Pythagoras Theorem:
hipotenuse²=leg₁²+leg₂²
First posible triangle:
hypotenuse=13 (13²=169)
leg₁=12 ( 12²=144)
leg₂=5 (5²=25)
13³=144 + 25
Answer:can be side lengths of a triangle
Second triangle:
hypotenuse=12.6 (12.6²=158.76)
leg₁=6.7 ( 6.7²=44.89)
leg₂=6.5 (6.5²=42.25)
leg₁²+leg₂²=44.89+42.25=87.14≠158.76
Answer: cannot be side lenghts of a triangle.
third triangle:
hypotenuse=13 (13²=169)
leg₁=12 ( 12²=144)
leg₂=11 (11²=121)
leg₁²+leg₂²=144+121=265≠169
Answer: cannot be side lenghts of a triangle.
fourth triangle:
hypotenuse=13 (13²=169)
leg₁=6 ( 6²=36)
leg₂=4 (4²=16)
leg₁²+leg₁²=36+16=52≠169
Answer: cannot be side lenghts of a triangle.
Answer:
x-6 = -4
Step-by-step explanation:
3x - 8= -2
Add 8 to each side
3x - 8+8= -2+8
3x = 6
Divide by 3
3x/3 = 6/2
x =2
We want to find x-6
2-6 = -4
So a triangular prisim is like a triangle stacked on top of x number of identical triangles making a height, like a piece of paper made into a stack
so
basically find the area of each side and add
look at diagram/attachment
find area of triangle
1/2 b times h= area of 1 triangular face
multiply by 2 because 2 sides so 1/2 times 2 b times h=b times h
then 3 other sides
find area of each side and add
areas=(H times B)+(H times c)+(H times a)
so SA=(b times H)+(H times B)+(H times c)+(H times a)
Answer:
p+2=-15 step1:-2 on both sides
p=-17
p=-17
Answer:
i believe that it is 45
Step-by-step explanation: