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nadya68 [22]
3 years ago
12

Using a table of values, determine the solution to the equation below to the nearest fourth of a unit.

Mathematics
1 answer:
s344n2d4d5 [400]3 years ago
3 0

Answer:

answer is 1

Step-by-step explanation:

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2/4 plus 1/3 as a fraction
QveST [7]

Answer: 5/6

Step-by-step explanation: trust me on this

7 0
3 years ago
Read 2 more answers
The base of an aquarium with given volume V is made of slate and the sides are made of glass. If the slate costs seven times as
Olin [163]

Answer:

x = ∛(2V/7)

y = ∛(2V/7)

z = 3.5 [∛(2V/7)]

{x,y,z} = { ∛(2V/7), ∛(2V/7), 3.5[∛(2V/7)] }

Step-by-step explanation:

The aquarium is a cuboid open at the top.

Let the dimensions of the base of the aquarium be x and y.

The height of the aquarium is then z.

The volume of the aquarium is then

V = xyz

Area of the base of the aquarium = xy

Area of the other faces = 2xz + 2yz

The problem is to now minimize the value of the cost function.

The cost of the area of the base per area is seven times the cost of any other face per area.

With the right assumption that the cost of the other faces per area is 1 currency units, then, the cost of the base of the aquarium per area would then be 7 currency units.

Cost of the base of the aquarium = 7xy

cost of the other faces = 2xz + 2yz

Total cost function = 7xy + 2xz + 2yz

C(x,y,z) = 7xy + 2xz + 2yz

We're to minimize this function subject to the constraint that

xyz = V

The constraint can be rewritten as

xyz - V = 0

Using Lagrange multiplier, we then write the equation in Lagrange form

Lagrange function = Function - λ(constraint)

where λ = Lagrange factor, which can be a function of x, y and z

L(x,y,z) = 7xy + 2xz + 2yz - λ(xyz - V)

We then take the partial derivatives of the Lagrange function with respect to x, y, z and λ. Because these are turning points and at the turning point, each of the partial derivatives is equal to 0.

(∂L/∂x) = 7y + 2z - λyz = 0

λ = (7y + 2z)/yz = (7/z) + (2/y) (eqn 1)

(∂L/∂y) = 7x + 2z - λxz = 0

λ = (7x + 2z)/xz = (7/z) + (2/x) (eqn 2)

(∂L/∂z) = 2x + 2y - λxy = 0

λ = (2x + 2y)/xy = (2/y) + (2/x) (eqn 3)

(∂L/∂λ) = xyz - V = 0

We can then equate the values of λ from the first 3 partial derivatives and solve for the values of x, y and z

(eqn 1) = (eqn 2)

(7/z) + (2/y) = (7/z) + (2/x)

(2/y) = (2/x)

y = x

Also,

(eqn 1) = (eqn 3)

(7/z) + (2/x) = (2/y) + (2/x)

(7/z) = (2/y)

z = (7y/2)

Hence, at the point where the box has minimal area,

y = x,

z = (7y/2) = (7x/2)

We can then substitute those into the constraint equation for y and z

xyz = V

x(x)(7x/2) = V

(7x³/2) = V

x³ = (2V/7)

x = ∛(2V/7)

y = x = ∛(2V/7)

z = (7x/2) = 3.5 [∛(2V/7)]

The values of x, y and z in terms of the volume that minimizes the cost function are

{x,y,z} = {∛(2V/7), ∛(2V/7), 3.5[∛(2V/7)]}

Hope this Helps!!!

7 0
3 years ago
Pls help I have thirty minutes to get this done
Radda [10]

Answer:

1. D

2. F

3. B

4. A

5. A

6. F

Step-by-step explanation:

Solve for X

8 0
2 years ago
Add.<br><br> (6x3+3x2−2)+(x3−5x2−3)<br><br> Express the answer in standard form.
sergiy2304 [10]

Answer:

7x^3-2x^2-5

Step-by-step explanation:

We need to add the two terms.

(6x^3+3x^2-2)+(x^3-5x^2-3)

Solving,

Combine the like terms and adding those terms

(6x^3+3x^2-2)+(x^3-5x^2-3)\\=6x^3+3x^2-2+x^3-5x^2-3\\=6x^3+x^3+3x^2-5x^2-2-3\\=7x^3-2x^2-5

So, the answer is:

7x^3-2x^2-5

6 0
3 years ago
How would you describe domain
irina [24]

Answer:

The domain represents the x-axis, more specifically, what is happening on the x-axis. So when looking at a graph, if you are asked to find the domain think about what the x-axis looks like. I put an image in to show you an example of what the domain would be for a parabola:

So on the left side of the x-axis, we can see that the line stretches out into negative infinity, so the domain would begin at negative infinity.

On the right side of the x-axis, the parabola also stretches into positive infinity, so here the domain would be (negative infinity, positive infinity), because it goes to both ends forever.

6 0
3 years ago
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