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ella [17]
3 years ago
15

The wavelength​ w, in meters per​ cycle, of a musical note is given by w=r/f, where r is the speed of the sound in meters per se

cond and f is the frequency in cycles per second.
The speed of sound in air is 344 m/ sec. What is the wavelength of a note whose frequency in air is 26 cycles per​ second? Round to the nearest tenth of a meter per cycle

A) 318.0 meters per cycle
B) 13.2 meters per cycle
C) 0.1 meters per cycle
D) 8,944.0 meters per cycle
Mathematics
1 answer:
Serjik [45]3 years ago
6 0
B. Plug in the numbers and use common sense.
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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.
  • \displaystyle g_x = \frac{x}{2},
  • \displaystyle g_y = \frac{y}{8}.

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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What is the range of the function shown<br> on the graph above?
OverLord2011 [107]

Answer:

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Step-by-step explanation:

look at the endpoints of the segment; (7, 8) and (-2, -9)

the y-values represent the range and they go from -9 to 8, inclusive

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Kyle built a tree house 4 ft. by 6 ft. What was the area of the tree house?
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Answer:

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Step-by-step explanation:

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Ahat [919]
Largest doll = 12 inches
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3rd doll = (12 x 0.7) x 0.7
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So, the height of the nth doll can be expressed as: 12 x 0.7^(n-1)

Example: 4th doll = 12 x 0.7^(4-1) = 12 x 0.7^3
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Answer:

It is already simplified

Step-by-step explanation:

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