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:
IM ON THE SAME ONEEE let me know when you have it
Step-by-step explanation:
Compose the quadratic condition in standard shape, ax2 + bx + c = 0. Recognize the values of a, b, c. Write the Quadratic Equation. At that point substitute within the values of a, b, c. Simplify. Check the arrangements.
Answer:63.75 ez
Step-by-step explanation:
first you multiply 10 x 8.5 and then divide by 2=85, after that you you divide by 2 =42.5 and you see the little triangle thats 42.5/2 wich is equal to 21.25 so after all of that you add 42.5+21.25 and you get the answer
Answer:
t+4, t≠8
Step-by-step explanation:
The value of t must not be one that makes the denominator zero, hence must not be 8.
