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dmitriy555 [2]
2 years ago
8

Which step in the construction of copying a line segment ensures that the new line segment has

Mathematics
1 answer:
katovenus [111]2 years ago
5 0

Answer:

Here are the steps I think you probably will go through.

1. Draw a line that is longer than the segment but shorter than the width of the page.

1a. Make sure this line is  at least 1/2 inch from the left hand side.  

2. Use a compass to measure the length of the original segment. Never use a ruler. Rulers do not exist in pure geometry.

3. Measure out the distance on the line you just drew with the compass. One end is on the left hand edge and the other end (the pencil end) is marking the segment so it is the same length as the compass. You are done.

The key step either two or three.

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Which of the following is not one of the 8th roots of unity?
Anika [276]

Answer:

1+i

Step-by-step explanation:

To find the 8th roots of unity, you have to find the trigonometric form of unity.

1.  Since z=1=1+0\cdot i, then

Rez=1,\\ \\Im z=0

and

|z|=\sqrt{1^2+0^2}=1,\\ \\\\\cos\varphi =\dfrac{Rez}{|z|}=\dfrac{1}{1}=1,\\ \\\sin\varphi =\dfrac{Imz}{|z|}=\dfrac{0}{1}=0.

This gives you \varphi=0.

Thus,

z=1\cdot(\cos 0+i\sin 0).

2. The 8th roots can be calculated using following formula:

\sqrt[8]{z}=\{\sqrt[8]{|z|} (\cos\dfrac{\varphi+2\pi k}{8}+i\sin \dfrac{\varphi+2\pi k}{8}), k=0,\ 1,\dots,7\}.

Now

at k=0,  z_0=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 0}{8}+i\sin \dfrac{0+2\pi \cdot 0}{8})=1\cdot (1+0\cdot i)=1;

at k=1,  z_1=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 1}{8}+i\sin \dfrac{0+2\pi \cdot 1}{8})=1\cdot (\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};

at k=2,  z_2=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 2}{8}+i\sin \dfrac{0+2\pi \cdot 2}{8})=1\cdot (0+1\cdot i)=i;

at k=3,  z_3=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 3}{8}+i\sin \dfrac{0+2\pi \cdot 3}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};

at k=4,  z_4=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 4}{8}+i\sin \dfrac{0+2\pi \cdot 4}{8})=1\cdot (-1+0\cdot i)=-1;

at k=5,  z_5=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 5}{8}+i\sin \dfrac{0+2\pi \cdot 5}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};

at k=6,  z_6=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 6}{8}+i\sin \dfrac{0+2\pi \cdot 6}{8})=1\cdot (0-1\cdot i)=-i;

at k=7,  z_7=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 7}{8}+i\sin \dfrac{0+2\pi \cdot 7}{8})=1\cdot (\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};

The 8th roots are

\{1,\ \dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2},\ i, -\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2},\ -1, -\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2},\ -i,\ \dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2}\}.

Option C is icncorrect.

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