<em>Hexagonal</em> implies that the <em>traffic sign</em> has six straight sides. But <u>dilating</u> it changes its <em>length</em> of sides, but not its angles. So that the <em>statement</em> that is <u>true</u> is option B.
i.e B. The <u>dilated</u> sign will have corresponding line segments that are <em>proportional</em> to those of the pre-image and <u>corresponding</u> angles that are congruent to that of the pre-image.
Dilation involves <u>increasing</u> or <u>decreasing</u> the line segments of a given shape by a <em>scale factor</em>. This process do not affect the measure of the internal <u>angles</u> of the shape.
Given a <u>hexagonal</u> traffic sign in the question, this implies that the traffic sign has six sides. <u>Dilating</u> it by a scale factor of 2.5 about its center would affect its <em>length</em> of sides. So that the <em>statement</em> that is <u>true</u> is option B.
i.e B. The dilated sign will have <u>corresponding</u> line segments that are proportional to those of the pre-image and <em>corresponding</em> angles that are congruent to that of the pre-image.
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Here we are given the expression:
![x^{2}+18](https://tex.z-dn.net/?f=x%5E%7B2%7D%2B18)
Now let us equate it to zero to find x first,
![x^{2}+18=0](https://tex.z-dn.net/?f=x%5E%7B2%7D%2B18%3D0)
Now subtracting 18 from the other side,
![x^{2}=-18](https://tex.z-dn.net/?f=x%5E%7B2%7D%3D-18)
taking square root on both sides,
So we will get two values of x as ,
![x=3\sqrt{-2}](https://tex.z-dn.net/?f=x%3D3%5Csqrt%7B-2%7D)
![x=-3\sqrt{-2}](https://tex.z-dn.net/?f=x%3D-3%5Csqrt%7B-2%7D)
Now we can write square root -1 as i,
So our factors become,
![x=3i\sqrt{2}](https://tex.z-dn.net/?f=x%3D3i%5Csqrt%7B2%7D)
![x=-3i\sqrt{2}](https://tex.z-dn.net/?f=x%3D-3i%5Csqrt%7B2%7D)
Answer:
The final factored form becomes,
![(x+3i\sqrt{2})(x-3i\sqrt{2})](https://tex.z-dn.net/?f=%28x%2B3i%5Csqrt%7B2%7D%29%28x-3i%5Csqrt%7B2%7D%29)
I think d but I’m not sure
Answer:
Step-by-step explanation:
(2u+3u)(u+v)-2u+3v
=2u(u+v)+3u(u+v)-2u+3v
=2u^2+2uv+3u^2+3uv-2u+3v
=5u^2+5uv-2u+3v