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SVETLANKA909090 [29]
3 years ago
15

A repeating pattern of figures that completely covers a plane, without gaps or overlaps, is called a _____ conjecture. transform

ation. tessellation. composition.
Mathematics
1 answer:
KonstantinChe [14]3 years ago
3 0

Answer:A repeating pattern of figures that completely covers a plane, without gaps or overlaps, is called a<u> tessellation</u>.

Step-by-step explanation:

  • A tessellation of a flat surface is the tiling of a plane using one or more geometric shapes are called tiles, without overlaps or gaps.
  • They can be generalized to higher dimensions and a variety of geometries.

Therefore, A repeating pattern of figures that completely covers a plane, without gaps or overlaps, is called a<u> tessellation</u>.

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Which equation represents an exponential function with an initial value of 500 ?
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Answer: f(x) = 500(2)^x

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The mathematics faculty at a college consists of 11 professors, 12 associate professors, 7 assistant professors, and 6 instructo
Nikolay [14]

Answer:

The probability of choosing a professor or an instructor is 47.22%.

Step-by-step explanation:

Given that the mathematics faculty at a college consists of 11 professors, 12 associate professors, 7 assistant professors, and 6 instructors, if a faculty member is selected, to find the probability of choosing a professor or an instructor the following calculation must be performed:

11 + 12 + 7 + 6 = 100

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3 years ago
Two points in the Cartesian plane are A(2.00 m, −4.00 m) and B(−3.00 m, 3.00 m). Find the distance between them and their polar
Brrunno [24]

The distance between the two points is d=8.6m

The polar coordinate of A is \left(4.47,296.57\right)

The polar coordinate of B is \left(4.24,135\right)

Explanation:

The two points are A(2,-4) and B(-3,3)

The distance between two points is given by,

d=\sqrt{(2+3)^{2}+(-4-3)^{2}}\\d=\sqrt{(5)^{2}+(-7)^{2}}\\d=\sqrt{25+49}\\d=8.6

Thus, the distance between the two points is d=8.6m

The polar coordinates of A can be written as (Distance, tan^{-1} \frac{y}{x} )

Distance = \sqrt{x^{2} +y^{2} }

Substituting A(2,-4), we get,

Distance = \sqrt{2^{2} +(-4)^{2} }=\sqrt{4+16 }=4.47

tan^{-1} \frac{y}{x} =tan^{-1} \frac{-4}{2}=-63.43

To make the angle positive, let us add 360,

\theta=360-63.43=296.57

The polar coordinate of A is \left(4.47,296.57\right)

Similarly, The polar coordinate of B can be written as (Distance, tan^{-1} \frac{y}{x} )

Distance = \sqrt{x^{2} +y^{2} }

Substituting B(-3,3), we get,

Distance = \sqrt{(3)^{2}+(3)^{2}}=4.24

tan^{-1} \frac{y}{x} =tan^{-1} \frac{3}{-3}=-45

To make the angle positive, let us add 360,

\theta=180-45=135^{\circ}

The polar coordinate of B is \left(4.24,135\right)

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