Ask question

Ask question

All categories

4 years ago

7

On a line are given n segments each pair of which intersect. prove that there is a point that belongs to all n segments

1 answer:

disa [49]4 years ago

5 0

Each segment forms a new piece of the line L that continue forever. This line L is defined by y=mx+b by definition, using two points on any segment. If P is a point on L, then it lies on one of the segment but again these segments are L so P is on L iff it belonfs to all n segments. did u make this questionup

You might be interested in

Circle the largest decimal in each of the following groups.

ExtremeBDS [4]

A. 0.370

B. 0.44

C. 0.1

D. 2.75

7 0

3 years ago

Read 2 more answers

The slope of EF¯¯¯¯ is −5/2 .

stellarik [79]

Answer:

LM and NP

Step-by-step explanation:

LM= 2/5

NP=2/5

GH= -5/2

JK= =5/2

4 0

3 years ago

3. A big league pitching coach tries to limit his pitchers to 110 pitches per game. If the pitcher has

krek1111 [17]

110 minus 52 is 58 so the answer would be 58 more pitches

5 0

3 years ago

Find a 15% tip on a restaurant meal costing $50.

Tema [17]

The 15% tip would be 50*0.15=$7.5

5 0

3 years ago

Read 2 more answers

Find the distance between the points (2, 7) and (-6, -2).

maw [93]

Answer:

Approximately 12.04 units.

Step-by-step explanation:

The find the distance between any two points, we can use the distance formula, which is:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2$

We have the points (2,7) and (-6,-2). Let's let (2,7) be (x₁, y₁) and let's let (-6, -2) be (x₂, y₂). Substitute:

$d=\sqrt{(-6-2)^2+(-2-7)^2$

Subtract:

$d=\sqrt{(-8)^2+(-9)^2$

Square:

$d=\sqrt{64+81}$

Add:

$d=\sqrt{145}$

Approximate

$d=\sqrt{145}\approx12.04$

So, the distance between (2,7) and (-6,-2) is approximately 12.04 units.

And we're done!

7 0

4 years ago