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Dmitry [639]
3 years ago
5

Suppose you make napkin rings by drilling holes with different diameters through two wooden balls (which also have different dia

meters). You discover that both napkin rings have the same height 5h. Use cylindrical shells to compute the volume V of a napkin ring of height 5 h created by drilling a hole with radius r through the center of a sphere of radius R and express the answer in terms of h .
Mathematics
1 answer:
Alexus [3.1K]3 years ago
5 0

Answer:

V = 1/6 π ( 5h)^3

Step-by-step explanation:

Height of napkin rings = 5h

<u>Compute the volume V of a napkin ring</u>

let a = 5

radius = r

express answer in terms of h

attached below is the detailed solution

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In the diagram below, circle O is circumscribed about quadrilateral DEFG. What is the value of x?
earnstyle [38]

Answer:

A. 61^{o}  

Step-by-step explanation:

We are told that circle O is circumscribed about quadrilateral DEFG.

Since we know that the opposite angles of quadrilateral inscribed inside a circle are supplementary.

Let us find value of x by equating the measure of x and its opposite angle with 180.

x+119^{o}=180^{o}

x=180^{o}-119^{o}

x=61^{o}

Therefore, the value of x is 61 degrees and option A is the correct choice.

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4 years ago
Read 2 more answers
Y=Cosx.arcsinx<br> What is the solution
Alinara [238K]

Step-by-step explanation:

Given: y = cos(sin^{-1}x))

As the domain for the inverse sine function is 1  ≤  x  ≤  1  as this is the range for the sine function.

The range for the function is the same as the range for the cosine function, 1  ≤  y  ≤  1

As

sin^{2}x +cos^{2}x = 1

So, using the identity cos(x) = \pm \sqrt{1-sin^{2}(x) }

y = \pm \sqrt{1-sin^{2}(sin^{-1}(x)) }

As the sin and inverse sin function do cancel each other, so only x² is left.

Hence,

y = \pm \sqrt{1-x^{2} ; -1\leq x\leq 1

<em>Keywords: trigonometric function </em>

<em>Learn more trigonometric function from brainly.com/question/12551115</em>

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4 years ago
Use the method of Lagrange multipliers to find the dimensions of the rectangle of greatest area that can be inscribed in the ell
Tanzania [10]

Answer:

Length (parallel to the x-axis): 2 \sqrt{2};

Height (parallel to the y-axis): 4\sqrt{2}.

Step-by-step explanation:

Let the top-right vertice of this rectangle (x,y). x, y >0. The opposite vertice will be at (-x, -y). The length the rectangle will be 2x while its height will be 2y.

Function that needs to be maximized: f(x, y) = (2x)(2y) = 4xy.

The rectangle is inscribed in the ellipse. As a result, all its vertices shall be on the ellipse. In other words, they should satisfy the equation for the ellipse. Hence that equation will be the equation for the constraint on x and y.

For Lagrange's Multipliers to work, the constraint shall be in the form: g(x, y) =k. In this case

\displaystyle g(x, y) = \frac{x^{2}}{4} + \frac{y^{2}}{16}.

Start by finding the first derivatives of f(x, y) and g(x, y)with respect to x and y, respectively:

  • f_x = y,
  • f_y = x.
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This method asks for a non-zero constant, \lambda, to satisfy the equations:

f_x = \lambda g_x, and

f_y = \lambda g_y.

(Note that this method still applies even if there are more than two variables.)

That's two equations for three variables. Don't panic. The constraint itself acts as the third equation of this system:

g(x, y) = k.

\displaystyle \left\{ \begin{aligned} &y = \frac{\lambda x}{2} && (a)\\ &x = \frac{\lambda y}{8} && (b)\\ & \frac{x^{2}}{4} + \frac{y^{2}}{16} = 1 && (c)\end{aligned}\right..

Replace the y in equation (b) with the right-hand side of equation (b).

\displaystyle x = \lambda \frac{\lambda \cdot \dfrac{x}{2}}{8} = \frac{\lambda^{2} x}{16}.

Before dividing both sides by x, make sure whether x = 0.

If x = 0, the area of the rectangle will equal to zero. That's likely not a solution.

If x \neq 0, divide both sides by x, \lambda = \pm 4. Hence by equation (b), y = 2x. Replace the y in equation (c) with this expression to obtain (given that x, y >0) x = \sqrt{2}. Hence y = 2x = 2\sqrt{2}. The length of the rectangle will be 2x = 2\sqrt{2} while the height will be 2y = 4\sqrt{2}. If there's more than one possible solutions, evaluate the function that needs to be maximized at each point. Choose the point that gives the maximum value.

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WITCHER [35]

Answer:

Area = 15.36 ft

Step-by-step explanation:

First find the area of the dotted line triangle using the formula: base x height x 1/2, which would be 1.2ft x 12.8 x 1/2 = 7.68ft.

Next, add another triangle at the top, mirroring the shown triangle, of the parallelogram.

Then, find the area of the shape as though it were a rectangle. (base x height) - 2.4 x 12.8 = 30.72.

Next, multiply the area of the invisible triangle by 2, 7.68 x 2 = 15.36ft

and Finally, subtract the area of the rectangle shape of the the sum of the two invisible triangles, 30.72 - 15.36 = 15.36ft

The area of the parallelogram is 15.36 ft.

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3 years ago
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