Sy123
This too shall pass
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- 2013
Wow......It's just backwards multiplication of two numbers. The original number ends in 6, and when multiplied by 4 gets the same number but with a 6 out the front instead of the end:
Multiply by the 4 and carry the 2 (since you multiply by the units and carry the tens)
Multiply by the 8 and carry the 1:
Multiply by the 3 and carry the 3:
Multiply by the 5 and carry the 1:
Multiply by the 1 and carry the 2:
Since we have just 6, we stop here, and so our number is all the numbers we multiplied - 648351 - but we reverse it as we did the multiplcation in the reverse order to get 153846.
I'll give someone else a go for this one, good question =)I was saving this question in case I felt like writing a trial paper, but that doesn't seem likely so here it is. Although it could go in the 3U thread, I thought the solution was quite nice and deserved to be here under "harder 3U" instead.
Consider the following diagram, where PQ is a tangent to the circle:
An alternative proof for the tangents from an external point theorem.
Part i):
LOL! do the other polynomial question i posted on the 3U marathon threadDo you even know how much u confused me with ur polynomials question, where u made it a^2 not a, that had me going for hours and hours trying to figure out how u got a^2 and it turns out it was a all along lol. Btw If U have a cool conics question can u plz give to me.
From the tangent-secant theorem:I was saving this question in case I felt like writing a trial paper, but that doesn't seem likely so here it is. Although it could go in the 3U thread, I thought the solution was quite nice and deserved to be here under "harder 3U" instead.
Consider the following diagram, where PQ is a tangent to the circle:
An alternative proof for the tangents from an external point theorem.
From tangent-secant theorem:I was saving this question in case I felt like writing a trial paper, but that doesn't seem likely so here it is. Although it could go in the 3U thread, I thought the solution was quite nice and deserved to be here under "harder 3U" instead.
Consider the following diagram, where PQ is a tangent to the circle:
An alternative proof for the tangents from an external point theorem.
<a href="http://www.codecogs.com/eqnedit.php?latex=\int \frac{1-x}{1-\sqrt{x}}\: dx" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int \frac{1-x}{1-\sqrt{x}}\: dx" title="\int \frac{1-x}{1-\sqrt{x}}\: dx" /></a>
Give it a bit of time, I'm working on it right nowHmm perhaps I should give hints for the second part of my question?
Well, no hint for you thenGive it a bit of time, I'm working on it right now
PQ is a tangent, so PQM is a right angle. Without stating the obvious, PM > PQ. PM = PQ when the radius = 0.I was saving this question in case I felt like writing a trial paper, but that doesn't seem likely so here it is. Although it could go in the 3U thread, I thought the solution was quite nice and deserved to be here under "harder 3U" instead.
Consider the following diagram, where PQ is a tangent to the circle:
An alternative proof for the tangents from an external point theorem.
Moving on from part (i)
We do not know that M is the centre.PQ is a tangent, so PQM is a right angle. Without stating the obvious, PM > PQ. PM = PQ when the radius = 0.
PQ is a tangent, so PQM is a right angle. Without stating the obvious, PM > PQ. PM = PQ when the radius = 0.
^^^We do not know that M is the centre.