All categories

3 years ago

A segment has endpoints V and X. What are two names for the segment?

Mathematics

1 answer:

wolverine [178]3 years ago

5 0

Given:

A segment has endpoints V and X.

To find:

The two names for the segment.

Solution:

A line segment is a small part of a line and it has two end points. The name of a segment is described by its end points.

For example: If A and B are two and points of a line segment, then the names for the segment are $\overline {AB}$ and $\overline {BA}$ .

It is given that the segment has endpoints V and X.

Therefore, the two names for the segment are $\overline {VX}$ and $\overline {XV}$ .

You might be interested in

3a-3/a+6=9/6 what is a

Gekata [30.6K]

Answer:

a=1/2

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Whichhhhh<br> oneeeeeeeeeeeee

Nostrana [21]

Answer:

Step-by-step explanation:

the slope of A is 1/2 which is greater than that of B

6 0

3 years ago

HELP ME WITH 61 <br>I'm freaking out!!!!!!!!!!!!<br>I don't know what to do

Brut [27]

4.25+1.25*m, 5.75+0.75*m
4.25+5.75=10 and 1.25+0.75 = 2 therefore the two expressions are 2m+10
0.5m-1.5 : 4.25+1.25+m-5.75+0.75*m =0.5m-1.5 ( I'm quite sure about this one but I can't think of another answer but this )

5 0

3 years ago

A lighthouse is located on a small island 2 km away from the nearest point P on a straight shoreline and its light makes four re

ddd [48]

Answer:

62.8 km per min

Step-by-step explanation:

Let the diagram of this situation is shown below,

In which $\theta$ is the angle made by light to the straight line joining the lighthouse and P,

∵ Light makes 4 revolutions per minute.

Also, 1 revolution = $2\pi$ radians

So, the change in angle with respect to t ( time ),

$\frac{d\theta}{dt}=8\pi\text{ radians per min}$

Let x be the distance of beam from P,

$\implies \tan \theta = \frac{x}{2}$

Differentiating with respect to x,

$\sec^2 \theta \frac{d\theta}{dt}=\frac{1}{2}\frac{dx}{dt}$

$(1+\tan^2 \theta) (8\pi) = \frac{1}{2}\frac{dx}{dt}$

$(1+\frac{x^2}{2^2})(8\pi) = \frac{1}{2}\frac{dx}{dt}$

If x = 1 km,

$(1+\frac{1}{4})8\pi =\frac{1}{2}\frac{dx}{dt}$

$\frac{5}{4}8\pi = \frac{1}{2}\frac{dx}{dt}$

$\implies \frac{dx}{dt}=\frac{80\pi}{4}= 20\pi\approx 62.8\text{ km per min}$

Hence, the beam of light moving along the shoreline with the speed of 62.8 km per min

7 0

4 years ago

How to solve 5(2b-3)-1(11b+1)=20

son4ous [18]

Answer:

b = -36

Step-by-step explanation:

Start by distributing both th 5 and the negative 1 into their corresponding parentheses.

((5*2b)-(5*3)) > (10b - 15)

((-1*11b)+(-1*1)) > (-11b - 1)

so you have:

10b - 15 - 11b - 1 = 20

Next you would Combine like terms to get:

-b - 16 = 20

Add the 16 to both sides

-b = 36

and divide each side by -1

b = - 36

hope this helps

3 0

3 years ago