Answer:
I would recommend using photo math its faster and easier I promise!! ;)
Answer:
4
Step-by-step explanation:
1 1/2 x 2 2/3
3/2 x 8/3
9/6 x 16/6
144/36
4
Answer:
9.V=480
10.V=25.1
11.V=377
12.V=160
13.V=179.6
14.V=452.4
Step-by-step explanation:
9.V=l*w*h
V=10*8*6
V=480
10.V=pi*1^2*8
V=8pi
V=25.1
11.V=1/3pi*6^2*10
V=1/3pi*36*10
V=1/3pi *360
V=120pi
V=377
12.V=l*w*h/3
V=10*8*6/3
V=480/3
V=160
13.V=4/3pi*3.5^3
V=4/3pi *42.875
V=171.5/3pi
V=179.6
14.V=2/3pi*6^3
V=2/3pi*216
V=432/3pi
V=144pi
V=452.4
Answer:
a) The length of the building that should border the dog run to give the maximum area = 25feet
b) The maximum area of the dog run = 1250 s q feet²
Step-by-step explanation:
<u><em>Step(i):-</em></u>
<em>Given function </em>
<em> A(x) = x (100-2x)</em>
<em> A (x) = 100x - 2x²...(i)</em>
<em>Differentiating equation (i) with respective to 'x'</em>
<em> </em>
<em></em>
<em> ⇒ </em>
...(ii)
<em>Equating zero</em>
⇒ 100 - 4x =0
⇒ 100 = 4x
Dividing '4' on both sides , we get
x = 25
<em>Step(ii):</em>-
Again differentiating equation (ii) with respective to 'x' , we get

Therefore The maximum value at x = 25
The length of the building that should border the dog run to give the maximum area = 25
<u>Step(iii)</u>
Given A (x) = x ( 100 -2 x)
substitute 'x' = 25 feet
A(x) = 25 ( 100 - 2(25))
= 25(50)
= 1250
<u><em>Conclusion</em></u>:-
The maximum area of the dog run = 12 50 s q feet²
<em> </em><em> </em>
<em></em>