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3 years ago

7

Which sequence of transformations on △ABC will result in a figure that is congruent to △ABC?

2 answers:

ryzh [129]3 years ago

6 0

Any sequence of transformations that include a “dilation” changes the overall size of the shape, so your answer would actually be “B” since it’s only being rotated and reflected. In other words, the location of the shape is the only thing being changed.

solong [7]3 years ago

3 0

Answer: A.) c is -1 -2 b is -4 -2 a is -3 -3

Step-by-step explanation: sry but thats all i know i hoped i helped a lil bit :)

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Which student correctly wrote the algebraic expression for the following statement?

Leno4ka [110]

Haileys work is correct so B should be the right answer

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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

Point M is the midpoint of segment AB. Find the X Coordinate of the midpoint M if A (5.1,4.6) and B (7.9,2.5). Round to the near

JulsSmile [24]

Answer:

6.5

Step-by-step explanation:

5.1+7.9=13

13 divided by 2

= 6.5

4 0

3 years ago

Solve and reduce to lowest terms 4\7 x 5\8

erastovalidia [21]

Answer:

5/14

Step-by-step explanation:

4/7 x 5/8 = 20 /56

Then divide by 4

20/56 x 4/4 = 5 /14

8 0

2 years ago