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I meant perimeter. I thought perimeter but typed diameter.

None of that compares to the photo I still have of a year 7 test. The question gave a square and said the side length was 5mm, and asked for the diameter. I gave 2cm, the correct answer but the teacher marked it incorrect because it should have been 20mm.

If there's only one chemistry class in my school's grade, doesn't that make me first in the grade?

Just got results from yearlies today, 90%, first in the class :). Pretty happy.

Yeah.

I get to drop two weeks after the beginning of next term. Dropping business and engineering, picking up 4u. It seems I can drop what I want at my school, but they do pressure you to keep 12 units because screw-ups in test / keeps you occupied / people always regret it. All the usual excuses.

Ok, here's the deal. I've been trying to graph the function x^2+8x+y^2-8y+2xy+16=0, and on a graphing calculator such as this here, and I've noticed the graph is a circle, with centre (-4,4) and radius 4. x^2+8x+y^2-8y+2xy+16 seems to be equivalent to x^2+8x+y^2-8y+16, which is also a circle...

I literally have never known this rule, that transversals cut parallel lines in the same ratio. I thought I'd have to prove that the two trapezoids were similar. The answer is 12 cm. Thanks guys.

Hey everyone, I was just doing a past paper, and I found this question. I feel like x should be equal to 12, but I don't know how to show it. If I could see working out for this, that would, be great.

You can intuit it out, but knowing how the formal method works is also important. It uses the distance formula. d=d\\\sqrt{(x-1)^2+(y+3)^2}=2\\(x-1)^2+(y+3)^2=4

Just had my prelim. physics yearly. I'm pretty happy with it, finished early, found some mistakes. I know there was still mistakes in there though.

\frac{12-4y}{y}\\=\frac{12}{y}-\frac{4y}{y}\\\frac{12}{y}-4

Yeah, there's this but not much else.

Does anyone have any good resources for circle geometry questions? I've done all of Maths in Focus, but the past papers I've been doing seem to have much harder questions (who knew?). The year 11 Cambridge textbook doesn't seem to cover it at all, and I can't find anywhere else.

Well, you can start by telling us what your AOS is, lol.

Awesome, thanks. Why do you use the triple-lined equal sign?

$Express $\cos{x}+7\sin{x}=5$ in the form $A\cos{(x+\alpha)}$. Hence, solve $\cos{x}+7\sin{x}=5$, 0^{\circ}\leq x\leq360^{\circ} I can express it in as a sine, but how do I get it to a cosine?

Lol, I just did that exactly today. Their questions are really hard.

Lol yeah, what a coincidence to find someone so close by on BoS... Sometimes I wonder if there are people at my school who have a BoS account.

Oh I just recognised the uniform, your school is really close to mine, lol. Infact, I went to the library which is near it just today.