Mathematics: From which set of dimensions could a triangle be constructed? a.side length of 7 side length of 8 side length of 15 side b.length of 8 side length of 5 side length of 14 c.side length of 6 side length of 3 side length of 10 d.side length of 2 side length of 6 side length of 7

Non-negative integers are positive integers or zero. 1. When then there are such possible cases for and In total 11 solutions for 2. For there are such possible cases for and In total 10 solutions for 3. This process gives youfor - 9 solutions;for - 8 solutions;for - 7 solutions;for - 6 solutions;for - 5 solutions;for - 4 solutions;for - 3 solutions;for - 2 solutions;for - 1 solution.4. Add all numbers of solutions: there are 66 possible solutions (with non-negative integer variables)...

From which set of dimensions could a triangle be

Ответ:

TheMadScientist8575

14.03.2022

Mathematics

The longest side of the triangle has to be less than the sum of the two other sides.

A ⇒ 7 + 8 = 15
B ⇒ 8 + 5 < 14
C ⇒ 6 + 3 < 10
D ⇒ 2 + 6 > 7

The answer is D, because the longest side length (7) is shorter than the two other side lengths (2 and 6).

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Ответ:

abdulspy786

16.10.2020

Mathematics

Non-negative integers are positive integers or zero.

1. When x_1=0, then there are such possible cases for x_2 and x_3:

$x_2=0,\ x_3=10;$ $x_2=1,\ x_3=9;$ $x_2=2,\ x_3=8;$ $x_2=3,\ x_3=7;$ $x_2=4,\ x_3=6;$ $x_2=5,\ x_3=5;$ $x_2=6,\ x_3=4;$ $x_2=7,\ x_3=3;$ $x_2=8,\ x_3=2;$ $x_2=9,\ x_3=1;$ $x_2=10,\ x_3=0.$

In total 11 solutions for x_1=0.

2. For x_1=1, there are such possible cases for x_2 and x_3:

$x_2=0,\ x_3=9;$ $x_2=1,\ x_3=8;$ $x_2=2,\ x_3=7;$ $x_2=3,\ x_3=6;$ $x_2=4,\ x_3=5;$ $x_2=5,\ x_3=4;$ $x_2=6,\ x_3=3;$ $x_2=7,\ x_3=2;$ $x_2=8,\ x_3=1;$ $x_2=9,\ x_3=0.$

In total 10 solutions for x_1=1.