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If it was 10a+2 then the statement would be true, but if you are sure that was the question you got then it is probably is wrong. At least I think.

I don't think that statement is true in the first place. say a=1 (assuming a is an integer) then 2(1)+1=3 which is divisible by 3. but 10(1)-2=8 which is not evenly divisible by 3 idk if there is something other to this question or @=)(= copied it wrong

part iii) Now we just have to find the roots of the equation, so lets just use the identity from part i and the product of roots to simultaneously solve for \beta and \gamma. \\\alpha+\beta+\gamma=0\\2i+\beta+\gamma=0\\\beta+\gamma=-2i\\\beta=-\gamma-2i \longrightarrow \textcircled{1} Product...

Alternatively if you wanted do it the hard way... using I_n=\frac{-2n}{2n+1}I_{n-1} we can just continually evaluate I_{n-1},\ I_{n-2},\ I_{n-3},\ ... I_{1} to get...

You were really close. \\(a^n-b^n)(a-b)\geq{0}\\\\a^n\cdot{a}-a^nb-b^na+b^n\cdot{b}\geq{0}\\\\a^{n+1}+b^{n+1}\geq{a^nb+b^na}

I think I get it Since y>0, this means that the body is still above the ground despite 2T seconds passing. Since it took T seconds to go up that means the other T seconds were spent going down but it wasn't able to cover the same distance as going up. Therefore the downward journey takes...

How do I do part iv?

All prime numbers are odd, except for 2 I think. So if 4n-1 is odd, 2 cannot be a factor hence all prime factors should be odd.

You can solve this question like finding the point of intersection between two graphs. The relationship between the horizontal run and vertical rise of the incline plane is \tan(10)=\frac{s_y}{s_x} For the projectile we know that u_x=15\cos(32) and u_y=15\sin(32). Then to find the horizontal...

Use E=\frac{V}{d} to find the strength of the magnetic field. E=\frac{200}{0.15} Then use E=\frac{F}{q} to find out the force that is acting on the electron \\\frac{200}{0.15}=\frac{F}{1.602\times{10^{-19}}}\\\\F=\frac{200\times1.602\times{10^{-19}}}{0.15} Now using F=ma we can find the...

For image 1: part i) you just sub in x = 1 and \beta into the equation and show that it equals to zero. part ii) \\\alpha+\gamma=1+\beta\\\alpha^2+2\alpha\gamma+\gamma^2=(1+\beta)^2\\\alpha^2-2\alpha\gamma+\gamma^2=(1+\beta)^2-4\left(\frac{1}{2}(1+\beta^2)\right)\ \text{from the identity...

wdym, I only select the most relevant tags

Thankyou!

I need help with part a.

When you use the formula E=\frac{V}{d}, d is meant to be the distance between the charged plates, not the distance between a charged plate and the particle. Also you should realise that a characteristic of a magnetic field between two charged plates is that it is equal at all points between the...

I think the question would have explicitly stated to use part (iv) if they meant for you to use it. But it I guess the question was kinda building up to that point through all the previous parts, so your way would have been correct too and in timed conditions my method would take way to long for...

Yeah that's true, I was trying to use part ii cause the question asked to.

I think this might be the solution to part v We know that z^2-2z\cos\left(\frac{2\pi}{5}\right)+1=\left(z-\text{cis}\left(\frac{2\pi}{5}\right)\right)\left(z-\text{cis}\left(\frac{-2\pi}{5}\right)\right) \\\therefore...

Thanks guys, I get it now