Asked by PJ
Is (5, 7, 74−−√ ) a Pythagorean triple? (1 point)
Responses
Yes, it is a Pythagorean triple.
Yes, it is a Pythagorean triple.
No, it is not a Pythagorean triple.
No, it is not a Pythagorean triple.
It cannot be determined.
Responses
Yes, it is a Pythagorean triple.
Yes, it is a Pythagorean triple.
No, it is not a Pythagorean triple.
No, it is not a Pythagorean triple.
It cannot be determined.
Answers
Answered by
GPT-4o mini
To determine if the set of numbers \( (5, 7, \sqrt{74}) \) forms a Pythagorean triple, we need to check if the following equation holds:
\[
a^2 + b^2 = c^2
\]
where \( c \) is the largest number. From the given set, we identify \( a = 5 \), \( b = 7 \), and \( c = \sqrt{74} \).
Now, we compute:
1. \( a^2 = 5^2 = 25 \)
2. \( b^2 = 7^2 = 49 \)
3. \( c^2 = (\sqrt{74})^2 = 74 \)
Now we add \( a^2 \) and \( b^2 \):
\[
a^2 + b^2 = 25 + 49 = 74
\]
And compare it with \( c^2 \):
\[
c^2 = 74
\]
Since \( a^2 + b^2 = c^2 \) holds true, we conclude that:
**Yes, it is a Pythagorean triple.**
\[
a^2 + b^2 = c^2
\]
where \( c \) is the largest number. From the given set, we identify \( a = 5 \), \( b = 7 \), and \( c = \sqrt{74} \).
Now, we compute:
1. \( a^2 = 5^2 = 25 \)
2. \( b^2 = 7^2 = 49 \)
3. \( c^2 = (\sqrt{74})^2 = 74 \)
Now we add \( a^2 \) and \( b^2 \):
\[
a^2 + b^2 = 25 + 49 = 74
\]
And compare it with \( c^2 \):
\[
c^2 = 74
\]
Since \( a^2 + b^2 = c^2 \) holds true, we conclude that:
**Yes, it is a Pythagorean triple.**
There are no human answers yet. A form for humans to post answers is coming very soon!