Note

Illustrator has a lot of power, especially when it comes to digital painting (see the box on The Artistic Dimension: Creating Vector Illustration). You can certainly produce professional-looking results in Photoshop or Illustrator, but both programs lack certain key capabilities, such as 2D shapes and rigid object placement. The most powerful of the Adobe drawing tools is Visio, which we discuss in VisioDraw.

As shown in the illustration on the next page, you can also open and save files in Illustrator in the Portable Document Format (PDF) file. So, if you don't have an Adobe graphics program installed on your computer or you don't want to run the risk of losing something in the middle of the work you're doing, you can export the work you've created to a file and open it with a graphics program that supports PDF files. However, you'll need to make sure that the program you're using can open and read what the program you're opening saved with, for example, PDF.

* **InDesign**. This industry-standard publishing program works with text, graphics, and photos to design and build print books. You can save publications in the Portable Document Format (PDF).

The next time you want to edit your images in Photoshop and see different effects you’ve never experienced before you will most likely remember this tutorial. This was the first edit I made to a photo that has been tagged as "patchwork" on the beautiful online photo editor, Adobe Stock. Many photographers prefer using this tool to Adobe Photoshop and sometimes I remember seeing people recommending this tool to others when I was in the process of making this tutorial.

I have tried to explain how to use the various tools in Photoshop Elements in an easy-to-understand way in a short tutorial. More details can be found here.







Things to Consider Before you Begin the Tutorial

There are two things to consider before starting with the tutorial. The first is whether you want to use Photoshop Elements or the full version of Photoshop.

A second important thing to consider is the size of the image. Photoshop Elements is a good alternative if your image is smaller than 2 gigabytes (GB) in size, that is smaller than 300 megapixels. Any larger size means that Photoshop Elements might not have the right tools to edit or work with the image. I recommend creating smaller images as smaller images get loaded faster on a PC.

I am a big fan of using Photoshop and I use it almost every day to edit my images. However, for my daily workflow, I don't use the full version of Photoshop. I use Photoshop Elements, so it is important that you know what you’re getting into.











Digital editing has advanced to a level where we can now make quick edits using an easy-to-use tool that makes it easy to create new effects. With the right tools and knowledge you can even make cool edits in Photoshop Elements.

There are two good online sites where you can see more examples of edits made in Photoshop Elements.

The first one is the fun, free community of Photoshop users that have created this tutorial, the Geektalk forum. You can find it here. Another fun site is Pixlr Editor.

When you use Photoshop Elements, you need to know how to use the different tools in Photoshop Elements. Below you can find the different tools in Photoshop Elements.

The tools are labeled with the little gray disks at the top of the tool.

Both tools are found in the same menu, but they appear in different places.

Buttons (red)

The buttons are
388ed7b0c7

Q:

Initializing a struct within a struct

This issue has been driving me nuts and as always, I'm looking for a better way. What is the best way to initialise a struct value within a struct?
Let's say we have a three class hierarchy.

User

id
Name
Address

Room

id
Name
UserID

User has a list of Rooms.
I want to be able to pass a room to User, then check that room's id and apply any custom settings to the room.
I can't change the way rooms are stored in my database.
I could pass the User and room id into a constructor for User and instantiate the User with the room id.
Or I could instantiate User with the Room and pass that into User. However, I'm not sure how I would create the instance of User within User.
I could also store User and Room in the same struct, but then I would have to do some intricate struct property setting, and be very careful that the Room doesn't get set to null.
I could also pass the Room into the constructor for User, but then my users will often pass the Room into the constructors of other objects for free - which would mean a lot of'setRoom' calls on every property.
What is the best way to initilise this type of scenario?
Thanks :)

A:

What is the best way to initialise this type of scenario?

Make User and Room their own classes and have properties inside of the classes to reference each other.

User has a list of Rooms. I want to be able to pass a room to User, then check that room's id and apply any custom settings to the room.

If User has a list of Rooms, you should put the rooms in the list and not create an instance of Room within a User. What if you wanted to get the room based on the id? Or create the object later? Would you have to go through all the rooms in the user and replace the id in every instance?

A:

The best way to initialize a struct within a struct, is to use a constructor that takes a reference to a struct. This way you do not have to do anything special.
E.g.
struct Room {
Room(User & user) : user_(user) {};

Q:

Integral $\int \frac{1}{(x^2+2x+9)}$

I need to find $$\int \frac{1}{(x^2+2x+9)}$$ I tried using substitution like $t=x+2$ but $x$ was outside the integral and I couldn't get $u$ and $dv$

A:

For the keyhole contour, starting on the real axis, moving right, moving down, moving back to the right, we have
$$\oint_C dz\,f(z)=\int_0^{2\pi} d\theta\,\Re\left(f(i\theta)\right)=2\pi\Im\left(f(i\theta)\right)$$
so that
$$\int_0^\infty dx\,\frac{1}{x^2+2x+9}=-i\oint_C dz\,\frac{z}{(z^2+1)(z-3i)}=2\pi i\left(z^2+1\right)\frac{z-3i}{z^3-6z^2-z+12i}|_0^\infty=-\frac{\pi}{6}(17-15i)\approx -0.011731990588$$
This general method works even for more complicated singularities (e.g., the one I use above) than poles.

A:

The substitution $x+2 = u$ transforms the integral into
$$\int_2^\infty \frac{u-2}{u^2-3u+12}\,du,$$
but this is clearly a standard integral and its value is $\frac{\pi}{6}$, so that the original value is also $\frac{\pi}{6}$.

Working mum Arianna Norton saw a training video that showed an Apache helicopter running out of control in Iran and became a bit of a dreamer; her husband was involved in a computer crash.

"How could I not look to see if we could go to Iran, which is the same country as Iraq, and get into the computer stuff and maybe bring the stuff back?" she told the Melbourne Age.

In this video you can see a helicopter, possibly an Apache, going

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