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amid [387]
2 years ago
6

An aircraft on a reconnaissance mission takes off from its home base and flies 550 miles at a bearing of S 46° E to a location i

n the sea. It then flies 483 miles from the sea at a bearing of S 55° W to another location. Finally, the aircraft flies straight back from the second location to its base. What is the total distance it flies, rounded to the nearest mile?
A.
550 miles
B.
1,033 miles
C.
1,583 miles
D.
1,692 miles
E.
2,210 miles
Mathematics
1 answer:
Alexandra [31]2 years ago
3 0
<h3>Answer: D) 1692 miles</h3>

============================================

Explanation:

A diagram is very handy for this type of problem. Refer to the diagram below (attached image). Here are the steps used to form the diagram

  1. Plot point A and make that the home base. Plot point D directly south of point A. I'm skipping B and C because those will be used to form the triangle later.
  2. Imagine you are standing at point A and looking south at point D. Turn 46 degrees toward the east direction (this is what the notation "bearing of S 46° E" means). Then move 550 miles from A to land on point B. This is the location where the plane makes its first turn.
  3. Plot the points E and F, which are north and south of point B.
  4. Imagine we are at point B and looking directly south at point F. Turn 55 degrees toward the west, which is due to the "S 55° W" and then move 483 miles to arrive at point C. This is the plane's location where it turns around to go back to home base.

The diagram below shows all of this visually summarized. We have the segments

  • AB = 550 miles
  • BC = 483 miles
  • AC = x miles

And we have these angles

  • Angle DAB = 46 degrees and Angle ABE = 46 degrees (blue)
  • Angle ABC = 79 degrees (red)
  • Angle ACB = 55 degrees and Angle FBC = 55 degrees (green)

Angle ABC comes from the fact that the blue and green angles add with the red angle to get 180. So we basically need to solve 46+y+55 = 180 which leads to y = 79.

------------------------------

Once the diagram is set up, we'll use the law of cosines to find x.

Focus on triangle ABC. Ignore points D,E,F. Ignore any extra lines as well.

The interior angles of triangle ABC are

  • A = 46 degrees
  • B = 79 degrees
  • C = 55 degrees

The sides opposite the angles are

  • a = 483 (opposite angle A)
  • b = x (opposite angle B)
  • c = 550 (opposite angle C)

Now we can apply the law of cosines

b^2 = a^2 + c^2 - 2*a*c*cos(B)

x^2 = 483^2 + 550^2 - 2*483*550*cos(79)

x^2 = 434412.180756441

x = sqrt(434412.180756441)

x = 659.099522649229

x = 659

Side AC, or CA, is roughly 659 miles long.

------------

To find the total distance the plane travels, we add up the three sides of the triangle (ie we find the perimeter)

Distance Traveled = AB + BC + CA

Distance Traveled = 550 + 483 + 659

Distance Traveled = 1692 miles

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Let φ(x, y) = arctan (y/x) .
Alexandra [31]

Answer:

a) \large F(x,y)=(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})

b) \large \mathbb{R}^2-\{(0,0)\}

c) the points of the form (x, -x) for x≠0

Step-by-step explanation:

a)

If φ(x, y) = arctan (y/x), the vector field F = ∇φ would be

\large F(x,y)=(\frac{\partial \phi}{\partial x},\frac{\partial \phi}{\partial y})

On one hand we have,

\large \frac{\partial \phi}{\partial x}=\frac{\partial arctan(y/x)}{\partial x}=\frac{-y/x^2}{1+(y/x)^2}=-\frac{y/x^2}{1+y^2/x^2}=\\\\=-\frac{y/x^2}{(x^2+y^2)/x^2}=-\frac{y}{x^2+y^2}

On the other hand,

\large \frac{\partial \phi}{\partial y}=\frac{\partial arctan(y/x)}{\partial y}=\frac{1/x}{1+(y/x)^2}=\frac{1/x}{1+y^2/x^2}=\\\\=\frac{1/x}{(x^2+y^2)/x^2}=\frac{x}{x^2+y^2}

So

\large F(x,y)=(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})

b)

The domain of definition of F is  

\large \mathbb{R}^2-\{(0,0)\}

i.e., all the plane X-Y except the (0,0)

c)

Here we want to find all the points such that

\large (-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})=(k,k)

where k is a real number other than 0.

But this means

\large -\frac{y}{x^2+y^2}=\frac{x}{x^2+y^2}\Rightarrow y=-x

So, all the points in the line y = -x except (0,0) are parallel to the vector field F, that is, the points (x, -x) with x≠ 0

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3 years ago
HELP PLEASE 50 points !!! Given a polynomial function describe the effects on the Y intercept, region where the graph is incre
Gwar [14]

Even function:

A function is said to be even if its graph is symmetric with respect to the , that is:

Odd function:

A function is said to be odd if its graph is symmetric with respect to the origin, that is:

So let's analyze each question for each type of functions using examples of polynomial functions. Thus:

FOR EVEN FUNCTIONS:

1. When  becomes  

1.1 Effects on the y-intercept

We need to find out the effects on the y-intercept when shifting the function  into:

We know that the graph  intersects the y-axis when , therefore:

So:

So the y-intercept of  is one unit less than the y-intercept of

1.2. Effects on the regions where the graph is increasing and decreasing

Given that you are shifting the graph downward on the y-axis, there is no any effect on the intervals of the domain. The function  increases and decreases in the same intervals of

1.3 The end behavior when the following changes are made.

The function is shifted one unit downward, so each point of  has the same x-coordinate but the output is one unit less than the output of . Thus, each point will be sketched as:

FOR ODD FUNCTIONS:

2. When  becomes  

2.1 Effects on the y-intercept

In this case happens the same as in the previous case. The new y-intercept is one unit less. So the graph is shifted one unit downward again.

An example is shown in Figure 1. The graph in blue is the function:

and the function in red is:

So you can see that:

2.2. Effects on the regions where the graph is increasing and decreasing

The effects are the same just as in the previous case. So the new function increases and decreases in the same intervals of

In Figure 1 you can see that both functions increase at:

and decrease at:

2.3 The end behavior when the following changes are made.

It happens the same, the output is one unit less than the output of . So, you can write the points just as they were written before.

So you can realize this concept by taking a point with the same x-coordinate of both graphs in Figure 1.

FOR EVEN FUNCTIONS:

3. When  becomes  

3.1 Effects on the y-intercept

We need to find out the effects on the y-intercept when shifting the function  into:

As we know, the graph  intersects the y-axis when , therefore:

And:

So the new y-intercept is the negative of the previous intercept shifted one unit upward.

3.2. Effects on the regions where the graph is increasing and decreasing

In the intervals when the function  increases, the function  decreases. On the other hand, in the intervals when the function  decreases, the function  increases.

3.3 The end behavior when the following changes are made.

Each point of the function  has the same x-coordinate just as the function  and the y-coordinate is the negative of the previous coordinate shifted one unit upward, that is:

FOR ODD FUNCTIONS:

4. When  becomes  

4.1 Effects on the y-intercept

In this case happens the same as in the previous case. The new y-intercept is the negative of the previous intercept shifted one unit upward.

4.2. Effects on the regions where the graph is increasing and decreasing

In this case it happens the same. So in the intervals when the function  increases, the function  decreases. On the other hand, in the intervals when the function  decreases, the function  increases.

4.3 The end behavior when the following changes are made.

Similarly, each point of the function  has the same x-coordinate just as the function  and the y-coordinate is the negative of the previous coordinate shifted one unit upward.

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2. Use (&lt;, &gt;, =) to compare 23 and 34.
lara31 [8.8K]

Answer:

Step--step explanation:

2/3 = 0.67

3/4= 0.75

So 2/3 < 3/4

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Step-by-step explanation:

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