are doing calculation, the reference state chosen is irrelevant as long as we use values in a single consistent array of tables and charts.
Hence, The reference point selected for the properties of a substance does't have any effect on thermodynamic analysis.
Perimeter (P) = 2L + x
4900 = 2L + x ⇒ 4900 - x = 2L ⇒ 
= L
Area (A) = L · x
A = () (x)
= 
Answer:
2
Step-by-step explanation:
The inequality is 8-1/4x>27. The solution of the inequality is b<-76.
Given that,
The inequality is 8-1/4x>27
We must determine how to address the inequity.
Take,
8-1/4x>27
Multiply the inequality's two sides by its lowest common denominator,
4×8-4×1/4b>27×4
Reduce the expression to the lowers term,
4×8-b>4×27
Calculate the product or quotient,
32-b>4×27
Calculate the product or quotient,
32-b>108
Rearrange unknown terms to the left side of the equation,
-b>108-32
Calculate the sum or difference,
-b>76
Divide the inequality's two sides by the variable's coefficient,
b<-76
Therefore, the solution of the inequality is b<-76.
To learn more about inequality visit: brainly.com/question/28823603
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y=2(3)x+1
−
y=2(3)x
y-y=2(3)x+1−2(3)x
the answer would be 0=1
but if it is asking u if it is false or true it is
0=1 is false, therefore the system of equations has no solutions
