I have an older specialized bike from 2001 that I ride and race on with a gearing of Shimano 53/39 and 9-speed cassette 12/26.  On my regular weekly rides I end up maxing out on my gears and can't get any more speed. How do I get more speed out of the bike with only 9 gears?  I don't know if an 11/? cassette will make a difference, if it is possible to get a bigger chainring, or if my technology is so old I should just get a new bike?
I enjoy going fast and my current bicycle tops too early, what kind of upgrades would I need to make to achieve a taller final gear?
Height of colorbox How do I set up shared folders in a Windows Virtualbox guest? I have installed VirtualBox OSE on Ubuntu 11.04 and Windows XP pro as the virtual machine. Now I want to have a shared folder which can be accessed on both, my host Ubuntu machine, and my WinXP guest virtual machine.  How can I do that? Check a string to see if all characters are hexadecimal values PC Blue Screens and then instantly reboots. How do I stop this or find the error later? Distance from a point to a plane in normed spaces How can we calculate the distance from a point to a plane in normed spaces ? where we are not inner product, for example:  Calculate the distance in $(C[0,1],||\cdot||)$ endowed with the supremum norm $$||f||=\sup_{t\in[0,1]}|f(t)|$$ from the function $f(t)=t^2$ to the linear hull of the functions $g(t)=t$  and  $h(t)=\sin(t)$  Any hints would be appreciated. Can \includegraphics be used to change an image color? I'm including a number of PDF images in my document using \includegraphics.  I'm wondering if there's a way for LaTeX to adjust their color profile, hopefully the same way it can flip images upside down, resize, and crop, as here:   Let me detail a little more my desired outcome, in case there's another way to achieve this: The images are brightly colored---I'd like to reduce color levels, or possibly render them grayscale, and I'd prefer to not have to do external pre-processing, if this is possible.  I'm using XeTeX. The PHB on p. 195 (or the ) says:     When you take the Attack action and attack with a light melee weapon that you’re holding in one hand, you can use a bonus action to attack with a different light melee weapon that you’re holding in the other hand. You don’t add your ability modifier to the damage of the bonus attack, unless that modifier is negative.   The rules do not make a distinction between "main-hand" and "off-hand" weapons.  My character dual wields a shortsword with her right hand and a dagger with her left hand.  Could my character attack with her left hand dagger first, doing 1d4 + STR/DEX mod damage, then attack with her right hand shortsword, doing 1d6 damage?  Does it matter which weapon I attack with first when two-weapon fighting? Equidistant points across Bezier curves Renaming user name I made a mistake in the username at the time of installation. So I changed the name of the user after logging in by creating a guest account with SUDO rights. But when I checked the list of users using ls /home command I am not seeing the new name for the user instead Old name is appearing. But I use the new name to login.   Is it like I need to change the name inside Home directory separately? Also , if there is any other changes needs to be done .. ? Buying from Play store using a credit card with foreign billing address Is it possible to use a credit card with a foreign billing address to buy items from a local Play store? I have a Visa credit card from a German bank with a billing address in Germany registered in my Google Wallet. I have successfully used my Wallet to pay for online purchases from several online service providers in different foreign currencies so I am sure the credit card is correctly registered in my Wallet account. However, while physically located in the UK, I cannot complete any purchase from the UK Play store. At the point when I need to select my preferred payment method from my Wallet account, it shows my registered credit card greyed out with the text "Unsupported Payment method" next to it. Is it possible to tell a Valyrian steel sword from a very well crafted steel sword, by sight alone How to correctly add a path to PATH? "Was going to" versus "would be" as the past tense of "will" Do you think that "was going to be" or "would be" could be used as past tense of will?     I thought this book was going to be your choice.        I thought this book would be your choice.   Do these two sentences basically mean the same thing or are they slightly different? Why do RAW images in Darktable have a lot of noise? File.Copy is giving exception How can I host my own Minecraft server? Sum of the series:$\frac{1}{1\cdot 2}+\frac{1\cdot3}{1\cdot2\cdot3\cdot4}+\cdots$ I have a question where i couldn't find any clue. The question is  $$\frac{1}{1\cdot 2}+\frac{1\cdot3}{1\cdot2\cdot3\cdot4}+\frac{1\cdot3\cdot5}{1\cdot2\cdot3\cdot4\cdot5\cdot6}+\cdots$$  I could get the general term as $t_n=\frac{1\cdot3\cdot5\cdot7\cdots(2n-1)}{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdots2n}$  I have also tried it to form the sequence in the telescopic form.But couldn't get. Any hint will be appreciated. I've been investigating the Cartesian graph of $x^y=y^x$. Obviously, part of the graph is comprised of the line $y=x$ but there is also a curve that is symmetrical about the line $y=x$. (We can prove this symmetry by noting that the function $x^y=y^x$ is self-inverse; all self-inverse functions are symmetrical about the line $y=x$.) An image of the graph is shown below:  I decided to find the intersection point and arrived at an intriguing result: the intersection point between the two curves is at $(e,e)$. The following is my method: If the gradient of the line $y=x$ is $1$, the gradient of the curve at the intersection point must be $-1$ as it's a normal to the line (as it's symmetrical about the line). This means that at that point $\frac{dy}{dx}=-1$. Now to find $\frac{dy}{dx}$. We have $x^y=y^x$. I then used a very powerful tehnique for differentiating these sorts of functions. We know that eg $$x^y=e^{\ln{x^y}}=e^{y\ln{x}}$$ and $$\frac{d}{dx}e^{f(x)}=f'(x)e^{f(x)}$$ Applying it to our function $x^y=y^x$ and using implicit differentiation and the product rule gives us: $$(\frac{dy}{dx}\times \ln x +\frac{y}{x})x^y=(\ln y +\frac{dy}{dx}\times \frac{x}{y})y^x$$ So $$\frac{dy}{dx}x^y\ln x +yx^{y-1}=\frac{dy}{dx}xy^{x-1}+y^x \ln y$$ Extensively rearranging gives: $$\frac{dy}{dx}=\frac{y^x \ln y -yx^{y-1}}{x^y \ln x -xy^{x-1}}$$ Let $\frac{dy}{dx}=-1$: $$y^x \ln y -yx^{y-1}=xy^{x-1}-x^y\ln x$$ But we know $x=y$ since we are at the intersection point with the line $y=x$: $$x^x \ln x -x^x=x^x-x^x\ln x$$ So $2x^x \ln x -2x^x=0$ $$2x^x(\ln x -1)=0$$ This means either $2x^x=0$ or $\ln x -1=0$ but we know $x^x$ is always greater than $0$ so $\ln x =1$, leaving us with: $$x=y=e$$ So I have my answer, but is there any other method of getting it? I have heard there is. Any help will be very welcome. Thanks in advance Getting JavaScript object key list HTML element does not fade in using jquery