1. Open the compass to a little more than halfway across the line segment XY. Draw an arc centered at the first endpoint X across the line segment XY. Without changing the width of the compass, place the compass tip on the
second endpoint Y. Draw a second arc across the line segment XY.
2. Line up a straightedge with the intersection of the arcs above the line XY,
and the intersection of the arcs below the line. Draw a line connecting
these two points. The line you draw is a perpendicular bisector. It
bisects the line XY at a right angle.
3. Use a compass and straightedge to construct the bisectors of the line YZ as you did with the first line segment. Extend the bisectors long enough that they intersect. The point of their intersection is the center of the circle.
4. The radius of a circle is the distance from the center to any point on the circle’s edge.
To set the width, place the tip of the compass on the center of the
circle, and open the compass to any one of your original points.Swing the compass around 360 degrees so that it draws a complete circle. The circle should pass through all three points.
So you subtract 180 from 90 you get you answer then you will divide it by 55 and your remaining answer which you will get will be x
A combination is an unordered arrangement of r distinct objects in a set of n objects. To find the number of permutations, we use the following equation:
n!/((n-r)!r!)
In this case, there could be 0, 1, 2, 3, 4, or all 5 cards discarded. There is only one possible combination each for 0 or 5 cards being discarded (either none of them or all of them). We will be the above equation to find the number of combination s for 1, 2, 3, and 4 discarded cards.
5!/((5-1)!1!) = 5!/(4!*1!) = (5*4*3*2*1)/(4*3*2*1*1) = 5
5!/((5-2)!2!) = 5!/(3!2!) = (5*4*3*2*1)/(3*2*1*2*1) = 10
5!/((5-3)!3!) = 5!/(2!3!) = (5*4*3*2*1)/(2*1*3*2*1) = 10
5!/((5-4)!4!) = 5!/(1!4!) = (5*4*3*2*1)/(1*4*3*2*1) = 5
Notice that discarding 1 or discarding 4 have the same number of combinations, as do discarding 2 or 3. This is being they are inverses of each other. That is, if we discard 2 cards there will be 3 left, or if we discard 3 there will be 2 left.
Now we add together the combinations
1 + 5 + 10 + 10 + 5 + 1 = 32 choices combinations to discard.
The answer is 32.
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Note: There is also an equation for permutations which is:
n!/(n-r)!
Notice it is very similar to combinations. The only difference is that a permutation is an ORDERED arrangement while a combination is UNORDERED.
We used combinations rather than permutations because the order of the cards does not matter in this case. For example, we could discard the ace of spades followed by the jack of diamonds, or we could discard the jack or diamonds followed by the ace of spades. These two instances are the same combination of cards but a different permutation. We do not care about the order.
I hope this helps! If you have any questions, let me know :)
