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Using the Pencil Tool The _Pencil Tool_, shown in Figure 7-1, is the most powerful tool in Photoshop. FIGURE 7-1: You can use the Pencil tool to manipulate every pixel in an image. Like a traditional pencil, this tool is at your disposal. But unlike a pencil, the Pencil tool is nondestructive, enabling you to draw on your image, edit, and then erase the line with a single click. After you create your image, you can always go back to erase the path, restore the original image, or share the final result. The Pencil tool is extremely versatile, enabling you to * Draw paths with straight, curved, and bezier lines. * Transform your shape using the Pen tool. * Draw new paths from previous points. * Make selections from the Pen tool. * Draw pixel sized squares and rectangles. * Erase individual pixels. This tool enables you to create and edit almost every shape. The Pen tool also serves as a vector-editing tool in which you can stretch or shrink the shape, add corners, and so on, as opposed to a pixel-editing tool. You need to keep in mind, however, that you use the Pen tool like a traditional pencil, though you use the Pen tool to draw paths (which you often call lines) instead of the traditional x marks for drawing. (Illustrator places the x marks automatically when you start the Pen tool. You can turn them off so you see the lines created by the Pen tool or Click here to learn more about shortcuts for selecting objects,

The five most useful features of Photoshop are included in this guide. 1. Keyboard shortcuts Any graphic designer will tell you that knowing keyboard shortcuts is essential to speed up your workflow. In Photoshop CC, the keyboard shortcuts are grouped by modules. Some of the most commonly used shortcuts are: Sorting – Ctrl + click Saving – Ctrl + S Channels – Ctrl + E Layers – Alt + Ctrl + Shift + N or Ctrl + Shift + N Filters – Ctrl + Alt + Shift + N or Ctrl + F Curves – Ctrl + D Clone – Ctrl + Alt + Shift + J Blur – Shift + Alt + E Gamma – Ctrl + I Adjustment – Shift + F These keyboard shortcuts can be easily found online or in this PDF guide. 2. Layer Composition The composition of layers can be very complex. You can adjust the opacity of a layer by clicking it and choosing Image > Adjustments > Opacity or Fade > Opacity. You can increase the value of a layer by dragging its selection and clicking on the image. If you want to decrease the opacity of one or more layers, drag them to the layer mask icon. You can add new layers by clicking on the Add Layer icon. You can add the same layer more than once by clicking on this icon and dragging an existing layer. You can delete layers by clicking on the trash icon. 3. HSL and RGB When you load an image, the HSL palette will appear, which enables you to adjust the saturation, lightness and hue of the image. You can adjust the HSL and RGB values of any color by clicking on it and choosing from the appropriate palette. The black, white and gray tiles at the bottom of the screen are useful when you are adjusting the color of an image and want to keep the image black and white. 4. Masking Masking allows you to lighten or darken any part of an image. This includes highlights, shadows, the entire image, or only a portion of it. You can mask a layer or multiply a layer by pressing Shift + Ctrl + I or Alt + Shift + I. Masking is easy because it consists of a paint bucket with an ellipse which you can click on and drag over the area of the image where 05a79cecff

Q: Reducing $ \sum_{n=1}^\infty \frac{1}{n^4+n} $ using contour integration. Using contour integration, the sum $$ \sum_{n=1}^\infty \frac{1}{n^4+n} $$ can be calculated to be $-\frac{3 \pi ^3}{4 \times 32}$. I have proven this with the substitution $x= \tan(\theta/2)$. However, I have also been asked to use contour integration, and I have never used it before. What would be the best way to go about calculating this sum using contour integration? A: Let $f(z) = 1/(z+1)$, $U = \mathbb{C} \setminus [1,\infty)$, $V = (-\infty,1)$ and $R$ be any contour around the real line connecting $V$ and $U$. Using the Residue Theorem, we get that $$\sum_{n=1}^{\infty} \frac{1}{n^4+n} = \frac{1}{2\pi i} \int_{\frac{1}{2}-i\infty}^{\frac{1}{2}+i\infty} \frac{1}{x^3(x+1)} \, dx.$$ Notice that $f(z)$ is analytic and vanishes on the contour $R$. Thus, the only singularity inside the contour is at $x=0$, and since $x+1$ is analytic at $x=0$, we can apply Cauchy’s Integral Theorem, which says that $$\frac{1}{2\pi i} \int_{\frac{1}{2}-i\infty}^{\frac{1}{2}+i\infty} \frac{1}{x^3(x+1)} \, dx = \frac{1}{2\pi i} \int_{\gamma} \frac{1}{x^3(x+1)} \, dx.$$ This is an $n$-fold integral of a function $h(x)$ analytic in $U$ and vanishes on

Riceol’s mortar Drimiel When you roll on the Drimiel mortar tile, you can roll a new tile equal to the cost of this Drimiel tile. If there are no other customers present, the Doctor may offer to treat the user (by spending 1 dust point on a roll). If the user pays the Doctor for treatment, the Doctor instantly spends 1 dust point on Drimiel’s treatment. The Dust Drimiel is a herb of strength, vitality, and mental acuity, used to enhance people’s mental and physical health. All doctors, when creating treatment packs, have a 15% chance to include a dust packet containing Drimiel. The dust packet is added to the treatment pack in the same fashion as other Dust Tokens. The user can pay either to receive the strength and vitality of Drimiel, or to acquire the mind-fogginess of phanaephobia. \otimes {{\cal L}}\right)\otimes E(\alpha)$. Recall that $E_r({{\cal E}}) = E_r({{\cal E}}^\vee)$ is also a direct sum of line bundles and $g^\vee$ is a surjection of vector bundles, so we can assume that our linearisation ${{\cal L}}$ is locally free. Now the result follows from Lemma \[lem:contr\]. Generalised eigencurve {#sub:geneigenc} ———————- \[defn:geneigenc\] The *generalised eigencurve* $X_G$ is the moduli space of ${{\Bbb G}}_m$-equivariant semistable logarithmic connections (as in definition \[def:sslogconn\]) on $X_{/k}$ equipped with level $n$ structure, where $n\in {{\Bbb N}}$, such that $\exp(2\pi \sqrt{ -1}/n)$ is not an eigenvalue of the connection at any point. As before, the universal family is a vector bundle ${\mathcal{F}}$ over the scheme $X_G^{ss}$ of semistable objects. The structure of ${\mathcal{F}}$ is identical to that of $E_r({\mathcal{F}}_\bul

In The Game This guide will show you how to create an ARTS collection in Civilization VI. The guide includes how to properly set up your game and a guide on how to add graphics, models and sound to the collection in addition to tips and tricks for editing and using the collection. Create the collection Note: You can use any version of Civ VI. When you first start up the collection wizard, you will be greeted with a large list of resources you have access to. There are several collections available

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