Answer:
it is number 3 (-1/5)
Step-by-step explanation:
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
Answer:
Step-by-step explanation:
Lets say there's a number 'x' and there's an expression 
where 'a' & 'b' are also numbers , <u>the simplified form of the expression is </u>
<u>.</u>
So ,
d = r * t
d/t = r
-35.75 / 3.25 = r
-11 = r
r = -11 ft / min
d = r*t
d = -11 ft/min * 1 min
d = -11 ft
The probe is 11 feet below sea level after 1 minute
I'm answer for u is 240 cubic units=b
