For the triangles shown, we can express their congruence with the statement $\mathrm{\u25b3}ABC\stackrel{\sim}{=}\mathrm{\u25b3}FED$ . By reordering the vertices, express this congruence with a different statement.

iohanetc
2020-12-09
Answered

For the triangles shown, we can express their congruence with the statement $\mathrm{\u25b3}ABC\stackrel{\sim}{=}\mathrm{\u25b3}FED$ . By reordering the vertices, express this congruence with a different statement.

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2abehn

Answered 2020-12-10
Author has **88** answers

Step 1

Given- The triangles ABC and FED are congruence with$\mathrm{\u25b3}ABC\stackrel{\sim}{=}\mathrm{\u25b3}FED$ from the given figure,

To express- The congruence with a different statement by reordering the vertices.

Concept Used- Two triangles are said to be congruent if the three sides and the three angles of both the angles are equal in any orientation.

Step 2

Explanation- As we know that two triangles are said to be congruent if the three sides and the three angles of both the angles are equal in any orientation.

Now, from the figure, as we see that the corresponding sides and corresponding angles are equal, so using the congruence law , we can write in correct order as,

$\mathrm{\u25b3}ABC\stackrel{\sim}{=}FED$ .

So, the correct order of congruence can be written as$\mathrm{\u25b3}ABC\stackrel{\sim}{=}FED$ .

Answer- Hence, the congruence with a different statement by reordering the vertices can be writtena as$\mathrm{\u25b3}ABC\stackrel{\sim}{=}FED$ .

Given- The triangles ABC and FED are congruence with

To express- The congruence with a different statement by reordering the vertices.

Concept Used- Two triangles are said to be congruent if the three sides and the three angles of both the angles are equal in any orientation.

Step 2

Explanation- As we know that two triangles are said to be congruent if the three sides and the three angles of both the angles are equal in any orientation.

Now, from the figure, as we see that the corresponding sides and corresponding angles are equal, so using the congruence law , we can write in correct order as,

So, the correct order of congruence can be written as

Answer- Hence, the congruence with a different statement by reordering the vertices can be writtena as

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I'd like to compute

${A}^{\ast}=\underset{{\scriptstyle \begin{array}{c}A\in {\mathbb{R}}^{d\times k}\\ {A}^{T}A=I\end{array}}}{\mathrm{a}\mathrm{r}\mathrm{g}\phantom{\rule{thinmathspace}{0ex}}\mathrm{m}\mathrm{a}\mathrm{x}}det({A}^{T}\mathrm{\Lambda}A)$

where $k\le d$, $\mathrm{\Lambda}=\mathrm{diag}({\lambda}_{1},\dots ,{\lambda}_{d})$ and ${\lambda}_{1}\ge \cdots \ge {\lambda}_{d}\ge 0$.

I strongly suspect that ${A}^{\ast}=[{\u03f5}_{1}\cdots {\u03f5}_{k}]$, but after various attempts to prove it I gave up.

${A}^{\ast}=\underset{{\scriptstyle \begin{array}{c}A\in {\mathbb{R}}^{d\times k}\\ {A}^{T}A=I\end{array}}}{\mathrm{a}\mathrm{r}\mathrm{g}\phantom{\rule{thinmathspace}{0ex}}\mathrm{m}\mathrm{a}\mathrm{x}}det({A}^{T}\mathrm{\Lambda}A)$

where $k\le d$, $\mathrm{\Lambda}=\mathrm{diag}({\lambda}_{1},\dots ,{\lambda}_{d})$ and ${\lambda}_{1}\ge \cdots \ge {\lambda}_{d}\ge 0$.

I strongly suspect that ${A}^{\ast}=[{\u03f5}_{1}\cdots {\u03f5}_{k}]$, but after various attempts to prove it I gave up.

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2) Is the same principle allow us to maximize /minimize $\mathrm{log}g(x)$ instead of $g(x)$ itself?

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