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There is a very sharp solution here you see how the difference between 42 degrees and 27 degrees makes 15 degrees right. Time to get creative and use what you know which is \cos{2\theta}=1-2\sin^{2}\theta or in this case \sin{\theta}=\pm{\sqrt{\frac{1-\cos{2\theta}}{2}}} use this to your...

Contraposition comes from the words contradict and position meaning opposite position so then you can say if ab is odd then both a and b are odd because the statement "at least one of a and b is even" means either a or b is even or both a and b are even.

To prove part c for the maximum do the second derivative. Note that if the result is negative then you have proven that it’s the maximum.

To start off we need to know the rate of change in volume per the rate of change in height which is written as \frac{dV}{dh}=\pi\left(R^{2}-\frac{3h^{2}}{4}\right). Once the foundation is set we want to find the maximum or minimum and we need to show that by stating the rate of change is zero...

I think here somewhere along your journey you have to apply SWOT analysis which is very useful in all aspects of life.

@tk8 Here is the solution. I want you to check your work to see where you tripped yourself in the path of solving these two parts. Part c To start off V=10\pi{r^{2}}-2\pi^{2}r^{3} Then, \frac{dV}{dr}=20\pi{r}-6\pi^{2}r^{2} Note this means the rate of volume divided by the rate of radius. So if...

I mean from here you can say x\left(\left(-x+2\right)\left(-x^{2}\right)-x+2\right) x\left(-x+2\right)\left(1-x^{2}\right) x\left(1+x\right)\left(1-x\right)\left(2-x\right)\leq0 Then use @Drongoski's working.

If it isn't then it is in the enrichment section of the Cambridge textbook.

For this one, you see that xe^{x} is two terms multiplied together right so thus you need to use the product rule to obtain xe^{x}+e^{x}. If you are integrating then you are going to notice a very important fact \frac{d}{dx}\left(xe^{x}\right)=xe^{x}+e^{x}...

What is the full question from part ii?

You mean part ii

Most steps are correct however I find this to be cleaner. \left(x+y\right)^{2}\geq{4xy} \left(x+y\right)\left(x+y\right)\geq{4xy} \frac{x+y}{xy}\geq{\frac{4}{x+y}} \frac{1}{y}+\frac{1}{x}\geq{\frac{4}{x+y}} That is what I would do if I were you @lmao1010

Negation is just the opposite of what is shown here and using your example the negation of x^{2}+y^{2}<1 is x^{2}+y^{2}\geq{1}. Doing that then you can say \left(x, y\right) is not a point inside the circlex^{2}+y^{2}=1 In this case the original statement is true and the negation is also true.

The reason why it does not matter if a, b, c, d are positive or negative is that you are for the most part working with even powers and in that case, if you have negative numbers in the numerals they will become positive numbers as noted because simply \left(...\right)^{2n} is going to give us...

That is what I think the question is asking.

Remove everything that is the left of the y-axis. As you are finding the region between the curve and tangent and the y-axis. This is a vital part for part c.

Okay, so the idea is that your shape will resemble a cone or in many of these cases the outside surface of the cone. The tangent from Q20a is y=3x. Now the concept of integrating is to find the section under the curve. Combining Q20a and what we know you are integrating y=x^{3}-3x+2. Now when...

Here is my working out but there are a few things to note A few common missteps are 1. Not simplifying when given the chance. 2. Stating that \sqrt{c}=0 because 0 is a rational number and that \sqrt{c} is irrational. Note I used 1 on the other side instead of \frac{p}{q} because if things are...

The laughing stock in the Bored of Studies community for students who do maths. Also, the book that people try to steer you away from and go into Cambridge and past papers written by Margaret Grove.

Alright did you jig because all your classes are filled with casual teachers? If so go for it.