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maria [59]
2 years ago
14

WILL CHOOSE BRAINLIEST!!! Triangle FGH has vertices F(−2, 0), G(0, 2), and H(2, 0). A dilation, centered at the origin, is appli

ed to this triangle. The image has vertices F′(−14, 0), G′(0, 14), and H′(14, 0).
What is the scale factor of this dilation?

Enter your answer in the box.
Mathematics
2 answers:
kenny6666 [7]2 years ago
5 0

Answer:

Its 3

Step-by-step explanation:

Salsk061 [2.6K]2 years ago
3 0
I think it is 7 because it is multiplied by 7
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How to write standard form for: (please answer this question soon )
professor190 [17]
Put the terms in order of decreasing degree, all on the left. When convenient, it is nice to have mutually prime integer coefficients with the leading one positive.

a) x² -2x +1 = 0

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or, better, divide out the common factor of 4.
  x² -3 = 0

d) 3x² -x -5 = 0

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6 0
3 years ago
Find e^cos(2+3i) as a complex number expressed in Cartesian form.
ozzi

Answer:

The complex number e^{\cos(2+31)} = \exp(\cos(2+3i)) has Cartesian form

\exp\left(\cosh 3\cos 2\right)\cos(\sinh 3\sin 2)-i\exp\left(\cosh 3\cos 2\right)\sin(\sinh 3\sin 2).

Step-by-step explanation:

First, we need to recall the definition of \cos z when z is a complex number:

\cos z = \cos(x+iy) = \frac{e^{iz}+e^{-iz}}{2}.

Then,

\cos(2+3i) = \frac{e^{i(2+31)} + e^{-i(2+31)}}{2} = \frac{e^{2i-3}+e^{-2i+3}}{2}. (I)

Now, recall the definition of the complex exponential:

e^{z}=e^{x+iy} = e^x(\cos y +i\sin y).

So,

e^{2i-3} = e^{-3}(\cos 2+i\sin 2)

e^{-2i+3} = e^{3}(\cos 2-i\sin 2) (we use that \sin(-y)=-\sin y).

Thus,

e^{2i-3}+e^{-2i+3} = e^{-3}\cos 2+ie^{-3}\sin 2 + e^{3}\cos 2-ie^{3}\sin 2)

Now we group conveniently in the above expression:

e^{2i-3}+e^{-2i+3} = (e^{-3}+e^{3})\cos 2 + i(e^{-3}-e^{3})\sin 2.

Now, substituting this equality in (I) we get

\cos(2+3i) = \frac{e^{-3}+e^{3}}{2}\cos 2 -i\frac{e^{3}-e^{-3}}{2}\sin 2 = \cosh 3\cos 2-i\sinh 3\sin 2.

Thus,

\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2-i\sinh 3\sin 2\right)

\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2\right)\left[ \cos(\sinh 3\sin 2)-i\sin(\sinh 3\sin 2)\right].

5 0
3 years ago
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