Hint: it will help to change all denominators to the same one using gcf or lcm
From the description we can infer that we have the expression:
.
Now, to write our expression as a as a root, we are going to apply the law of exponents:
first
Next, we are going to apply the law about fractional exponents:
![x^{ \frac{m}{n}}= \sqrt[n]{x^m}](https://tex.z-dn.net/?f=x%5E%7B%20%5Cfrac%7Bm%7D%7Bn%7D%7D%3D%20%5Csqrt%5Bn%5D%7Bx%5Em%7D%20%20)
![\frac{1}{41^{ \frac{2}{5} }}= \frac{1}{ \sqrt[5]{41^2} }](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B41%5E%7B%20%5Cfrac%7B2%7D%7B5%7D%20%7D%7D%3D%20%5Cfrac%7B1%7D%7B%20%5Csqrt%5B5%5D%7B41%5E2%7D%20%7D%20)
We can conclude that the
value of B is 2.
Answer: There are 24 traffic lights on Berry Boulevard.
Step-by-step explanation:
Since we have given that
Total length of road = 
Number of roads on every 
= 2
So, According to unitary method,
In every , number of roads = 2
In every 1 mile, number of roads = 
So, in every 
the number of roads = 
Hence, there are 24 traffic lights on Berry Boulevard.
L
=
∫
t
f
t
i
√
(
d
x
d
t
)
2
+
(
d
y
d
t
)
2
d
t
. Since
x
and
y
are perpendicular, it's not difficult to see why this computes the arclength.
It isn't very different from the arclength of a regular function:
L
=
∫
b
a
√
1
+
(
d
y
d
x
)
2
d
x
. If you need the derivation of the parametric formula, please ask it as a separate question.
We find the 2 derivatives:
d
x
d
t
=
3
−
3
t
2
d
y
d
t
=
6
t
And we substitute these into the integral:
L
=
∫
√
3
0
√
(
3
−
3
t
2
)
2
+
(
6
t
)
2
d
t
And solve:
=
∫
√
3
0
√
9
−
18
t
2
+
9
t
4
+
36
t
2
d
t
=
∫
√
3
0
√
9
+
18
t
2
+
9
t
4
d
t
=
∫
√
3
0
√
(
3
+
3
t
2
)
2
d
t
=
∫
√
3
0
(
3
+
3
t
2
)
d
t
=
3
t
+
t
3
∣
∣
√
3
0
=
3
√
3
+
3
√
3
=6The arclength of a parametric curve can be found using the formula:
L
=
∫
t
f
t
i
√
(
d
x
d
t
)
2
+
(
d
y
d
t
)
2
d
t
. Since
x
and
y
are perpendicular, it's not difficult to see why this computes the arclength.
It isn't very different from the arclength of a regular function:
L
=
∫
b
a
√
1
+
(
d
y
d
x
)
2
d
x
. If you need the derivation of the parametric formula, please ask it as a separate question.
We find the 2 derivatives:
d
x
d
t
=
3
−
3
t
2
d
y
d
t
=
6
t
And we substitute these into the integral:
L
=
∫
√
3
0
√
(
3
−
3
t
2
)
2
+
(
6
t
)
2
d
t
And solve:
=
∫
√
3
0
√
9
−
18
t
2
+
9
t
4
+
36
t
2
d
t
=
∫
√
3
0
√
9
+
18
t
2
+
9
t
4
d
t
=
∫
√
3
0
√
(
3
+
3
t
2
)
2
d
t
=
∫
√
3
0
(
3
+
3
t
2
)
d
t
=
3
t
+
t
3
∣
∣
√
3
0
=
3
√
3
+
3
√
3
=
6
√
3
Be aware that arclength usually has a difficult function to integrate. Most integrable functions look like the above where a binomial is squared and adding the two terms will flip the sign of the binomial.
Be aware that arclength usually has a difficult function to integrate. Most integrable functions look like the above where a binomial is squared and adding the two terms will flip the sign of the binomial.
The answer is A. The rest of the answers are not necessarily correct with the operations.
