Answer:
The fourth graph is the graph of 
Step-by-step explanation:
As this inequality has a
, we know two things:
The graph must have a solid line as that signifies that it includes the value.
The shaded area must be less than 2.
The fourth graph is the only one that meets our criteria
We will use integration by substitution, as well as the integrals
∫
1
x
d
x
=
ln
|
x
|
+
C
and
∫
1
d
x
=
x
+
C
∫
x
3
x
2
+
1
d
x
=
∫
x
2
x
2
+
1
x
d
x
=
1
2
∫
(
x
2
+
1
)
−
1
x
2
+
1
2
x
d
x
Let
u
=
x
2
+
1
⇒
d
u
=
2
x
d
x
. Then
1
2
∫
(
x
2
+
1
)
−
1
x
2
+
1
2
x
d
x
=
1
2
∫
u
−
1
u
d
u
=
1
2
∫
(
1
−
1
u
)
d
u
=
1
2
(
u
−
ln
|
u
|
)
+
C
=
x
2
+
1
2
−
ln
(
x
2
+
1
)
2
+
C
=
x
2
2
−
ln
(
x
2
+
1
)
2
+
1
2
+
C
=
x
2
−
ln
(
x
2
+
1
)
2
+
C
Final answer
Answer:
See attachment
Step-by-step explanation:
Isolate y in the first inequality:

Now, with both x and y inequalities found, graph it.
Answer:
Option b is correct 175
Step-by-step explanation:
n = 7
k = 6
3k -2 ------1
put k = 6 in above eq. for finding first term
a1 = 3(6) - 2 = 18 - 2 = 16
put k = 7 in above eq. for finding first term
a2 = 3(7) - 2 = 21 - 2 = 19
a3 = 3 (8) - 2 = 24 - 2 = 22
16, 19 , 22, ... //Arithmetic series formation
a1 = 16 , a2 = 19
d = a2 - a1 = 19 - 16 = 3 //Difference of first two terms
Using sum forumula for arithmetic series
sum = 
= 
= 
=
=
= 7 * 25
= 175