Answer:
After 1 second, the ball will reach a maximum height of 16 feet
Step-by-step explanation:
The height of the ball after t seconds: h(t) = -16t^2 + 32t
The graph of this quadratic function is parabola which opens downwards. The vertex of a quadratic equation is the maximum or minimum point on the equation's parabola
t = -b/2a = -(32)/2(-16) = -32/-32 = 1 second
then
h(t) = -16(1)^2 + 32(1) = -16 + 32 = 16
After 1 second, the ball will reach a maximum height of 16 feet
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
75 hope this helps but I’m not sure what you’re asking
Answer:
16
Step-by-step explanation:
if x=23, 23-7=16
