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Vera_Pavlovna [14]
2 years ago
8

"which of the numbers below can be classified as an integer" and the options are . 10÷2 , -0.3 ,√20 ,. 0.4​

Mathematics
1 answer:
Ivahew [28]2 years ago
4 0

Answer:

The number 10/2 = 5 is an integer.

Step-by-step explanation:

An integer is a whole number which cannot be expressed in fractions. For example, 1, 2, 3...

Integer can be positive,  negative of zero.

\frac{10}{2}=5

It is a whole number so it is an integer.

- 0.3

It is not an integer as it has a fractional part.

\sqrt20

It also has a fractional part, so it is not an integer.

0.4

It also has a fractional part, so it is not an integer.

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  c.  quadrilateral

Step-by-step explanation:

All of the sides are different lengths, so the quadrilateral cannot be a parallelogram, rhombus, or square.

Its best descriptor is <em>parallelogram</em>.

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A <em>parallelogram</em> has opposite sides parallel and congruent. A <em>rhombus</em> also has adjacent sides congruent. A <em>square</em> is a special case of rhombus in which the corner angles are right angles.

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Which series of transformations will not map figure H onto itself
7nadin3 [17]

Answer:

D

Step-by-step explanation:

Given a square with vertices at points (2,1), (1,2), (2,3) and (3,2).

Consider option A.

1st transformation (x+0,y-2) will map vertices of the square into points

  • (2,1)\rightarrow (2,-1);
  • (1,2)\rightarrow (1,0);
  • (2,3)\rightarrow (2,1);
  • (3,2)\rightarrow (3,0).

2nd transformation = reflection over y = 1 has the rule (x,2-y). So,

  • (2,-1)\rightarrow (2,3);
  • (1,0)\rightarrow (1,2);
  • (2,1)\rightarrow (2,1);
  • (3,0)\rightarrow (3,2)

These points are exactly the vertices of the initial square.

Consider option B.

1st transformation (x+2,y-0) will map vertices of the square into points

  • (2,1)\rightarrow (4,1);
  • (1,2)\rightarrow (3,2);
  • (2,3)\rightarrow (4,3);
  • (3,2)\rightarrow (5,2).

2nd transformation = reflection over x = 3 has the rule (6-x,y). So,

  • (4,1)\rightarrow (2,1);
  • (3,2)\rightarrow (3,2);
  • (4,3)\rightarrow (2,3);
  • (5,2)\rightarrow (1,2)

These points are exactly the vertices of the initial square.

Consider option C.

1st transformation (x+3,y+3) will map vertices of the square into points

  • (2,1)\rightarrow (5,4);
  • (1,2)\rightarrow (4,5);
  • (2,3)\rightarrow (5,6);
  • (3,2)\rightarrow (6,5).

2nd transformation = reflection over y = -x + 7 will map vertices into points

  • (5,4)\rightarrow (3,2);
  • (4,5)\rightarrow (2,3);
  • (5,6)\rightarrow (1,2);
  • (6,5)\rightarrow (2,1)

These points are exactly the vertices of the initial square.

Consider option D.

1st transformation (x-3,y-3) will map vertices of the square into points

  • (2,1)\rightarrow (-1,-2);
  • (1,2)\rightarrow (-2,-1);
  • (2,3)\rightarrow (-1,0);
  • (3,2)\rightarrow (0,-1).

2nd transformation = reflection over y = -x + 2 will map vertices into points

  • (-1,-2)\rightarrow (4,3);
  • (-2,-1)\rightarrow (3,4);
  • (-1,0)\rightarrow (2,3);
  • (0,-1)\rightarrow (3,2)

These points are not the vertices of the initial square.

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