Just use Photomath it helps a lot
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:
The fourth one.
Step-by-step explanation:
I am pretty sure.
Answer:
If ‘A’ can finish a work in ‘n’ days then part of work finished in 1 day
will be
.
Step-by-step explanation:
From the question, it is clear that
- If ‘A’ can finish a work in ‘n’ days, then
- we have to determine the part of work finished in 1 day.
So
let '
' be the number of days
takes to complete the whole work
Let the whole job be denoted as '
'
Thus, the part of work finished in 1 day will be:
![\:\frac{Total\:work}{Time\:taken\:by\:A\:to\:complete\:the\:work}\:=\frac{1}{n}](https://tex.z-dn.net/?f=%5C%3A%5Cfrac%7BTotal%5C%3Awork%7D%7BTime%5C%3Ataken%5C%3Aby%5C%3AA%5C%3Ato%5C%3Acomplete%5C%3Athe%5C%3Awork%7D%5C%3A%3D%5Cfrac%7B1%7D%7Bn%7D)
Therefore, If ‘A’ can finish a work in ‘n’ days then part of work finished in 1 day will be
.
Keywords: work, word problem
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