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kupik [55]
3 years ago
6

In isosceles triangle FGH, FG is congruent to GH. If Elijah draws a line segment from the vertex G to the midpoint of FH, labeli

ng the midpoint K, what can he prove about triangles FGK and HGK?
A= The triangle are neither congruent nor similar
B= The triangles are congruent, but not similar
C= The triangles are similar, but not congruent
D= The triangles are both congruent and similar

Mathematics
2 answers:
Crank3 years ago
7 0

Answer:

<h2>B) The triangles are congruent, but not similar.</h2>

Step-by-step explanation:

Givens

  • \triangle FGH is isosceles.
  • FG \cong GH.

In the image attached you can observe what Elijah did. The vertical line goes from vertex G to the midpoint K. When this happens, the resulting right triangles are congruents, that is, because we already know that FG is congruent to GH, GK is common for both triangles and the pair of angles in the base of the isosceles are congruent, by definiton.

Therefore, the triangles are congruent, but no similar. Choice B is correct.

Remember, when we talk about similarity, it refers to proportional corresponding sides of the triangles. When we talk about congruence, it refers to the exact equivalence between corresponding sides of both triangles. That's the difference. So, as you can deduct, a triangle cannot be similar and congruent at the same time.

miskamm [114]3 years ago
6 0
B= The triangles are congruent, but not similar
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With convolution theorem the equation is proved.

According to the statement

we have given that the equation and we have to evaluate with the convolution theorem.

Then for this purpose, we know that the

A convolution integral is an integral that expresses the amount of overlap of one function as it is shifted over another function.

And the given equation is solved with this given integral.

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\mathscr{L} \left( \int_{0}^{t} e^{-\tau} \cos \tau d \tau \right) = \frac{ \mathscr{L} (e^{-\tau} \cos \tau ) }{s} \\\mathscr{L} \left( \int_{0}^{t} e^{-\tau} \cos \tau d \tau \right) = \frac{\frac{s+1}{(s+1)^2+1}}{s} \\\mathscr{L} \left( \int_{0}^{t} e^{-\tau} \cos \tau d \tau \right) = \frac{1}{s}\left (\frac{s+1}{(s+1)^2+1} \right).

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