Asked by Anonymous
I am having trouble with a pre-cal problem. If you could explain how to do it that would be great!
((3^3)^(cosx+2/3sinx))^2cosx+3sinx all divided by ((3^sinx)^13)^cosx = x
((3^3)^(cosx+2/3sinx))^2cosx+3sinx all divided by ((3^sinx)^13)^cosx = x
Answers
Answered by
Reiny
I do not follow your order of exponents
You will have to be much more specific by using brackets
e.g.
2^3^4
done on my calculator by punching it in that way
gives me 4096
(The calculator appears to evaluate from left to right and simply does (2^3)^4 = 2^12= 4096
But, what about
(2^(3<sup>^4</sup>)
= 2^81
= huge !
on calculator , enter
2
y<sup>x</sup>
(
3
y<sup>x</sup>
4
)
=
to one humongous number (2.41... x 10^24)
Please clean up your expression
You will have to be much more specific by using brackets
e.g.
2^3^4
done on my calculator by punching it in that way
gives me 4096
(The calculator appears to evaluate from left to right and simply does (2^3)^4 = 2^12= 4096
But, what about
(2^(3<sup>^4</sup>)
= 2^81
= huge !
on calculator , enter
2
y<sup>x</sup>
(
3
y<sup>x</sup>
4
)
=
to one humongous number (2.41... x 10^24)
Please clean up your expression
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