6.) D (x+7)(x+1)
7.) B (x-9)(x+8)
8.) D
9.) A
10.)A
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.
9r=3r+6
Subtract 3r from both sides
6r=6
Divide both sides by 6
r=1
X/9 - 7 is the answer. note x/9 is in fraction form
