Which numerical expression correctly
1 answer:
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We can apply some rules backwards
first
f'(x)=
(x-4)(-2x+4)
-2x^2+12x-16
we know that
so therefor maybe
-2x^2=rnx^{n-1}
2=n-1
n=3
rn=-2
3r=-2
r=-2/3
one is
second part
12x
12x=
x^1, 1=n-1
n=2
rn=12
2r=12
r=6
is the second bit
last part
-16
-16x^0=
0=n-1
n=1
rn=-16
1r=-16
r=-16
-16x^1
so therfor f(x)=
first
f'(x)=
(x-4)(-2x+4)
-2x^2+12x-16
we know that
so therefor maybe
-2x^2=rnx^{n-1}
2=n-1
n=3
rn=-2
3r=-2
r=-2/3
one is
second part
12x
12x=
x^1, 1=n-1
n=2
rn=12
2r=12
r=6
last part
-16
-16x^0=
0=n-1
n=1
rn=-16
1r=-16
r=-16
-16x^1
so therfor f(x)=
Answer:
8w^2 I don't know why it is not on here, but I put it in an algebra calculator and everything. Good luck. :)
Step-by-step explanation:
Simply the radicals by multiplying and taking square roots.
2 *
= 2 *2w
= 4w
=4w(2w)
=