Answer:
yes i can but will have to get back to you in about half a hour just doing maths atm with the teacher
Step-by-step explanation:
Answer:
6/1 · 11/4 = 66/4 = 16 1/2
-10/3 · (-17/5) = 170/15 = 11 1/3
-9/2 · 26/3 = -234/6 = -39
11/6 · (-9/1) = -99/6 = -16 1/2
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:
The height of wall is 3 feet and length of wall is 5 feet
Step-by-step explanation:
Given as :
The Area of retaining wall = 15 square feet
Let The width of wall = W feet
The Height of wall = ( W - 2 ) feet
Now, Area of Rectangle = Height width
Or , 15 ft² = ( W - 2 ) ft × W ft
Or, 15 = W² - 2 W
Or, W² - 2 W - 15 = 0
Or, W² - 5 W + 3 W - 15 = 0
Or, W ( W - 5 ) + 3 ( W - 5 ) = 0
So , ( W - 5 ) ( W + 3 ) = 0
So, W = 5 , - 3
∴ The width of wall = W = 5 feet
And The Height of wall = ( W - 2 ) = 5 - 2 = 3 feet
Hence The height of wall is 3 feet and length of wall is 5 feet. Answer
