All categories

3 years ago

In hyperbolic geometry, a triangle could potentially have ___ degrees.

Mathematics

2 answers:

ludmilkaskok [199]3 years ago

7 0

Answer:

In hyperbolic geometry, a triangle could potentially have less than 180 degrees.

Step-by-step explanation:

Hyperbolic geometry is the geometry, where we consider all postulates of Euclid except for the fifth one. We take the fifth postulate's negative form.

So, in this geometry, the angle sum of a triangle is less than 180°.

asambeis [7]3 years ago

6 0

It could possibly be 180° or less

You might be interested in

What is expanded form for 4,408,730?

olganol [36]

4,000,000
400,000
00,000
8,000
700
30
0

7 0

3 years ago

I need help with my homework I’m struggling and don’t understand the equation

Elodia [21]

= -2

y= 2x+6. y= -2x-2
2x+6= -2x-2
+2x. +2x
____________
4x+6= -2
4x= -8
x=-2

5 0

2 years ago

What did the butcher say to the piece of meat

katrin2010 [14]

Do the daaaaaMn work yourself
.
.
.
.
.
.
.
.
.
.
.
.
.

I said do it yourselff

5 0

2 years ago

I ONLY NEED HELP ON NUMBER 2

REY [17]

So, the area of a circle is a=pi r^2
so do that and get a=pi(4)
and since the circle touches the edge of the rectangle, the rectangle is 4 cm tall, so 4*8 is 32 and pi(4) is about 12.56, so 32-12.56 is 19.44, and I believe that is the answer. (19.4)

4 0

2 years ago

Read 2 more answers

An NFL coach sometimes uses a defense that utilizes 3 defensive linemen, 4 linebackers, and 4 defensive backs. His roster (the p

Alexandra [31]

Answer:

68,600

Step-by-step explanation:

The order of the players is not important. For example, a defensive line of Shaq Lawson, Ed Oliver and Jerry Hughes is the same as a defensive line of Ed Oliver, Shaq Lawson and Jerry Hughes. So we use the combinations formula to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

Defensive Lineman:

3 from a set of 8. So

$C_{8,3} = \frac{8!}{3!5!} = 56$

56 combinations of defensive lineman

Linebackers:

4 from a set of 7. So

$C_{7,4} = \frac{7!}{4!3!} = 35$

35 combinations of linebackers

Defensive backs:

4 from a set of 7. So

$C_{7,4} = \frac{7!}{4!3!} = 35$

35 combinations of defensive backs

How many different ways can the coach pick the 11 players to implement this particular defense?

56*35*35 = 68,600

68,600 different ways can the coach pick the 11 players to implement this particular defense

7 0

3 years ago