Answer:
Step-by-step explanation:
Use the info given in the exponential equation to find the value of b, the rate of decay.
where v(t) is the value of the car after a certain number of years, t, have gone by, a is the initial value, and b is the rate of decay. We have everything we need but b:
a = 20000
v(t) = 16000 after t = 1 year:
so
b = .8 Taken in context, this means that the car depreciates 20% each year. Now we can solve the problem being asked of us, which is to find the value of the car after t = 5 years:
which simplifies down a bit to
v(t) = 20000(.32768) so
v(t) = 6553.60, choice C.
B is less than or equal to -11
This is F(n) = 3n
The X is an interger which is holding a value
The relationship between Y and X is 3n as we multiply X by 3 to get the value of Y and is proven yet so.
Answer:
10x² - 48y² - 11x + 2y - 74xy
Step-by-step explanation:
(-x+8y)+(-10x-6y)(−x+8y)+(−10x−6y)
Rearranging,
=> - x - 10 x + 8y - 6y + (-10x-6y)(−x+8y)
=> - 11x + 2y + 10x² - 48y² + 6xy - 80xy
=> 10x² - 48y² - 11x + 2y - 74xy
Answer:
4x+2h
Step-by-step explanation:
The average rate of change of a continuous function,
f
(
x
)
, on a closed interval
[
a
,
b
]
is given by
f
(
b
)
−
f
(
a
)
b
−
a
So the average rate of change of the function
f
(
x
)
=
2
x
2
+
1
on
[
x
,
x
+
h
]
is:
A
r
o
c
=
f
(
x
+
h
)
−
f
(
x
)
(
x
+
h
)
−
(
x
)
=
f
(
x
+
h
)
−
f
(
x
)
h
...
.
.
[
1
]
=
2
(
x
+
h
)
2
+
1
−
(
2
x
2
+
1
)
h
=
2
(
x
2
+
2
x
h
+
h
2
)
+
1
−
2
x
2
−
1
h
=
2
x
2
+
4
x
h
+
2
h
2
−
2
x
2
h
=
4
x
h
+
2
h
2
h
=
4
x
+
2
h
Which is the required answer.
Additional Notes:
Note that this question is steered towards deriving the derivative
f
'
(
x
)
from first principles, as the definition of the derivative is:
f
'
(
x
)
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
This is the function we had in [1], so as we take the limit as
h
→
0
we get the derivative
f
'
(
x
)
for any
x
, This:
f
'
(
x
)
=
lim
h
→
0
4
x
+
2
h
=
4
x
