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a is a positive number and 1/a^2
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21 Apr 2025, 03:24
21 Apr 2025, 03:24
Expert Reply
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Question Stats:
0% (00:00) correct
100% (00:58) wrong based on 1 sessions
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a is a positive number
Quantity A |
Quantity B |
\(\frac{1}{a^2} \left[ \begin{array}{cc|r}[a-d][a][a+d]+a d^2\end{array} \right]\) |
\(a\) |
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Re: a is a positive number and 1/a^2
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25 Apr 2025, 04:30
25 Apr 2025, 04:30
Expert Reply
The value of Column
$$
\(\begin{aligned}
\mathrm{A} & =\frac{1}{a^2}\left[a\left(a^2-d^2\right)+a d^2\right] \\
& =\frac{1}{a^2}\left[a^3-a d^2+a d^2\right]=a
\end{aligned}
$$\)
and the value of Column $\(\mathrm{B}=a\)$.
Therefore, Column $\(\mathrm{A}=$ Column $B\)$.
Thus, the correct option is $C$.
$$
\(\begin{aligned}
\mathrm{A} & =\frac{1}{a^2}\left[a\left(a^2-d^2\right)+a d^2\right] \\
& =\frac{1}{a^2}\left[a^3-a d^2+a d^2\right]=a
\end{aligned}
$$\)
and the value of Column $\(\mathrm{B}=a\)$.
Therefore, Column $\(\mathrm{A}=$ Column $B\)$.
Thus, the correct option is $C$.
Re: a is a positive number and 1/a^2
[#permalink]
02 May 2025, 10:47
02 May 2025, 10:47
1
i guess there is something wrong with how the brackets have been put
