The table below shows the measurements (in inches) from cubes with different side lengths
Check all that apply
![The table below shows the measurements in inches from cubes with different side lengths Check all that apply class=](https://us-static.z-dn.net/files/dea/715e59d33cbbfa23014661651bad9918.png)
The answers to your question is,
Side length and perimeter of 1 face AND area of 1 face and surface area.
-Mabel <3
Answer:
1st and 4th Option are correct.
Step-by-step explanation:
Given:
A table with multiple variables.
To find: Linear relationship between variables.
First, between side length and perimeter of 1 face.
Consider, [tex]\frac{side\:length}{Perimeter}=\frac{3}{12}=\frac{1}{4}[/tex]
[tex]\frac{4}{16}=\frac{1}{4}[/tex]
[tex]\frac{5}{20}=\frac{1}{4}[/tex]
[tex]\frac{6}{24}=\frac{1}{4}[/tex]
Since all the ratio are equal.
Thus, Both Variables are in Linear Relationship.
Second, between Perimeter of 1 face and Area of 1 face.
[tex]\frac{Perimeter}{Area}=\frac{12}{9}=\frac{4}{3}[/tex]
[tex]\frac{16}{16}=1[/tex]
Since first two ratio are not equal.
Thus, Both Variables are not in Linear Relationship.
Third, between Surface Area and Volume
[tex]\frac{Surface\:Area}{Volume}=\frac{54}{27}=\frac{2}{1}[/tex]
[tex]\frac{96}{64}=\frac{3}{2}[/tex]
Since first two ratio are not equal.
Thus, Both Variables are not in Linear Relationship.
Fourth, between Area of 1 face. and Surface Area
[tex]\frac{Area}{Surface\:Area}=\frac{9}{54}=\frac{1}{6}[/tex]
[tex]\frac{16}{96}=\frac{1}{6}[/tex]
[tex]\frac{25}{150}=\frac{1}{6}[/tex]
[tex]\frac{36}{216}=\frac{1}{6}[/tex]
Since all the ratio are equal.
Thus, Both Variables are in Linear Relationship.
Fifth, between Side Length and Volume
[tex]\frac{Side\:length}{Volume}=\frac{3}{27}=\frac{1}{9}[/tex]
[tex]\frac{4}{64}=\frac{1}{16}[/tex]
Since first two ratio are not equal.
Thus, Both Variables are not in Linear Relationship.
Therefore, 1st and 4th Option are correct.