What fraction is 1 degree?

Which step is the same when constructing an inscribed regular hexagon and an inscribed equilateral triangle?.

Naddik [55]

Step is the same when constructing an inscribed regular hexagon and an inscribed equilateral triangle is shown below:

<h3 /><h3>What is construction?</h3>

Geometric construction is the process of drawing a geometrical figure using two geometrical instruments, a compass, and a ruler.

Construction of an inscribed regular hexagon

1) Set compass width to the radius of the circle

2) Make an arc on the circle with the set radius.

3) From the previous arc draw another arc with the same radius.

4) Continue this process for the other vertex.

5) Connect all the vertices.

Construction of an inscribed equilateral triangle

1) Set compass width to the radius of the circle

2) Make an arc on the circle with the set radius.

3)From the previous arc draw another arc with the same radius.

4)Continue this process for the other vertex.

5) Connect the every other vertices point by drawing a line between every other adjacent vertex point to form an inscribed equilateral triangle.

Learn more about construction here:

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7 0

1 year ago

write in scientific notation 1. 0.0000364 2. 0.00751 3. 0.10005 4. 1,094 5. 0.00000099 6. 0.04101 7. 10,500 8. 8,9000

solmaris [256]

$1.\\0\underbrace{.00003}_{5\to}54=3.54\times10^{-5}\\2.\\0\underbrace{.007}_{3\to}51=7.51\times10^{-3}\\3.\\0\underbrace{.1}_{1\to}0005=1.0005\times10^{-1}\\4.\\1\underbrace{,094}_{\leftarrow3}=1.094\times10^3\\5.\\0\underbrace{.0000009}_{7\to}9=9.9\times10^{-7}$
$6.\\0\underbrace{.04}_{2\to}101=4.101\times10^{-2}\\7.\\1\underbrace{0,500}_{\leftarrow4}=1.05\times10^4\\8.\\8\underbrace{9000}_{\leftarrow4}=8.9\times10^4$

3 0

2 years ago

Suppose sarah is trying to design a metal box (a rectangular prism), which will hold 450 cubic centimeters of liquid. the length

schepotkina [342]

<span>So, the volume of a container can also be thought of as the amount of liquid that the container is capable of holding. It follows that 450=(length)(width)(height). The constraints of the problem say that the length of the box must be twice as large as the width, so it follows that j(height)=(2w)(w)(height)-450. This then implies that j(height)=(2w^2)(height)-450. In simpler terms, the height is defined as 450/(2w^2), or j(w)=450/(2w^2).</span>

5 0

2 years ago

Is the answer 3, 4, 5, or 6? need answer asap please! :)

-BARSIC- [3]

Don't touch the center. It is already even.
Start anywhere by connecting a dotted line from one vertex to the next. To keep things so we know what we are talking about, go clockwise. Now you have 2 points that are Eulerized that were not before.

Skip and edge and do the same thing to the next two vertices. Those two become eulerized. Skip an edge and do the last 2.

Let's try to describe this better. Start at any vertex and number them 1 to 6 clockwise.
Join 1 to 2
Join 3 to 4
Join 5 to 6

I think 3 is the minimum.
3 <<<< answer

4 0

2 years ago

Can someone help me with this?

Lerok [7]

Answer:

1 is C and 2 is B

Step-by-step explanation:

4 0

2 years ago