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Damm [24]
3 years ago
12

Help please?? Answer:

Mathematics
1 answer:
Sveta_85 [38]3 years ago
3 0
1. Yes, a regular hexagon can be drawn using rotations.
2.To find the answer, first find the number of sides of a hexagon. A hexagon has six sides.
Divide 360 by 6 = 360/6=60 degrees.
So the angle of rotation for a point on the circle for drawing a regular hexagon is 60 degrees.


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5 0
3 years ago
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How do you create a restriction for √z + 8=13 ?
Zinaida [17]
√z = 13 - 8
√z = 5
z = 5²
z = 25
In this case the z can be any number, however z cannot be less than 0 and that is the restriction. (for example; √-1 = irrational number, cannot be solved.)
6 0
3 years ago
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4) This results when you flip the numerator and denominator of a fraction.
Semenov [28]

Answer:

To divide one fraction by another one, flip numerator and denominator of the second one, and then multiply the two fractions. The flipped-over fraction is called the multiplicative inverse or reciprocal.

Step-by-step explanation:

I got this from google.

3 0
4 years ago
Consider an experiment with sample space S 5 50, 1, 2, 3, 4, 5, 6, 7, 8, 96 and the events A 5 {0, 2, 4, 6, 8} B 5 {1, 3, 5, 7,
Brut [27]

Answer with Step-by-step explanation:

S={0,1,2,3,4,5,6,7,8,9}

A={0,2,4,6,8}

B={1,3,5,7,9}

C={0,1,2,3,4}

D={5,6,7,8,9}

a.A'=S-A

A'={0,1,2,3,4,5,6,7,8,9}-{0,2,4,6,8}

A'={1,3,5,7,9}

b.C'=S-C

C'={0,1,2,3,4,5,6,7,8,9}-{0,1,2,3,4}

C'={5,6,7,8,9}

c.D'=S-D

D'={0,1,2,3,4,5,6,7,8,9}-{5,6,7,8,9}

D'={0,1,2,3,4}

d.A\cup B={0,2,4,6,8}\cup{1,3,5,7,9}

A\cup B={0,1,2,3,4,5,6,7,8,9}=S

e.A\cup C={0,2,4,6,8}\cup{0,1,2,3,4}

A\cup C={0,1,2,3,4,6,8}

f.A\cup D={0,2,4,6,8}\cup{5,6,7,8,9}

A\cup D={0,2,4,5,6,7,8,9}

8 0
3 years ago
Determine the line described by the given point and slope (0,0) and 2/3
AlekseyPX

Answer:

The  line is  y=\frac{2}{3} x.

Step-by-step explanation:

Given:

Point and slope (0,0) and 2/3.

Now, to determine the line.

Here the point is:

(x_{1},y_{1}) =(0,0)

And the slope is:

m=\frac{2}{3}

Now, putting the formula and substituting the value from above to determine the line:

y-y_{1} = m(x-x_{1})

y-0=\frac{2}{3}(x-0)

y=\frac{2}{3} x

Therefore, the  line is y=\frac{2}{3} x.

4 0
3 years ago
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