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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=

Respuesta :

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.

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