An open-topped glass aquarium with a square base is designed to hold ~62.5~ 62.5 space, 62, point, 5, space cubic feet of water. what is the minimum possible exterior surface area of the aquarium?

75 square feet

Step-by-step explanation:

The aquarium is designed to hold 62.5cubic feet

Since the aquarium has a square base, assume the aquaruim is a square prism

Let X be the lenght

Let y be the height

x^2.y = 62.5

y = 62.5x^2

The area of each side of the aquarium is xy. Since there are four sides, the total area of the side= 4xy

The area of the bottom of the aquarium = x^2

Total surface area = 4xy + x^2

Put y = 62.5/x^2 into the total surface area equation

A = 4x(62.5/x^2) + x^2

A= 250/x + x^2

Differentiate A with respect to x

dA/dx = -250(x^-2) + 2x

dA/dx = 0

0 = 2x - 250x^-2

2x = 250x^-2

2x =250/ x^2

2x^3 = 250

x^3 = 250/2

x^3 = 125

x = cbrt (125)

x = 5

Put x= 5 into A = x^2 + 250x^-1

A= 5^2 + 250(5^-1)

A= 25 + 250/5

A = 25 + 50

A = 75 square feet

The surface area of a solid object can be determine by measuring the total area that surface of the object cover.

The minimum possible exterior surface area of the aquarium is .

Given:

The aquarium designed hold .

Let assume the aquarium is square prism.

Let is height and is length.

The aquarium designed hold . Write the system equation.

The area of four side of aquarium is as follows,

The bottom area of aquarium is as follows,

The total area is as follows,

Put in above expression.

Differentiate with respect to .

For the minimum possible exterior surface area .

Substitute to find the area.

Thus, the minimum possible exterior surface area of the aquarium is .

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hey there!

12 is twice as much as 6 because 6 goes into 12 2 times.

since the rabbit is twice as tall, we should multiply the weight of the first rabbit, 0.25, by 2.

0.25 x 2 = 0.5

therefore, the 12 inch chocolate rabbit weighs 0.5 lb, or half a pound.

hope this !

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2