Equal set is equivalent set but equivalent set not need equal set? why
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:
Step-by-step explanation:
Your equation is:
You also know that x = -3 and y = 2. You can rewrite the equation by substituting x and y with their actual values:
First, you want to do the exponentiation (meaning you want to calculate and
parts).
equals 9 and
equals 16.
So the equation now is:
Now you want to do multiplication. equals 36 and
equals 32. So you're left with:
Which equals 68.