Plssssss help me! What is the area of the shaded portion to the nearest hundredth?

Mathematics

1 answer:

Katarina [22]4 years ago

7 0

Step-by-step explanation:

A1= area of the rectangle

A1= 4yd • 10yd

A1= 40 square yd

A2= area of the part of the circular section

A2= 180°• pi • 4 square yd/360°

A2= 180° • 16yd • pi/ 360°

A2= 16yd•pi/2

A2= 8yd•pi

A2= 25,12yd

A= areo of rhe shaded portion

A= A1-A2

A= 40yd-25,12yd

A= 14,88yd

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A certain transverse wave is described by y(x,t)=bcos[2π(xl−tτ)], where b = 5.90 mm , l = 29.0 cm , and τ = 3.90×10−2 s .

Vitek1552 [10]

Part A:

The general form of the equation of a transverse wave is given by:

$y(x,t)=A\cos\left[2\pi\left( \frac{x}{\lambda} - \frac{t}{T} \right)\right]$

where A is the amplitude, $\lambda$ is the wavelength, and T is the period.

Given that a certain transverse wave is described by:

$y(x,t)=bcos[2\pi(xl-t\tau)]$ , where b = 5.90 mm , l = 29.0 cm , and \tau = 3.90\times10^{-2} s

Thus, the amplitude is b = 5.90 mm = 5.9\times10^{-3} \ m

Part B:

The general form of the equation of a transverse wave is given by:

$y(x,t)=A\cos\left[2\pi\left( \frac{x}{\lambda} - \frac{t}{T} \right)\right]$

where A is the amplitude, $\lambda$ is the wavelength, and T is the period.

Given that a certain transverse wave is described by:

$y(x,t)=bcos[2\pi\left(\frac{x}{l}-\frac{t}{tau}\right)\right]$ , where b = 5.90 mm , l = 29.0 cm , and \tau = 3.90\times10^{-2} s

Thus,

$y(x,t)=bcos[2\pi\left(\frac{x}{l}-\frac{t}{tau}\right)\right\\ \\ \frac{1}{\lambda} = \frac{1}{l} \\ \\ \Rightarrow\lambda= l =28.0 \ cm=\bold{2.8\times10^{-1}}$

Part C:

The general form of the equation of a transverse wave is given by:

$y(x,t)=A\cos\left[2\pi\left( \frac{x}{\lambda} - \frac{t}{T} \right)\right]$

where A is the amplitude, $\lambda$ is the wavelength, and T is the period.

Given that a certain transverse wave is described by:

$y(x,t)=bcos[2\pi\left(\frac{x}{l}-\frac{t}{tau}\right)\right]$ , where b = 5.90 mm , l = 29.0 cm , and \tau = 3.90\times10^{-2} s

The wave's frequency, f, is given by:

 $f= \frac{1}{T} = \frac{1}{\tau} = \frac{1}{3.40\times10^{-2}} =\bold{29.4 \ Hz}$

Part D:

Given that the the wavelength is $2.8\times10^{-1} \ m$ and that the wave's frequency is 29.4 Hz

The wave's speed of propagation, v, is given by:

$v=f\lambda=29.4(2.8\times10^{-1})=8.232 \ m/s$