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The formula A=6(s+2)2 can be used to find the surface area of a cube whose side length is s + 2 units. Nathan made an error when solving the formula for s. In which step did his error occur?

A: A6=(s+2)2

B: ±A6−−√=s+4

C: ±A√6√=s+4

D: ±6A√6=s+4

E: −4±6A√6=s

Respuesta :

bawbdmo

In the second step he should have had s+2 again rather than 2+4. Other than that, it's fine. A corrected version would look like this:

[tex] A: \frac{A}{6}=(s+2)^2 \implies\\
B: \pm\sqrt{\frac{A}{6}}=s+2 \implies\\
C: \pm\frac{\sqrt{A}}{\sqrt{6}}=s+2\implies \\
D:\pm\frac{\sqrt{6A}}{6}=s+2 \implies\\
E:-2\pm\frac{\sqrt{6A}}{6}=s [/tex]

The plus or minus is irrelevant, you cannot have a negative area or volume, but I don't think that's the problem.

Answer:

Nathan made a mistake in step B.

Step-by-step explanation:

Given formula:

[tex]A=6(s+2)^2[/tex]

The correct steps to find the surface area of a cube whose side length is s + 2 units are

Step A : Divide both sides by 6.

[tex]\frac{A}{6}=(s+2)^2[/tex]

Step B : Taking square root both sides.

[tex]\pm \sqrt{\frac{A}{6}}=s+2[/tex]

Step C : Distribute square root.

[tex]\pm \frac{\sqrt{A}}{\sqrt{6}}=s+2[/tex]

Step D : Rationalize denominator.

[tex]\pm \frac{\sqrt{6A}}{6}=s+2[/tex]

Step E : Subtract 2 from both sides.

[tex]-2\pm \frac{\sqrt{6A}}{6}=s[/tex]

Therefore, Nathan made a mistake in step B.

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