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Alona [7]
3 years ago
5

Is (1,2) a solution of 6x-y>3

Mathematics
1 answer:
tangare [24]3 years ago
3 0

Answer:

Yes

Step-by-step explanation:

To determine if (1, 2 ) is a solution of the inequality, substitute the coordinates into the inequality and if satisfied then they are a solution, that is

(6 × 1) - 2 = 6 - 2 = 4 > 3

Since 4 > 3 then (1, 2 ) is a solution

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Regular hexagon ABCDEF has vertices at A(4, 4!3), B(8, 4!3), C(10, 2!3), D(8, 0), E(4, 0) and F(2, 2!3).
marissa [1.9K]

Since the given hexagon is a regular hexagon all it's sides will be of equal length. Now, we know that the Area of any regular hexagon is given by:

A=\frac{3\sqrt{3}}{2} a^2

Where A is the area of the regular hexagon

a is the side length of the regular hexagon

Also, it's Perimeter is given by:

P=6a

Thus, all that we need to do is to find the side length of any one of the sides and to do that let us have a look at at the data of vertices points given and find out which points are definitely adjacent to each other and are also easy to calculate.

A quick search will yield that D(8, 0) and E(4, 0) are definitely adjacent to each other.

Please check the attached file here for a better understanding of the diagram of the original regular hexagon. Points D and E indeed are adjacent to each other.

Let us now find the distance between the points D and E using the distance formula which is as:

d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}

Where d is the distance.

(x_1,y_1) and (x_2,y_2) are the coordinates of points D and E respectively. (please note that interchanging the values of the coordinates will not alter the distance d)

Applying the above formula we get:

d=\sqrt{(8-4)^2+(0-0)^2} =\sqrt{4^2}=4

\therefore d=4

We know that this distance is the side length of the given regular hexagon.

\therefore d=a=4

Now, if the sides of the given regular polygon are reduced by 40%, then the new length of the sides will be:

a_{small}=4-\frac{40}{100}\times 4=2.4

Thus, the area of the smaller hexagon will be:

A_{small}=\frac{3\sqrt{3}}{2} a_{small}^2=\frac{3\sqrt{3}}{2} (2.4)^2\approx14.96 unit squared

and the new smaller perimeter will be:

P_{small}=6a_{small}=6\times 2.4=14.4 unit

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3 years ago
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