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Verizon [17]
3 years ago
9

Using Euler’s Formula (F + V = E + 2), find the number of vertices.

Mathematics
2 answers:
Usimov [2.4K]3 years ago
7 0
I don't know euler's formula but I can guess that the one you gave is enough

F=18
V=?
E=33
so

18+?=33+2
18+?=35
minus 18 from both siddes
?=17

vertices is 17
Art [367]3 years ago
4 0

Answer:

Vertices: 17

Step-by-step explanation:

Given the statement:

Here, E represents the number of Edges , F represents the number of  Faces and V represents the number of  vertices

then;

Edges(E) = 33

Faces(F) =  18

We have to find  number of vertices.

Using  Euler’s Formula

F + V = E + 2:

then;

18+V= 33+2

⇒18+V = 35

Subtract 18 from both sides we have;

V = 17

Therefore, the number of vertices =  17

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It is known that the straight line l is tangent to the circle x^2+y^2=4 at a point on the x-axis, and intersects with the straig
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Step-by-step explanation:

The general equation of a circle is

                                         (x \ - \ h)^2 \ + \ (y \ - \ k)^2 \ = \ r^{2},

where <em>h</em> and <em>k</em> forms the coordinates of the centre of the circle.

When the circle has a centre at the origin, the equation reduces into

                        .                            x^2 \ + \ y^2 \ = r^2.

Now, we are interested in solving for the <em>x</em>-intercepts (the <em>x</em>-coordinates when the circle intersects the <em>x</em>-axis), of the circle

                                                      x^2 \ + \ y^2 \ = \ 4 .

Thus,

                                                     x^2 \ + \ (0)^2 \ = \ 4 \\ \\ \\ \-\hspace{1.3cm} x^{2} \ = \ 4 \\ \\ \\ \-\hspace{1.4cm} x \ = \ \pm \ 2.

Geometrically speaking, the tangent to the circle at the point defined by one of the <em>x</em>-intercepts of the circle is actually a vertical line, more specifically the lines x \ = \ \pm \ 2.

First and foremost, for the vertical line x \ = \ 2, it intersects the straight line y \ = \ \displaystyle\frac{1}{2}x  , giving the y-coordinate for point P,

                                                    y \ = \ \displaystyle\frac{1}{2}(2) \\ \\ \\ y \ = \ 1.

Hence, the coordinates of point P are (1, \ 1).

However, since there are no boundaries given in the question and a circle is symmetrical about its centre, thus, point P also exists when the vertical line x \ = \ -2 and interdects the straight line y \ = \ \displaystyle\frac{1}{2}x.

                                                      y \ = \ \displaystyle\frac{1}{2}(-2) \\ \\ \\ y \ = \ -1.

Therefore, the coordinates of point P are also (1, \ -1).

8 0
2 years ago
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