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Akimi4 [234]
3 years ago
9

What kinds of symmetry does the capital letter E have

Mathematics
2 answers:
Maslowich3 years ago
4 0
It has reflectional symmetry where the line of reflection (mirror line) is horizontal and it cuts through the middle of the letter
marusya05 [52]3 years ago
3 0
Answer[:]
The capital letter E has three kinds of symmetry(s):

1) Horizontal Line of Symmetry
2) One line of symmetry (NOT meaning they only have one type of symmetry, they have one LINE of symmetry)
3) Reflectional Symmetry (If you were to flip it on it's side, and cut it in half, then both sides would be the same)


[:] DustinBR [:]
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Find the components of the vertical force Bold Upper Fequalsleft angle 0 comma negative 8 right anglein the directions parallel
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Answer with Step-by-step explanation:

We are given that

F=<0,-8>=0i-8j=-8j

\theta=\frac{\pi}{3}

The component of force is divided into two direction

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2.Perpendicular to the plane

1.The vector parallel to the plane will be=r=cos\frac{\pi}{3}i-sin\frac{\pi}{3}j=\frac{1}{2}i-\frac{\sqrt 3}{2}j

By using cos\frac{\pi}{3}=\frac{1}{2},sin\frac{\pi}{3}=\frac{\sqrt 3}{2}

Force along the plane will be=\mid F_x\mid=F\cdot r

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By using i\cdot i=j\cdoty j=k\cdot k=1,i\cdot j=j\cdot k=k\cdot i=j\cdot i=k\cdot j=i\cdot k=0

Therefore, force along the plane=\mid F_x\mid(\frac{1}{2}i-\frac{\sqrt 3}{2}j)=4\sqrt 3(\frac{1}{2}i-\frac{\sqrt 3}{2}j)

2.The vector perpendicular to the plane=r=-sin\frac{\pi}{3}-cos\frac{\pi}{3}=-\frac{\sqrt 3}{2}i-\frac{1}{2}j

The force perpendicular to the plane=\mid F_y\mid=F\cdot r=-8j(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)

The force perpendicular to the plane=4N

Therefore, F_y=4(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)

Sum of two component of force=F_x+F_y=4\sqrt 3(\frac{1}{2}i-\frac{\sqrt 3}{2}j)+4(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)

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