You have to subtract $40 from $68 make sure the larger number is in the front. If you did that all you would get $28.
Answer:
12
Step-by-step explanation:
.......................
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:
I think it's the first one :?
