That is:
Forty-five thousand, five hundred seventy-two.
~Deceptiøn
Answer:
(4,-8)
Step-by-step explanation:
Plug in the answers for the equations.
From the right hand side, we will need to find a way to rewriting 3x²y in terms of cube roots.
We know that 27 is 3³, so if we were to rewrite it in terms of cube roots, we will need to multiply everything by itself two more twice. (ie we can rewrite it as ∛(3x²y)³)
Hence, we can say that it's:
![\sqrt[3]{162x^{c}y^{5}} = \sqrt[3]{(3x^{2}y)^{3}} * \sqrt[3]{6y^{d}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B162x%5E%7Bc%7Dy%5E%7B5%7D%7D%20%3D%20%5Csqrt%5B3%5D%7B%283x%5E%7B2%7Dy%29%5E%7B3%7D%7D%20%2A%20%5Csqrt%5B3%5D%7B6y%5E%7Bd%7D%7D)
![= \sqrt[3]{162x^{6}y^{3+d}}](https://tex.z-dn.net/?f=%3D%20%5Csqrt%5B3%5D%7B162x%5E%7B6%7Dy%5E%7B3%2Bd%7D%7D)
Hence, c = 6 and d = 2
Answer:
9x-2
Step-by-step explanation:
Since you are adding the two functions, you should first start adding like terms. 4x + 5x is 9x. -8 + 6 is -2. Thus the value of (f+g)(x) is 9x-2.
Answer:
If a+b+c=1,
a
2
+
b
2
+
c
2
=
2
,
a
3
+
b
3
+
c
3
=
3
then find the value of
a
4
+
b
4
+
c
4
=
?
we know
2
(
a
b
+
b
c
+
c
a
)
=
(
a
+
b
+
c
)
2
−
(
a
2
+
b
2
+
c
2
)
⇒
2
(
a
b
+
b
c
+
c
a
)
=
1
2
−
2
=
−
1
⇒
a
b
+
b
c
+
c
a
=
−
1
2
given
a
3
+
b
3
+
c
3
=
3
⇒
a
3
+
b
3
+
c
3
−
3
a
b
c
+
3
a
b
c
=
3
⇒
(
a
+
b
+
c
)
(
a
2
+
b
2
+
c
2
−
a
b
−
b
c
−
c
a
)
+
3
a
b
c
=
3
⇒
(
a
+
b
+
c
)
(
a
2
+
b
2
+
c
2
−
(
a
b
+
b
c
+
c
a
)
+
3
a
b
c
=
3
⇒
(
1
×
(
2
−
(
−
1
2
)
+
3
a
b
c
)
)
=
3
⇒
(
2
+
1
2
)
+
3
a
b
c
=
3
⇒
3
a
b
c
=
3
−
5
2
=
1
2
⇒
a
b
c
=
1
6
Now
(
a
2
b
2
+
b
2
c
2
+
c
2
a
2
)
=
(
a
b
+
b
c
+
c
a
)
2
−
2
a
b
2
c
−
2
b
c
2
a
−
2
c
a
2
b
=
(
a
b
+
b
c
+
c
a
)
2
−
2
a
b
c
(
b
+
c
+
a
)
=
(
−
1
2
)
2
−
2
×
1
6
×
1
=
1
4
−
1
3
=
−
1
12
Now
a
4
+
b
4
+
c
4
=
(
a
2
+
b
2
+
c
2
)
2
−
2
(
a
2
b
2
+
b
2
c
2
+
c
2
a
2
)
=
2
2
−
2
×
(
−
1
12
)
=
4
+
1
6
=
4
1
6
Extension
a
5
+
b
5
+
c
5
=
(
a
3
+
b
3
+
c
3
)
(
a
2
+
b
2
+
c
2
)
−
[
a
3
(
b
2
+
c
2
)
+
b
3
(
c
2
+
a
2
)
+
c
3
(
a
2
+
c
2
)
]
=
3
⋅
2
−
[
a
3
(
b
2
+
c
2
)
+
b
3
(
c
2
+
a
2
)
+
c
3
(
a
2
+
b
2
)
]
Now
a
3
(
b
2
+
c
2
)
+
b
3
(
c
2
+
a
2
)
+
c
3
(
a
2
+
b
2
)
=
a
2
b
2
(
a
+
b
)
+
b
2
c
2
(
b
+
c
)
+
c
2
a
2
(
a
+
c
)
=
a
2
b
2
(
1
−
c
)
+
b
2
c
2
(
1
−
a
)
+
c
2
a
2
(
1
−
b
)
=
a
2
b
2
+
b
2
c
2
+
c
2
a
2
−
(
a
2
b
2
c
+
b
2
c
2
a
+
c
2
a
2
b
)
=
−
1
12
−
a
b
c
(
a
b
+
b
c
+
c
a
)
=
−
1
12
−
1
6
⋅
(
−
1
2
)
=
0
So
a
5
+
b
5
+
c
5
=
6
−
0
=
6
Step-by-step explanation:
