Answer:
The box should have base 16ft by 16ft and height 8ft Therefore,dimensions are 16 ft by 16 ft by 8 ft
Step-by-step explanation:
We were given the volume of the tank as, 2048 cubic feet.
Form minimum weight, the surface area must be minimum.
Let the height be h and the lengths be x
the volume will be: V=x²h then substitute the value of volume, we have
2048=hx²
hence
h=2048/x²
Since the amount of material used is directly proportional to the surface area, then the material needs to be minimized by minimizing the surface area.
The surface area of the box described is
A=x²+4xh
Then substitute h into the Area equation we have
A= x² + 4x(2048/x²)
A= x² + 8192/x
We want to minimize
A
dA/dx = -8192/x² + 2 x= 0 for max or min
when dA/dx=0
dA/dx= 2x-8192/x²=0
2x=8192/x²
Hence
2x³=8192
x³=4096
x=₃√(4096)
X=16ft
Then h=2048/x²
h=2048/16²
h=8ft
The box should have base 16ft by 16ft and height 8ft
Hence the dimensions are 16 ft by 16 ft by 8 ft
Answer:
<em>About 900 cans not stamped in 3 hours</em>
Step-by-step explanation:
<u>Proportions</u>
Sally found that 150 cans were not stamped in the 1/2 hour trial run.
Following the same proportion in time, we can say:
150 cans not stamped in 1/2 hour. Multiplying by 2:
300 cans not stamped in 1 hour. Multiplying by 3:
900 cans not stamped in 3 hours
About 900 cans not stamped in 3 hours
Answer:
3 miles
Step-by-step explanation:
1/2 hour is 30 minutes so 1.5 x 2 is 3
Answer:
100
Step-by-step explanation:
We have the sum of first n terms of an AP,
Sn = n/2 [2a+(n−1)d]
Given,
36= 6/2 [2a+(6−1)d]
12=2a+5d ---------(1)
256= 16/2 [2a+(16−1)d]
32=2a+15d ---------(2)
Subtracting, (1) from (2)
32−12=2a+15d−(2a+5d)
20=10d ⟹d=2
Substituting for d in (1),
12=2a+5(2)=2(a+5)
6=a+5 ⟹a=1
∴ The sum of first 10 terms of an AP,
S10 = 10/2 [2(1)+(10−1)2]
S10 =5[2+18]
S10 =100
This is the sum of the first 10 terms.
Hope it will help.
