Mathematics
14 Online
OpenStudy (anonymous):
(x*3)^2 * (x*5)^4
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OpenStudy (anonymous):
Use:
\[\large (x^a)^b = x^{a \times b}\]
\[\large x^a \times x^b = x^{a+b}\]
OpenStudy (anonymous):
so would it be x^6+^20?
hartnn (hartnn):
where did u get + from??
it would be x^6*x^20
then use the 2nd formula given by waterineyes
OpenStudy (anonymous):
\[\large x^{6+20}\]
OpenStudy (anonymous):
you can add 6 + 20 here..
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OpenStudy (anonymous):
thats what i meant... i cant use the small script so i added the 6 and the 20
OpenStudy (anonymous):
So finally what did you get ??
OpenStudy (anonymous):
x^26
hartnn (hartnn):
correct :)
OpenStudy (anonymous):
thanks guys! i appreciate that!
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OpenStudy (anonymous):
Sorry for late..
yes \(x^[26}\) is correct..
OpenStudy (anonymous):
\(x^{26}\) is correct..
OpenStudy (anonymous):
Welcome dear..
OpenStudy (anonymous):
Thanks @hartnn for being on time here...
hartnn (hartnn):
no need of thanks...i was just looking around...u actually solved it....
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OpenStudy (anonymous):
f^2*g^7
OpenStudy (anonymous):
you cannot solve it further as f and g are two different variables..
OpenStudy (anonymous):
they have me over here doing these ridiculous math problems at 2 am!
OpenStudy (anonymous):
So you will leave it as such:
\[f^2 \cdot g^7\]
OpenStudy (anonymous):
-2x^8*-2x^7 (i got 4x^15 for my final answer is that correct?
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hartnn (hartnn):
yup :)
OpenStudy (anonymous):
ok.. i had to make sure! thanks!
OpenStudy (anonymous):
Yep..
OpenStudy (anonymous):
4^3*4^7*4^4 it wants me to simplify and i got 4^14 for my final answer do you think i worked it out right?
OpenStudy (anonymous):
Yeah it looks good to me..
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OpenStudy (anonymous):
thanks
OpenStudy (anonymous):
Welcome..
OpenStudy (anonymous):
8/s^-9