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What is Isabella's error?

A. Isabella should have divided 12 by 2 before squaring to get [tex][tex]$V=\frac{1}{3}(B)^2(9) \pi$[/tex][/tex].

B. Isabella should have squared 9 instead of 12 to get [tex][tex]$V=\frac{1}{3}(9)^2(12) \pi$[/tex][/tex].

C. Isabella should have divided the product of 12 and 9 by 3 to get [tex][tex]$V=\frac{12(9)}{3}$[/tex][/tex].

D. Isabella should have divided the product of [tex][tex]$12 \pi$[/tex][/tex] and 9 by 3 to get [tex][tex]$V=\frac{12 \pi(9)}{3}$[/tex][/tex].


Sagot :

To solve this problem, let's break down the options given and explain why the correct one involves dividing the product of 12 and 9 by 3.

1. Isabella should have divided 12 by 2 before squaring to get [tex]\( V = \frac{1}{3}(B)^2(9) \pi \)[/tex]:
- This option does not align with our problem since we are working with the numbers 12 and 9 directly and not with some intermediate division and squaring steps.

2. Isabella should have squared 9 instead of 12 to get [tex]\( V = \frac{1}{3}(9)^2(12) \pi \)[/tex]:
- Squaring 9 or 12 is not mentioned in the context of our original problem.

3. Isabella should have divided the product of 12 and 9 by 3 to get [tex]\( V = \frac{12(9)}{3} \)[/tex]:
- This aligns with the correct approach. To find the correct volume [tex]\( V \)[/tex], Isabella should multiply 12 and 9 first to get 108, and then divide by 3, resulting in 36.

4. Isabella should have divided the product of [tex]\( 12 \pi \)[/tex] and 9 by 3 to get [tex]\( V = \frac{12 \pi(9)}{3} \)[/tex]:
- This option unnecessarily complicates the formula as [tex]\(\pi\)[/tex] is not part of our original needs for 12 and 9.

Given this, the correct step-by-step solution to identify Isabella's error is:

1. Calculate the product of 12 and 9:
[tex]\[ 12 \times 9 = 108 \][/tex]

2. Divide the product by 3:
[tex]\[ \frac{108}{3} = 36 \][/tex]

Therefore, the correct statement is:
- Isabella should have divided the product of 12 and 9 by 3 to get [tex]\( V = \frac{12(9)}{3} \)[/tex].

By doing so, Isabella would obtain the correct volume as 36.