Cosx= sin(90-x)
From the RHS,
sin(90-x)= sin90cosx-cos90sinx
Since sin90=1, cos90=0
sin90cosx-cos90sinx=cosx-0=cosx (proven)
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
Multiply both sides by 8. Then divide both sides by 3. Answer is 8/3
Answer:
The answer is 3+(-6) = -9, because -9 is 6 units to the left of 3.
Answer: SECOND OPTION.
Step-by-step explanation:
For this exercise it is important to remember "Rigid transformations".
Rigid transformations are those transformation that preserves length. These are the following:
- Reflection.
- Tanslation.
- Rotation.
According to the information given in the exercise, the triangle JKL was translated 12 units vertically and then reflected over the y-axis to create triangle J'K'L'. Therefore, JKL and J'K'L' are congruent.
Knowing the above, you can set up the following equation:
The final step is to solve for "x" in order to find its value. You get that this is:
