The lengths of three sides of a triangle are m units, n units, and p units, respectively. Which inequality must be true?

Mathematics

2 answers:

Bogdan [553]2 years ago

4 0

Answer:

n < m + p

Step-by-step explanation:

If ABC is a triangle then the sum of any two sides of ABC will be greater than the third side i.e. AB + BC > CA or BC + CA > AB or CA + AB > BC.

Now, if the lengths of three sides of a triangle are m units, n units, and p units respectively.

Then the inequality must be true is n < m + p, where m + p is the sum of any two sides which is greater than the third side of n length. (Answer)

alexdok [17]2 years ago

3 0

The Triangle Inequality Theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. Therefore, the inequality that must be true is n < m + p.

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Find the perimeter of triangle ABC.

Harlamova29_29 [7]

By definition, the perimeter of the triangle is the sum of its sides.
We must then use the formula of distance between points:
$d = \sqrt{(x2-x1) ^ 2 + (y2-y1) ^ 2}$
We now look for the longitus for each of the sides:

For L1:
$L1 = \sqrt{(6-2)^2 + (6-2)^2}$
$L1 = \sqrt{32} 
$
$L1 = 4\sqrt{2}$

For L2:
$L2 = \sqrt{(7-2)^2 + (-3-2)^2}$
$L2 = \sqrt{50}$
$L2 = 5\sqrt{2}$

For L3:
$L3 = \sqrt{(7-6)^2 + (-3-6)^2}$
$L3 = \sqrt{82}$

Then, the perimeter is given by:
P = L1 + L2 + L3
Substituting values we have:
$P = 4\sqrt{2} + 5\sqrt{2} + \sqrt{82} 

P = 9\sqrt{2} + \sqrt{82}$

Answer:
the perimeter of triangle ABC is:
none of the above.

5 0

3 years ago

Read 2 more answers

BRAINLIEST BRAINLIEST BRAINLIEST PLEASE ANDSER IT IS TIMED THANK YOU SO MUCCH

Doss [256]

Answer:

x=-b/2a

Step-by-step explanation:

hope that helps :)

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8 0

2 years ago