#is this just 3x^2?
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i said it wrong 1 sec
$\frac{1}{3x^2} = (3x^2)^{-1}$
Ludwig
basically it says make it in the form kx^n
What is the question?
so if you times both the bottom and the top by the bottom then it would make it 3x^12
ill screenshot the whole thing
i thought abt this as a way to do the question but i pretty much know its stupid
ik i was being stupid
so you write my stuff above in a different way
if i use this then just work out the bracket then it would make 3x^-2
which would work i think
where did you get the power of 3 from?
What is $(2 \cdot 3)^{-1}$
Ludwig
and dont use $2 \cdot 3 = 6$
Ludwig
cdot is times
ok
does $\frac{1}{3x^{2}} = 3x^{-2}$?
!𒐪 ɹɐupoɯ⇂ㄥ8𝟝 𒐪!
thats what i was thinking
well test a value of x
idk what that means haha
plug in x=1
well note that 1 to the power of anything is still 1
so we’re left with 1/3=3 which isn’t true
so what should our value of k be
i dont understand where 1/3=3 came from
alright, what’s $1^{2}$ and $1^{-2}$
!𒐪 ɹɐupoɯ⇂ㄥ8𝟝 𒐪!
yup both are 1
1
now use that for this
idk how that can be used for the equation im so sorry i havent done maths for 3 months
alright well
we can replace all the x powers with 1 because we know they’re just 1 right?
so 1/3^2 cus the 3*1^2
i figured out it wouldnt be 3x^-2 but ik its (3x^2)^-1 but i need it in a form that looks like kx^n. you asked what k meant but k is just another number for n.
yup, try expanding (3x^2)^-1
ive been trying to but i just cant find it
cus (3x^2)^-1 = 1/3x^2 but i need a number that is like 3x^2 but with the - in it so it can still make 1/3x^2 if you know what i mean
cus i need it in the form of kx^n but i have it as (kx^n)^n
so im missing something very small
but very important
I got it