PLEASE HELP WIL MARK BRAINLIEST MY DUDE

A rectangular tank that is 2048 ft cubed with a square base and open top is to be constructed of sheet steel of a given thicknes

ira [324]

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

8 0

2 years ago

How many solutions does the nonlinear system of equations graphed below have?

FrozenT [24]

Answer:

A) Zero

Step-by-step explanation:

A good way to tell how many solutions there are in a system of equations, is to look at how many times the equations intersect. In this case, the equations don't intersect so there are no solutions.

If this answer is correct, please make me Brainliest!

4 0

3 years ago

1 point) Use Stoke's Theorem to evaluate ∫CF⋅dr∫CF⋅dr where F(x,y,z)=xi+yj+1(x2+y2)kF(x,y,z)=xi+yj+1(x2+y2)k and CC is the bound

Drupady [299]

Answer:

0

Step-by-step explanation:

Thinking process:

$\int\limits^a_b {cF} \, .dr$

= $\int\limits^a_b {x} \, dx \int\limits^a_b {x} \, dx curlFdS$ by Stoke's Theorem

= $\int\limits^a_b {} \, \int\limits^a_b {} \, < 12y, -12x, 0 > .< z_x,-z_y, 1 > dA\\= \int\limits^a_b {} \, \int\limits^a_b {} \, < 12y, -12x, 0 > .< 2x, 2y, 1 > dA$ since z = $25-x^{2} -y^{2} \\= 0$

3 0

3 years ago

When a breeding group of animals is introduced into a restricted area such as a wildlife reserve, the population can be expected

jasenka [17]

Answer:

A. Initially, there were 12 deer.

B. N(10) corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.

C. After 15 years, there will be 410 deer.

D. The deer population incresed by 30 specimens.

Step-by-step explanation:

$N=\frac{12.36}{0.03+0.55^t}$

The amount of deer that were initally in the reserve corresponds to the value of N when t=0

$N=\frac{12.36}{0.33+0.55^0}$

$N=\frac{12.36}{0.03+1} =\frac{12.36}{1.03} = 12$

A. Initially, there were 12 deer.

B. $N(10)=\frac{12.36}{0.03 + 0.55^t} =\frac{12.36}{0.03 + 0.0025}=\frac{12.36}{y}=380$

B. N(10) corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.

C. $N(15)=\frac{12.36}{0.03+0.55^15}=\frac{12.36}{0.03 + 0.00013}=\frac{12.36}{0.03013}= 410$

C. After 15 years, there will be 410 deer.

D. The variation on the amount of deer from the 10th year to the 15th year is given by the next expression:

ΔN=N(15)-N(10)

ΔN=410 deer - 380 deer

ΔN= 30 deer.

D. The deer population incresed by 30 specimens.

8 0

3 years ago

Alma is changing the shape of her backyard from 150 feet long by 62 feet wide to a square that has the same area. What is the pe

umka2103 [35]

Hey there!
Here's your answer: the perimeter of the redone yard would be ~193 feet

Hope this helps!

6 0

2 years ago