Ask
your own question, for FREE!
Mathematics
13 Online
Is this true: 'The only way you can get from one perfect square, q^2, to another, p^2, by multiplying by a constant [i.e. p^2=a^2 q^2] is if q IS that factor [i.e. q^2=a^2]'?
Still Need Help?
Join the QuestionCove community and study together with friends!
'the first square IS that factor' what is 'that' here? factor of whose ? \(b^2=ma^2\)
\[q^2=49p^2\]Must \[q=49\]?
I've edited it. Is that still ambiguous?
then i don't think its always true.... q equal 49, is not necessary
Of course apart from the trivial case where p=1
Still Need Help?
Join the QuestionCove community and study together with friends!
Counterexample: 6^2= 4 * 2^2 Damn, it's wrong!
yea, its false....
This popped up in http://www.komal.hu/verseny/feladat.cgi?a=honap&h=201210&t=mat&l=en if anyone's interested...
Can't find your answer?
Make a FREE account and ask your own questions, OR help others and earn volunteer hours!
Join our real-time social learning platform and learn together with your friends!
Join our real-time social learning platform and learn together with your friends!
Latest Questions
VERNEISH:
Two objects are moving along separate linear paths where each path is described by position, d, and time, t.
Maci:
what are the best stock apps for people who want to start investing?
Twaylor:
genuienly how do i get better self control i keep watcing youtube and theres this
IzzyCookie:
|dw:1792074056138:dw| Uhm, I don't know where to ask this, but I'm in college welding at my community college, and I need help learning vertical welding for
thehearken:
not sure if this is considered spam, im sorry if it is, but does anyone have tips
1 day ago
1 Reply
0 Medals
5 days ago
3 Replies
2 Medals
1 week ago
3 Replies
0 Medals
3 days ago
9 Replies
1 Medal
1 week ago
2 Replies
1 Medal