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2 years ago

Which undefined geometric term is described as a two-dimensional set of points that has no beginning or end?

Mathematics

2 answers:

Svetach [21]2 years ago

7 0

The answer is Line. Hope this helps :D

sertanlavr [38]2 years ago

3 0

There are three undefined terms in geometry .These words are point, line and plane.Let us look at these :

Point :we define a point as a location that has no length ,width or height.

Line:A line is defined as something that extends infinitely in either direction but has no width and is one dimensional.

Plane : A plane extends infinitely in two dimensions.

The geometric term described as a two-dimensional set of points that has no beginning or end is a line.

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Trigonometry Problem <br><br> Find the length of the missing side.

Alinara [238K]

Answer:

29.187

Step-by-step explanation:

Cos38°= base/hypotenuse

Cos38°= 23/w

w= 23/Cos38°

w= 23/0.788

w= 29.187

Hope it helps :-)

6 0

2 years ago

EXPERTS/ACE/TRUSTED HELPERS

snow_tiger [21]

the figure moved 7 to the right and 7 down

answer is C

5 0

2 years ago

Solve for C: 20 (greater than or equal too sign) C-11

podryga [215]

Answer:

C (is greater than it equal to) 31

Step-by-step explanation:

Isolate the c by adding 11 to both sides (cancels out the 11 and turns 20 into 31). Keep the sign the same!! Hope this helped! <33

5 0

1 year ago

Prove that the segments joining the midpoint of consecutive sides of an isosceles trapezoid form a rhombus.

sergiy2304 [10]

Answer:

See explanation

Step-by-step explanation:

a) To prove that DEFG is a rhombus, it is sufficient to prove that:

All the sides of the rhombus are congruent: $|DG|\cong |GF| \cong |EF| \cong |DE|$
The diagonals are perpendicular

Using the distance formula; $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$|DG|=\sqrt{(0-(-a-b))^2+(0-c)^2}$

$\implies |DG|=\sqrt{a^2+b^2+c^2+2ab}$

$|GF|=\sqrt{((a+b)-0)^2+(c-0)^2}$

$\implies |GF|=\sqrt{a^2+b^2+c^2+2ab}$

$|EF|=\sqrt{((a+b)-0)^2+(c-2c)^2}$

$\implies |EF|=\sqrt{a^2+b^2+c^2+2ab}$

$|DE|=\sqrt{(0-(-a-b))^2+(2c-c)^2}$

$\implies |DE|=\sqrt{a^2+b^2+c^2+2ab}$

Using the slope formula; $m=\frac{y_2-y_1}{x_2-x_1}$

The slope of EG is $m_{EG}=\frac{2c-0}{0-0}$

$\implies m_{EG}=\frac{2c}{0}$

The slope of EG is undefined hence it is a vertical line.

The slope of DF is $m_{DF}=\frac{c-c}{a+b-(-a-b)}$

$\implies m_{DF}=\frac{0}{2a+2b)}=0$

The slope of DF is zero, hence it is a horizontal line.

A horizontal line meets a vertical line at 90 degrees.

Conclusion:

Since $|DG|\cong |GF| \cong |EF| \cong |DE|$ and $DF \perp FG$ , DEFG is a rhombus

b) Using the slope formula:

The slope of DE is $m_{DE}=\frac{2c-c}{0-(-a-b)}$

$m_{DE}=\frac{c}{a+b)}$

The slope of FG is $m_{FG}=\frac{c-0}{a+b-0}$

$\implies m_{FG}=\frac{c}{a+b}$

5 0

2 years ago

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MissTica

Answer:

Step-by-step explanation:

7 0

2 years ago