Search results

The question introduced (a-2)x^2+(1-3b)x+(5-2c) as simply a polynomial (which it is, if we assume a, b, c are constants). It never claimed this polynomial was a quadratic and in fact it is not, because quadratic equations must be expressible in the form \alpha x^2 + \beta x + \gamma where...

"Linear Algebra" is a massive field of study, so a comprehensive formula sheet would be at least the size of a textbook. You would have to explicitly give the course outline for your linear algebra course - and at that point you could just take notes on lectures and make your own. Also, your...

Where did you get this question? You probably wouldn't be able to get this result from just e^n \ge {(n+1)^n}/{n!} as \frac{(n+1)^n}{n!} \stackrel{?}{\ge}1+n^2 is not actually true, even taking n \ge 0. You'd have to at least start a bit further back. I doubt this is the intended answer...

Usually I'd expect a mistake like this to only incur a 1/2 mark penalty. If it was really really minor (e.g. you wrote the wrong units) then you might even get away for free.

In more concrete words, you are asked to show n = \cos^{-1} \frac 35, given the conditions a \cos \alpha = 1, a \cos (n+\alpha) = 5, a \cos (2n+\alpha) = 5. So, try to solve for n . You have three equations with as many variables, so intuitively this should be solvable. Bonus points -...

Actually, the simplest way is to just get the answer from the back of the textbook, ask someone else who knows how to do it, or to search it up online. And this is what we would do if we only cared about the number we get at the end. So what is the point of even learning calculus and...

\binom{n}{2} \left(\frac{9}{10}\right)^{n-2} < \binom{n}{3}\left(\frac{9}{10}\right)^{n-3} \frac 1{10} Divide by (9/10)^{n-3}: \binom{n}{2} \left(\frac{9}{10}\right) < \binom{n}{3} \frac 1{10} Now use \binom{n}{k} = \frac{n!}{(n-k)! \, k!} so \frac{n!}{(n-2)! \, 2!}...

Your reasoning is correct. Another way to look at it is the following: If O is any point that is not \pi/2 , then the particle will start off with a positive velocity, which means its displacement will increase. But when its displacement increases, its velocity ( v = \cos^2x ) gets smaller...

When we discuss real-valued functions it's often convenient or more practical for us to take "continuous" to mean "continuous on \mathbb R". Most people accept this because whatever meaning we take is usually clear to the reader from context. In this case it is (probably) not, so you should be...

If your fraction stack is growing out of control, you can just get rid of the fraction entirely - write \frac{\frac{1}{x}+e^x}{1+\frac{1}{\sin^{-1} x}} as \left( \frac{1}{x}+e^x \right) \left( 1+\frac{1}{\sin^{-1} x} \right) ^{-1}. Another useful trick is e^{\frac{1}{x} + \sin x} \to...

If one wanted to actually find examples of M , here is an approach that works for M \in \mathbb R . I'll use the hyperbolic function \cosh and its inverse, which aren't in the MX2 syllabus but their definitions are quite simple to understand: \cosh x = \frac 12 \left(e^x+e^{-x}\right) and...

Is the condition M \notin \mathbb Z really necessary? The only case of this I can see is 1 + 1/1 = 2 \in \mathbb Z, so it would be simpler to leave it unmentioned.

If the queues are distinct, then we should have 17280 = \underbrace{\left( \binom{8}{4} \times 4! \times 4!\right)}_{\text{no restrictions}} - \underbrace{\left(\binom{6}{3} \times 4!\ \times 4!\right)\times 2}_{\text{Sean and Liam in diff. queues}}, or 17280 = \left( \binom{6}{4} \times 4...

For both balls we have \begin{cases} y_1(t) &= -\frac{9.8}{2}t^2+100\sin(\pi/6)t+h \\ x_1(t) &=100\cos(\pi/6)t\end{cases} \begin{cases} y_2(t) &= -\frac{9.8}{2}t^2+h \\ x_2(t) &=100t\end{cases} Solve y_1(t) = 0 and y_2(t) = 0 to get the time of flight for each ball (call these t_1 and...

Recall that for two vectors \bold x, \bold y we have \cos \theta = \frac{\bold x \cdot \bold y}{|\bold x||\bold y|}, where \theta is the angle between \bold x and \bold y .

Unfortunately it's not that simple. You can't "break up" the absolute value function like that. If y = |x-3| , then y = \pm(x - 3). You can try graphing these to visualise the effect of the absolute value function.

That's called the absolute value function, and it's defined by |x| = \begin{cases} x, &\text{if }x \ge 0, \\-x, &\text{if }x < 0.\end{cases} Basically, you only want the magnitude of the number, not its sign. For example, |-1| = 1, |-2| = 2, |1| = 1, |0| = 0. If x = |y| , then...

Say you have a statement you're trying to prove by induction. A normal induction proof would go like "if this statement is true for k then it must also be true for k + 1." A strong induction proof would be more like "if this statement is true for all numbers less than or equal to k then it...

This is one of the more common differentiation mistakes. \frac{d}{dx} x^x \overset{?}=x \cdot x^{x-1} For trigonometry in general, people make a lot of mistakes. \cot x \overset{?}= \frac{1}{\tan x} \sin^{-1}( \sin x ) \overset{?}= x , or equivalently, \sin x = y \overset{?} \implies x...

The quantity \frac{k^2-RX^2}{2\cdot PX \cdot QX} can probably be simplified so as to remove k . The answer I gave holds numerically for all such equilateral and isosceles triangles formed and most likely for the rest of them too. The other solutions are definitely much nicer though. I...