Answer:
No, a triangle cannot be constructed with sides of 2 in., 3 in., and 6 in.
For three line segments to be able to form any triangle you must be able to take any two sides, add their length and this sum be greater than the remaining side.
2
in.
+
3
in.
=
5
in.
5
in.
<
6
in.
For a triangle with sides 3 in., 4 in. and 5 in. which can form a triangle:
3 + 4 = 7 which is greater than 5
3 + 5 = 8 which is greater than 4
4 + 5 = 9 which is greater than 3
Step-by-step explanation:
I’m pretty sure it should be ,B
Not exactly as the answer is 0.428571428571428571 and continues infinitely repeating the 6 digits "428571" in sequence
Hi friend,
Answer of 4th u need nah as u ticked on it-:
Yes then-:
8*x*x*y*z*z*z*z/16*x*y*y*z
=x*z^3/2y
Hope it helps..........
Answer: 1: k 2: 12r-20 3: 10n-13 4: -14x
Step-by-step explanation:
