Complete the equations:<br> 300 = 3 hundreds<br> ?

Brianna spots an airplane on radar that is currently approaching in a straight line,

yuradex [85]

Using the slope concept, it is found that the distance the plane traveled from point A to point B is of 14,475 feet.

<h3>What is a slope?</h3>

The slope is given by the <u>vertical change divided by the horizontal change</u>.

It's also the tangent of the angle of depression.

At point A:

Altitude of 6700 feet, which is the vertical change.

The horizontal position we want to find is $x_A$ .

The angle is of 16º.

Hence:

$\tan{16^{\circ}} = \frac{6700}{x_A}$

$x_A = \frac{6700}{\tan{16^{\circ}}}$

$x_A = 23365.7$

At point B:

Altitude of 6700 feet, which is the vertical change.

The horizontal position we want to find is $x_B$ .

The angle is of 37º.

Hence:

$\tan{37^{\circ}} = \frac{6700}{x_B}$

$x_B = \frac{6700}{\tan{37^{\circ}}}$

$x_B = 8891.2$

Then, the distance is of:

$d = x_A - x_B = 23365.7 - 8891.2 \approx 14475$

You can learn more about the slope concept at brainly.com/question/26342863

8 0

2 years ago

Can someone please help me!!

grandymaker [24]

Answer:

rotation 90° CW about the origin we defined

Step-by-step explanation:

The image has the same orientation (clockwise order of vertices) as the pre-image, so an even number of reflections is involved. These cannot be reflections across the x- or y-axes. The longest edge has an east-west orientation in the pre-image and a south-north orientation in the image. Any reflections must have a net effect of a rotation 90° CW.

Equivalently, the image is rotated 90° clockwise from the pre-image.

With a suitable choice of center of rotation, the image can be obtained with a single 90° CW rotation transformation.

If we define the lower left corner of the pre-image as having coordinates (-1, 2), then a 90° CW rotation about (0, 0) will yield the image.

3 0

3 years ago

Use a surface integral to find the general formula for the surface area of a cone with height latex: h and base radius latex: a(

BlackZzzverrR [31]

We can parameterize this part of a cone by

$\mathbf s(u,v)=\left\langle u\cos v,u\sin v,\dfrac hau\right\rangle$

with $0\le u\le a$ and $0\le v\le2\pi$ . Then

$\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|\,\mathrm du\,\mathrm dv=\sqrt{1+\dfrac{h^2}{a^2}}u\,\mathrm du\,\mathrm dv$

The area of this surface (call it $\mathcal S$ ) is then

$\displaystyle\iint_{\mathcal S}\mathrm dS=\sqrt{1+\frac{h^2}{a^2}}\int_{v=0}^{v=2\pi}\int_{u=0}^{u=a}u\,\mathrm du\,\mathrm dv=a^2\sqrt{1+\frac{h^2}{a^2}}\pi=a\sqrt{a^2+h^2}\pi$

6 0

3 years ago

The salary of teachers in a particular school district is normally distributed with a mean of $50,000 and a standard deviation o

andrey2020 [161]

Answer:

D. 54,900

Step-by-step explanation:

We have been given that the salary of teachers in a particular school district is normally distributed with a mean of $50,000 and a standard deviation of $2,500.

To solve our given problem, we need to find the sample score using z-score formula and normal distribution table.

First of all, we will find z-score corresponding to probability $0.975(1-0.025)$ using normal distribution table.