how do you solve 2x^4/3-2=160?

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how do you solve 2x^4/3-2=160?

Mathematics
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2.99
Oops that's not right sorry
hahhah yeah...

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Other answers:

its like solving square roots and radical equations
27
^^ That's the answer sorry about that haha
okay but like how do u do it....hahah
explllain :)
2x^(4/3) = 162 x^(4/3) = 81 x = 81^(3/4) x = 27
This is because (x^(4/3))^(3/4) = (81)^(3/4)
\[2x ^{4/3} - 2 = 160\] \[x ^{4/3} = 81\] \[\sqrt[3]{x^{4}} = 81\] \[x ^{4} = 81^{3}\] x = 27
Make me a lifesaver :)
I beat krb686
thankss guys!!!
thankss guys!!!
i have more questions hold on....hahah
i have more questions hold on....hahah
okay so the next section is inverse relations and functions...
wait forgett that haha..okay so how do u graph square roots and radical functions like wtff i dont get itt.... one problem is like y=-\[y=-\sqrt{x+2}\]
Do you know how to find the domain of a function?
yeaahh
So the domain of the function here is [-2,infinity)
If we were looking at y=(x+2)^(1/2) the function would be increasing since the bigger the x the bigger the y.
But we have y=-(x+2)^(1/2), so the function is getting smaller for big x
If you can graph the function y = root(x), then take that graph and shift it to the left by 2, and flip it upside down
ohh so thats all?
thats easyy
Yep! When there is an additive factor inside of the function with the x, a positive # shifts to the left and negative shifts to the right. And a negative multipler on the outside flips it upside down.
No technically the domain includes values less than negative 2, which would be considered i values (or in this case imaginary), this is very useful in electrical engineering where there is an imaginary and real domain. Very useful in euler's method as well. The values don't exist if they are indeterminant or non-existent in general, and krb is correct in the fact that it will be shifted and the values will flip.
what about if its like... \[y=\sqrt{x+2}+1\]?
No the domain includes values greater than -2
Okay so the inside 2 shifts it to the left 2, and the outside 1 shifts it UP 1
Shift left 2, and increased y intercept by 1
mathgenius101 is technically correct, if the domain is allowed to include all numbers not just reals.
He didn't specify if the domain included complex values. So technically there could be a domain x<-2, imaginary is still technically a complex number.
im pretty sure we are only talking about real numbers here
True
hes in college algebra
You should never be an electrical engineer.
or some type of algebra
wait okay so..
see you confused him
what about if theres a number (either neg or pos) before the square root?
umm im not a him if youre talking about me...hahahha
Multiplied or added/subtracted? Multiplied flips it either up or down, added or subtracted shifts it up or down.
Yeah, stop assuming like how you're assuming it's a real number. She's a girl.
hahha love ya math genius <3 <3
r u guys like math nerds....
I guess you could say that. Soon-to-be engineers.
i gotta go study more then sleep. BYEEE NEW FRIENDSSSSS THANKS FOR THE HELP :) <3
i gotta go study more then sleep. BYEEE NEW FRIENDSSSSS THANKS FOR THE HELP :) <3
dont miss me too muchhhhhh!!! hahahhaha
I'll try not to ;)
i like math genius the best...youre the only one who responds to my pointless comments
i like math genius the best...youre the only one who responds to my pointless comments
k bye...for reall..

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