n < 0, is another way to say "n is negative", so let's check
![\bf ~~~~~~~~~~~~\textit{negative exponents} \\\\ a^{-n} \implies \cfrac{1}{a^n} \qquad \qquad \cfrac{1}{a^n}\implies a^{-n} \qquad \qquad a^n\implies \cfrac{1}{a^{-n}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ a^n~\hspace{10.5em}\stackrel{n = -n}{a^{-n}}\implies \cfrac{1}{a^n} \\\\\\ a^{-n}~\hspace{10em}\stackrel{n=-n}{a^{-(-n)}}\implies a^{+n}\implies a^n](https://tex.z-dn.net/?f=%20%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bnegative%20exponents%7D%0A%5C%5C%5C%5C%0Aa%5E%7B-n%7D%20%5Cimplies%20%5Ccfrac%7B1%7D%7Ba%5En%7D%0A%5Cqquad%20%5Cqquad%0A%5Ccfrac%7B1%7D%7Ba%5En%7D%5Cimplies%20a%5E%7B-n%7D%0A%5Cqquad%20%5Cqquad%0Aa%5En%5Cimplies%20%5Ccfrac%7B1%7D%7Ba%5E%7B-n%7D%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0Aa%5En~%5Chspace%7B10.5em%7D%5Cstackrel%7Bn%20%3D%20-n%7D%7Ba%5E%7B-n%7D%7D%5Cimplies%20%5Ccfrac%7B1%7D%7Ba%5En%7D%0A%5C%5C%5C%5C%5C%5C%0Aa%5E%7B-n%7D~%5Chspace%7B10em%7D%5Cstackrel%7Bn%3D-n%7D%7Ba%5E%7B-%28-n%29%7D%7D%5Cimplies%20a%5E%7B%2Bn%7D%5Cimplies%20a%5En%20)
4/10
Hope this helped ᕕ(ᐛ)ᕗ
Answer:
The set of transformations has been performed on triangle ABC is dilation by a scale factor of 2 followed by reflection about the x-axis
Step-by-step explanation:
Let us write the vertices of Δ ABC
∵ Vertex A = (-2, -1)
∵ Vertex B = (0, 0)
∵ Vertex C = (1, -3)
Let us Write the vertices of Δ A'B'C'
∵ Vertex A' = (-4, 2)
∵ Vertex B' = (0, 0)
∵ Vertex C' = (2, 6)
Let us find the scale factor k between the x-coordinates of the points and their images.
∵ -4 = k (-2) ⇒ Divide both side by -2
∵
= 
∴ 2 = k
∴ The scale factor of a dilation is 2
∵ The signs of the y-coordinates of the images are the opposite of
the signs of the y-coordinates of the points
∴ The triangle is reflected about the x-axis
∴ The set of transformations has been performed on triangle ABC
is dilation by a scale factor of 2 followed by reflection about the x-axis
Method 1.
Look at the picture 1.
We have:

Picture 2.

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Method 2. (Picture 3)
ΔLOM and ΔMON are isosceles triangles. Therefore the two angles opposite the legs are equal.
The sum of the internal angles in each triangle is 180°. Thereofre in ΔLMN we have:


Approximately 10 or x= ln(1024)/ln(2)