Search results

welp, probably gonna get E1 if I ever did it

Oops, thought this was maths Ext 2 :oops: Probably should have read what forum it was in

1 byte = 8 bits (1B=8b), therefore 495MB=3960Mb Time taken to download would be {3960}\div{82.7}\approx48 seconds Not sure if it is in the syllabus though. Edit: Thanks for CM_Tutor and jimmysmith560 for pointing out my error. \begin{align*}1\ \text{megabyte} = 2^{23}\...

[Blank] when he needs some extra soldiers for his army.

Grind...

Since the ball was launched horizontally u_y=0, we know that a_y=-9.81 and s_y=-1.3 and we are tying to solve for t. possible methods are: Finding velocity using the equation v^2=u^2+2as then subbing velocity into v=u+at to find time or subbing the values into s=ut+\frac{1}{2}at^2, and then...

We know that the passengers have to feel weightlessness at the top of the loop, therefore the centripetal force must be equal to the gravitational force. \\\frac{mv^2}{r}=mg\\\\\frac{v^2}{8}=9.81\\\\v=\sqrt{78.48}ms^{-1} Then we need to use conservation of energy to find a relationship between...

First we have to solve the two equations simultaneously so that the zeros of that equation are the points of intersection. \\y=x^2-k\longrightarrow\textcircled{1}\\x^2+y^2=1\longrightarrow\textcircled{2} Sub \textcircled{1} into \textcircled{2}...

\\\frac{\frac{dT}{dt}}{T}=\frac{\frac{d}{dt}(25+70(1.5)^{-0.4t})}{25+70(1.5)^{-0.4t}}\\\\\\\frac{\frac{dT}{dt}}{25+70(1.5)^{-0.4t}}=\frac{\frac{d}{dt}(25+70(1.5)^{-0.4t})}{25+70(1.5)^{-0.4t}}\\\\\\\frac{dT}{dt}=\frac{d}{dt}(25+70(1.5)^{-0.4t}) We kind of end up going back to where we started...

I'm assuming you mean to take the ln of both sides and then differentiate. \\T=25+70(1.5)^{-0.4t}\\\ln(T)=\ln(25+70(1.5)^{-0.4t})\\\\\frac{dT}{dt}(\ln(T))=\frac{dT}{dt}(\ln(25+70(1.5)^{-0.4t})) On the LHS I'm not even sure how to derive \ln(T), and on the RHS you are going to have to derive...

You use the identity a=e^{\ln(a)} \\T=25+70(1.5)^{-0.4t}\\T=25+70e^{\ln(1.5^{-0.4t})}\\T=25+70e^{-0.4t\ln(1.5)} Then you differentiate from there.

If you look at the derivative \frac{dx}{dt}=3-2e^{-2t} First looking at the starting velocity at t=0 \frac{dx}{dt}=3-2e^{0}=1 Then we look at where velocity approaches as t approaches to infinity As t\longrightarrow\infty \\\frac{dx}{dt}\longrightarrow{3-2e^{-\infty}}\longrightarrow{3}...

First we should find the general term of (1+ax)^{10} \text{General Term}\\\\={10 \choose k}\cdot{a}^{k}\cdot{x}^{k} Now since they told us that the coefficient of x^6 is 0, we can find a by finding the coefficient of x^6 in terms of a. Since the general term is being multiplied by (1-2x) we...

You basically just sub in x=1 \\\text{LHS}=(1+1)^n\\=2^n\\\\\text{RHS}=\binom{n}{0}+\binom{n}{1}(1)+\binom{n}{2}(1)^2+\binom{n}{3}(1)^3\text{...}+\binom{n}{n}(1)^n\\\\=\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\binom{n}{3}\text{...}+\binom{n}{n}\\\\=\sum_{k=0}^{n} \binom{n}{k}\\\\\\\therefore...

I've tried to solve it, but I haven't learnt any Ext 2 level projectile motion stuff yet. I'll try to learn it over the next few weeks if possible.

\\\text{General Term}\\\\{7 \choose k}\cdot5^{7-k}\cdot(2x^2)^k\\\\={7 \choose k}\cdot5^{7-k}\cdot2^k\cdot{x^{2k}} Since x is to the power of 6, k has to equal three \\{7 \choose 3}\cdot5^{7-3}\cdot2^3\cdot{x^{2(3)}}\\\\=35\cdot625\cdot8\cdot{x^6}\\=175000x^6 Therefore the coefficient of x^6...

Let gradient be m First we will sub the point (1,-2) and m into point gradient formula \\y-(-2)=m(x-1)\\y=mx-(m+2) Since the line and the parabola touch each other, we have to solve them simultaneously \\y=mx-(m+2)\longrightarrow \textcircled{1}\\y=x^2\longrightarrow \textcircled{2} Sub...

You get the value of m by subbing the equation of l into y=x^3-x+3, then using the products and sum of roots to find the value of m.

You use the point gradient formula y-y_1=m(x-x_1)

For part a: momentum is calculated by p=mv therefore momentum would be... \\p=1500\times20\\\\p=30000kgms^{-1} For part b: We have the values for initial velocity, final velocity and displacement and we need to calculate acceleration, therefore the most suitable equation would be v^2=u^2+2as...