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:
<h3>6 days</h3>
Step-by-step explanation:
Given the inequality expression of the total cost (c) in dollars of renting a car for n days as c ≥ 125 + 50n
To get the maximum number of days for which a car could be rented if the total cost was $425, substitute c = 425 into the expression and find n
425 ≥ 125 + 50n
Subtract 125 from both sides
425 - 125 ≥ 125 + 50n - 125
300≥ 50n
Divide both sides by 50
300/50≥50n/50
6 ≥n
Rearrange
n≤6
<em>Hence the maximum number of days for which a car could be rented if the total cost was $425 is 6days</em>
<em></em>
I wish you were using parentheses here. I must assume (without knowing for sure) that y ou mean
a - 3/7
----------
3 - a/21
The LCD here is 21, since 7 divides into 21 without a remainder.
Thus, multiply numerator and denominator by 21:
21a - 9
-----------
63 - a
You could leave this result as is, or you could factor out 3:
7a-3
3 ---------
63-a
Answer:0.2588
Step-by-step explanation:
Answer:
x=10 (22.5x+6.50=250)
Step-by-step explanation:
