The function you seek to minimize is
()=3‾√4(3)2+(13−4)2
f
(
x
)
=
3
4
(
x
3
)
2
+
(
13
−
x
4
)
2
Then
′()=3‾√18−13−8=(3‾√18+18)−138
f
′
(
x
)
=
3
x
18
−
13
−
x
8
=
(
3
18
+
1
8
)
x
−
13
8
Note that ″()>0
f
″
(
x
)
>
0
so that the critical point at ′()=0
f
′
(
x
)
=
0
will be a minimum. The critical point is at
=1179+43‾√≈7.345m
x
=
117
9
+
4
3
≈
7.345
m
So that the amount used for the square will be 13−
13
−
x
, or
13−=524+33‾√≈5.655m
Factor and group common ones
60=2*2*3*5*x*x*x*x*y*y*y*y*y*y*y
45x^5^5=3*3*5*x*x*x*x*x*y*y*y*y*y
75x^3y=3*5*5*x*x*x*y
the commmon gropu to all is 3*5*x*x*x*y=15x^3y
<h2>
Answer:</h2>
Option: C is the correct answer.
C. 5
<h2>
Step-by-step explanation:</h2>
We will solve the following problem with the help of the Venn diagram.
Based on the Venn diagram we are asked to find the value of v i.e. those who have parakeet only.
From the information that:
x+y+10=35
i.e.
x+y=25
Also, based on the information :
u+z+10=28
i.e.
u+z=18
Also, 42 had neither a cat nor a dog nor a parakeet.
This means it cover the outer region of the three circles.
Total 100 families were surveyed it means that:
42+x+y+10+u+z+v=100
i.e. 42+25+10+18+v=100
i.e. 95+v=100
i.e. v=100-95
i.e. v=5
Hence, the number of families who have only parakeet are: 5