Search results

Complete the square in the denominator first.

I'll only do the inductive step: \textup{Assume the result is true for}\; n=k\; \textup{i.e.}\; \cos \left ( x+k\pi \right )=\left ( -1 \right )^{k}\cos x\\ \textup{Required to prove true for}\; n=k+1\; \textup{i.e.}\;\cos \left ( x+(k+1)\pi \right )=\left ( -1 \right )^{k+1}\cos x\\...

x''=0 x'=c x=ct+d y''=-10 y'=-10t+e y=-5t^2+et asuming launch from 'the ground' At the highest point of its path y-velocity = y'= 0, so x'=10 m/s. y'=0 when y=8 ----> 10t=e ----> t=e/10 ----->8=-5(e/10)^2+e^2/10 ---> e=12.649 y'(0)=e=12.649 x'(0)=x'=10 initial velocity = 16.1245...

Yeh, integrating sec is one of those you have to remember to multiply out by (sec + tan).

Equilibrium constant tells you 'how much' towards which side of the equation equilibrium is - it gives a number.

One connected to each end of the coil.

Domain for arctan is all R

HCl, HNO3, H2SO4 are all strong, ethanoic (CH3COOH) and citric (C6H8O7) acids are weak. They are the common ones.

I'll give you some hints, see if you can finish it. Get the molcular mass of H2 and N2, and figure out how much of the mass of 50kg is H2 and how much is N2 (and how many moels of each there is), then use the balanced equation of the haber process to determine the moles of ammonia.

Looks good ^^^.

\begin{align*} \textup{Let}\; y=f(x)=\frac{1}{4}\left [ \left ( x-1 \right )^{2}-7 \right ]\\ \textup{Assuming}\; x\geq 1\; \textup{we have:}\; y&=\frac{1}{4}\left [ \left ( x-1 \right )^{2}-7 \right ]\\ 4y&=\left ( x-1 \right )^{2}-7\\ \left ( x-1 \right )^{2}&=4y+7\\ x-1&=\pm \sqrt{4y+7}\\...

3C3 is valid, with nCr notation, n has to be a positive integer i.e. n=1,2,3,... and r has to be a non-negative integer up to (and can be including) n, i.e. r=0,1,2,3,...,n.

kx^2\geq 0\; \textup{for} \; k\geq 0\; \textup{so}\; \zeta +kx^2\geq \zeta \; \textup{no matter what}\; \zeta \; \textup{is.} i.e. adding a non-negative number to any number results in something that is at least as great as the original number. Fairly intuitive.

jetblack (Mathematics moderator) has a very useful guide stickied to these forums on how to use the LaTeX equation editor.

True - if you don't see it, it seems very daunting; but if you are exposed to this before and it comes up, the last line of my solution isn't too hard to see. It can and (almost always ) is.

I'm a Chemistry/Physics mod, but always happy to help on the maths forums. As for your last question, it is rather trivial (and indeed the answer shows you that), so i'll leave that to you.

\begin{align*} \textup{Let}\; y&=2^{\ln x} \\ \therefore \ln y&=\ln \left (2^{\ln x} \right )\\ \ln y&=\ln x\ln 2\\ \therefore y&=2^{\ln x}=e^{\ln x\ln 2}=\left (e^{\ln x} \right )^{\ln 2}=x^{\ln 2}\end{align*}\\ \begin{align*} \textup{Hence}\; \int 2^{\ln x}dx&=\int e^{\ln x\ln 2}dx \\ &=\int...

Disregard my previous post, the solution is correct: Now label the four (distinct) fingers A,B,C and D, and the nine (dsitinct) rings 1,2,3,...,9. What we are asked to do is find the total number of ways in which we can place the 9 rings on the four fingers, if the order of the rings on...

\begin{align*} \frac{\mathrm{d} }{\mathrm{d} x}\left \{ xe^x \right \}&=xe^x+e^x \\ \therefore \int \frac{\mathrm{d} }{\mathrm{d} x}\left \{ xe^x \right \}dx&=\int xe^x+e^xdx\\ \int d\left (xe^x \right )&=\int xe^xdx+\int e^xdx\\ xe^x+\alpha &=\int xe^xdx+e^x+\beta \; \; \; (\alpha \...

One (crucial) line is incorrect. \frac{4^{k+1}-1+12^{k+1}}{3} should be \frac{4^{k+1}-1+3.(4^{k+1})}{3}.