help please!!!! ASAP!!! Thank you

- anonymous

help please!!!! ASAP!!! Thank you

- schrodinger

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- anonymous

The arcs in the photo (will give later on below) to the right appear to be paths of stars rotating about the North Star. To produce this effect, the photographer set a camera on a tripod and left the shutter open for a long time. If the photographer left the shutter open for a full 24 hours, each arc would be a complete circle. You can model a star’s“rotation” in a coordinate plane. Place the North Star at the origin.Let P(1, 0) be the position of the star at the moment the camera’s shutter opens.Suppose the shutter is left open for 2 hours and 40 min.
(1)Find the angle of rotation that maps point P on to P’
(2)What are the x and y coordinates of point P’ to the nearest thousandths?
(3) Determine a translation rule that maps point P onto P

- mathstudent55

In 1 hour, the rotation is a full circle, 360 degrees of rotation
In 2 hours and 40 minutes, how many degrees of rotation is it?

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## More answers

- mathstudent55

\(\dfrac{2~ h ~40 ~mn}{24 ~h} \times 360^o\)
You can convert both measures to hours or to minutes.

- anonymous

@mathstudent55 there are 3 questions in this problem, can you help me solve?

- mathstudent55

I thought that's what I was doing above.
Above, I did part A. You just need to convert the numerator and denominator to the same units and do the calculations.
What do you get for part A?

- mathstudent55

\(\large =\dfrac{2~ h \times \frac{60~mn}{h} + ~40 ~mn}{24 ~h\times \frac{60~mn}{h}} \times 360^o\)
\(\large =\dfrac{120 ~mn + ~40 ~mn}{1440~mn} \times 360^o\)
\(\large =\dfrac{160 ~mn}{1440~mn} \times 360^o\)
\(\large = 40^o\)

- mathstudent55

For part B, use a right triangle with hypotenuse 1 unit long with an angle of 40 degrees.
Then find the lengths of the adjacent and opposite legs.
They are the x- and y-coordinates.

- anonymous

okay so how do i maps it? I bad at mapping

- anonymous

@mathstudent55 how do I maps? like graph it out?

- anonymous

@imqwerty I don't fully understand much for part B

- mathmate

To map by translation, you define a translation operator and add the difference of the respective coordinates.
For example, from A(2,4) to A'(7,3)
the translation operator is
\(t_{x,y}=(x+5, y-1)\)

- anonymous

Thank you so much @mathmate :)

- mathmate

You're welcome! :)

- anonymous

@AngusV part B only, I understand part A and C :)

- anonymous

Hold on, let me finish my game of Hearthstone lol

- anonymous

Ah well, there's two ways you can do this really. The easy way - or the one that you are likely intended to follow - would involve using a bit of trigonometry.
If what I understood from the problem is correct, here is the following situation:
|dw:1439767293119:dw|

- anonymous

That is what the guy meant by with "make a right angled triangle out of it".

- anonymous

The idea now is that you have a right angled triangle with the hypotenuse of 1 and an angle of 40 degrees. The side opposing the 40 degree angle is the y coordinate of the point - the other side is the x coordinate of the point.
To find the coordinates we must use sin and cos.
Remember how sin and cos are defines as in right angled triangles:
sin(angle) = side opposing the angle / the hypotenuse
cos(angle) = other side to the angle / the hypotenuse
Since you are allowed to use a calculator, we already know what sin(40) and cos(40) is and since the radius and the hypotenuse are one and the same (and equals 1 just like in the unit circle) - the sides are all a matter of multiplication now.

- anonymous

sin (40) = side opposing the angle / the hypotenuse = y_P' / 1
Therefore, y_P' = sin(40) * 1
Similarly, x_P' = cos(40) * 1

- anonymous

The second method (which spares you all of this trouble) involves using the notion of polar coordinates.

- anonymous

Long story short - what you are using now is the Cartezian system of coordinates - every point in space has a position defined by two coordinates: and X and an Y so to speak.
While this system is great for many things when you're dealing with lines, triangles, rectangles and so forth - it fails miserably when it comes to circles.

- anonymous

As you will probably notice from your calculations, you will end up with P' having reaaaaaaally nasty coordinates over the X and the Y axis (hence why the "round it up to the nearest whatever").

- anonymous

Instead - the better alternative in mathematics would be to use "polar coordinates".
With polar coordinates, we no longer define the position of a point in space through 2 coordinates on the X and Y axis respectively (in terms of origin) but instead define the position of a point in space according to radius of the circle and the angle it forms as that point "travels" on the circle in a trigonometric way, starting from the right side at (1,0) in Cartezian coordinates or (1,0 degrees) in Polar coordinates.

- anonymous

|dw:1439768260763:dw|

- anonymous

Let me redo the last one

- anonymous

|dw:1439768485346:dw|

- anonymous

Now, why am I telling you this

- anonymous

There is a simple way (it's easy to prove but take my word for it) to convert something from Cartezian coordinates into Polar coordinates.
x= radius * cos(angle)
y = radius * sin(angle)
So basically, in your problem you have the Polar coordinates ( radius + sin/cos of angle) and you need your Cartezian coordinates (the x and the y from the formula above).
If you want to impress your teacher simply write:
"We express the Cartezian coordinates of P' as Polar coordinates in a circle with a radius of 1 and the angle of 40 degrees.
x= radius * cos(angle)
y = radius * sin(angle)"
Done.

- anonymous

It's the same deal but you're adding a bit of flavor to it.

- anonymous

In depth description of polar coordinates here :
https://en.wikipedia.org/wiki/Polar_coordinate_system
But it is a tad bit more complex than what I just wrote. A similar system is used for spheres and such in 3D space called "spherical coordinates" but it will fry your brains out.

- anonymous

@AngusV okay :) thanks for explaining :) is more clearly now :) I better at numbers and functions. Btw, I like the way you answer :) " It's the same deal but you're adding a bit of flavor to it" by Angus V :)

- anonymous

You're welcome! Polar coordinates are fascinating.

- anonymous

@AngusV I always though it only exist in Science theory stuff, not calculating.

- anonymous

@AngusV is the radius is (1,0) I just want to make sure

- anonymous

Oh yeah, it says so in the picture as well.

- anonymous

@AngusV wait @.@ I kinda confuse a little bit: let me rewrite this:
so: 1 +cos 40= 1,766 (y)
1 + sin 40= 1,643 (x)
that's it?

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