Search results

Idk if this would even work but... Using Faraday's law \epsilon=-N\frac{{d}\phi}{dt} using chain rule \frac{{d}\phi}{dt}=\frac{{d}\phi}{d\theta}\times\frac{{d}\theta}{dt} using the magnetic flux equation \phi=BA\cos(\theta) \frac{{d}\phi}{d\theta}=-BA\sin(\theta) therefore...

say at t=0, N(0)=A therefore at t=x, N(x)=2A, since x days later the population must double. hence \\2A=Ae^{0.15x}\\\\2=e^{0.15x}\\\\0.15x=\ln(2)\\\\x=\frac{\ln(2)}{0.15}\\\\x=4.6

Try to derive it yourself. Use the relationship between \cos(2\alpha) and cos^2(\alpha) that @yanujw has included. In a sense the thought process is similar to the way you get the expression \cos(\theta)=\frac{e^{i\theta}+e^{-i\theta}}{2} Spoiler if you couldn't find it out

using the second equation \\3\cos^2(\alpha)+2\cos^2(\beta)=4\\\\\frac{3}{2}(1+\cos(2\alpha))+(1+\cos(2\beta))=4\\\\3+3\cos(2\alpha)+2+2\cos(2\beta)=8\\\\3\cos(2\alpha)=3-2\cos(2\beta)\\\\\cos(2\alpha)=\frac{3-2\cos(2\beta)}{3} using Pythagoras we can say that...

You can use this website https://www.integral-calculator.com/ if you want a fast way to check answers.

For this question Skipping a few steps we need to prove 1+4+...+(3(k+1)-2)=\frac{(k+1)(3(k+1)-1)}{2} subbing in n=k+1 \\LHS=1+4+...+(3k-2)+(3(k+1)-2)\\\\LHS=\frac{k(3k-1)}{2}+(3(k+1)-2) Using the inductive statement that would have been stated if it was worked out properly...

for part ii) we'll start with \\(1+i)^n+(1-i)^n\ \text{Where n is a multiple of 4}\\\\={n\choose0}+{n\choose1}i-{n\choose2}-{n\choose3}i+{n\choose4}-...+{n\choose{n}}+{n\choose0}+{n\choose1}(-1)i-{n\choose2}-{n\choose3}(-1)^3i+{n\choose4}+...+{n\choose{n}}\ \text{Binomial...

For the second attachment part i: \\LHS-RHS=\frac{1}{a}+\frac{1}{b}-\frac{2}{3}\\\\=\frac{3b+3a-2ab}{3ab}\\\\=\frac{3(b+a)-2ab}{3ab}\\\\=\frac{0.5(b+a)(b+a)-2ab}{3ab}\\\\=\frac{\frac{(b^2+2ab+a^2)}{2}-2ab}{3ab}\\\\=\frac{\frac{(b^2+a^2)}{2}-ab}{3ab}\\\\\geq\frac{\sqrt{a^2b^2}-ab}{3ab}\...

I would recommend finding the modulus and the argument separately. For the modulus you would use pythag OD^2=OA^2+AD^2 Given that since OB is a tangent to the circle \\OD=\sqrt{(OC^2-CA^2)+(R-r)^2}\ \text{pythag again as}\...

For part a) We need to get the DE v'=r(1-v) therefore we need an equation for \frac{dv}{dx} using the chain rule we can say \frac{dv}{dx}=\frac{dy}{dx}\times\frac{dv}{dy} the question already says that \frac{dy}{dx}=ry(1-y) to find \frac{dv}{dy} we can use the given identity...

Assuming the same domain applies as the previous equation -180^{\circ}\leq{x}\leq180^{\circ}\implies-630^{\circ}\leq3(x-30^{\circ})\leq450^{\circ} Thus solving for \cos(3(x-30^{\circ}))=-1 will get us \\3(x-30^{\circ})=\ -540^{\circ},\ -180^{\circ},\ 180^{\circ}\\\\x=-150^{\circ},\...

You could shift the domain to something that is easier for questions for example -\pi\leq{x}\leq\pi\implies{0}\leq{x+\pi}\leq{2}\pi Say for question a originally you would have to solve 2\cos(2x)=1 for -\pi\leq{x}\leq\pi but by shifting the domain by pi forward you get 2\cos(2(x+\pi))=1 for...

oh now I see, thankyou!

How do I use the perpendicular distance formula in part iv, I don't really know which values I need to sub in.

for part e) \omega=\frac{\Delta\theta}{\Delta{}t} \Delta\theta=\frac{\Delta{d}}{r} where d is the distance traveled during time t. \therefore\omega=\frac{\Delta{}d}{\Delta{}t}\times\frac{1}{r}=\frac{v}{r} where v is the tangential velocity. hence v=r\omega Using a_c=\frac{v^2}{r}...

for part (i) you could use \left(\frac{1}{x}-\frac{1}{y}\right)^2\geq0

But for this specific question there is no minimum turning point, cause f'(x)\neq0.

Since the derivative is f'(x)=\frac{3}{x^2}\implies f'(x)>0 for all real x, since square numbers are positive and x\neq0. Therefore for all values of x within the domain of f(x) would be increasing.

Thank you! It was double ibp all along

How do I find the reduction formula of I_n=\int{(\sin^{-1}x)^ndx. Thanks