Can someone please answer these 2 questions?

$\triangle ABC\ \text{and}\ \triangle ACD\ \text{are similar. Therefore the sides of triangles}\\\text{ are in proportion:}\\\\\dfrac{y}{9+3}=\dfrac{9}{y}\qquad\text{cross multiply}\\\\y^2=(9)(12)\\\\y^2=108\to y=\sqrt{108}\\\\y=\sqrt{36\cdot3}\\\\y=\sqrt{36}\cdot\sqrt3\\\\\boxed{y=6\sqrt3}$

4 0

3 years ago

Shawn drew a rectangle that had a width of 4.2 inches and a length of 8.1 inches. Find the perimeter of Shawn's rectangle

MArishka [77]

Answer: 24.6 inches

Step-by-step explanation: Since it's a rectangle, and not a rectangular prism, that means it's 2D. Length is the long sides on the top and bottom, and width is the shorter sides on it's left and right. So since there's two of each, just do 8.1 + 8.1 + 4.2 + 4.2 and you have your answer!

3 0

3 years ago

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:

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

4 0

3 years ago