Mathhemathh

Using distance formula, the distance of the parabola from its focus is \(\sqrt{(x+1)^2+(y-5)^2}\) and the distance from the directrix is \(\sqrt{(y-6)^2}\). On aparabola, these distances are always equal, so:

\(y-6=\sqrt{(x+1)^2+(y-5)^2}\)

\(y^2-12y+36=(x+1)^2+y^2-10y+25\)

\(-2y+11=(x+1)^2\)

\(y=-{(x+1)^2-11\over2}\)

And that is your equation.

Denary is a less common term for decimal, or base 10. Assuming that number is in base 10, 100000 in denary is still 100000. 

This is equal to \(14+(14+3)+(14+3*2)...\)

Separate the 14s: \(14*102+3+3*2+...+3*101\)

Factor: \(14*102+3(1+2+...+101)\)

Evaluate: \(1428+3(1+2...+101)\)

Sum of consecutive integers = \({n(n+1)\over2}\)

Evaluate again: \(1428+3({101(102)\over2})\)

\(1428+3*5151\)

\(1428+15453\)

Final answer = \(16881\) 

My interpretation: The shortest side of the right triangle is 6cm and the hypotenuse is 18cm.

Pythagorean Theorem: \(leg1^2+leg2^2=hyp^2\)

\(6^2+?^2=18^2\)

\(36+?^2=324\)

\(?^2=288\)

\(?=\sqrt{288}\)

Simplifying, we get \(?=12\sqrt{2}\) cm. 

1 with 1000000000000 zeroes after it. Remember: \(10^n\) = 1 with \(n\) zeroes after it. 

Not sure, but... for the 120 case, I guess it's 239? {120, 1, 1..., 1, 1}? Anyway, ask CPhill.

Malis verbis meis...

First, note that \(\tan({\pi\over4})=1\). Imagine this on the unit circle. Since we need to find \(x\) when its tangent is negative 1, we have to get to the 2nd and 4th quadrant while retaining the tangent's absolute slope. To do that, we add \(90^o\) or \({\pi\over2}\) and \(270^o\) or \({3\pi\over2}\) to the angle \({\pi\over4}\). When we do that, we get the angles \({3\pi\over4}\) and \({7\pi\over4}\).

Q.E.D. 

Let \(A\) be the first pack of 400g, and \(B\) be the second pack of 400g+25%.

First, we calculate the total weight of pack \(B\) with the 25%:

\(400g*25\%=100g\)

\(400g+100g=500g\) in total

Then, we calculate the unit price (price per gram) for each pack: \(unitPrice={price\over quantity}\)

A). \({\$2.98\over400g}=\$0.00745\)

B). \({\$3.28\over500g}=\$0.00656\)

As we can clearly see, the unit price of \(A\) is greater than the unit price of \(B\), therefore pack \(B\) is a better buy, but only by a fraction of a cent!

Nice! Not how I did it, but nice. (I didn't know this theorem existed )