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Question
1.)mary is 24 years old.mary is twice as old as ana was when mary was as old as ana as now.how old is ana?
2.)the sum of the prent's age is twice the sum of their children's ages.in 15 years,the sum of the parent's ages will be equal to the sum of their children ages.how many children were in the family
3.)3.)lot ABCDEFA is a closed traverse in the form of a regular hexagon w/ each side equal to 100m..the bearing of AB is N 25 degrees E.what is the bearing of CD.?

4.)a non-square rectangle is inscribed in a square so that each vertex of the rectangle is at the trisection point of the diff. sides of the square.Find the ratio of the area of the rectangle to the area of the square.?

5.)a square of edge "a" revolves about a line through one vertex,making an angle "A" w/ an edge and not crossing the square .Find the volume generated.?


Answer
1) If Mary is twice as old, she has lived as long after Ana’s birth as before.  That means that when she was Ana’s age, Ana was just born.  Ana’s age is 0.


2) Let's look at it a simple way.

If they had two kids, in 15 years, there is no way the kids could add up.

If they had 3 kids, in 15 years the kids age would be 45 bigger and the parents would only be 30 bigger.  

Let P=sum of the parents age.
Let K=sum of the kids age.

3 kids implies P=2K, P+2(15) = K+3(15) =>
2K + 30 = K + 45 => K = 15 => P+30 = 60 => P=30,
which makes no sense, since the kids could be 4, 5, and 6,
but that would mean the parents started at 9!

4 kids implies P=2K, P+2(15) = K+4(15) =>
2K+30 = K+60 => K = 30 => P = 2K = 60, so they could both be 30.
That makes sense since the kids could be 6, 7, 8, and 9,
which means the parents started at 21.  This is the most likely.

5 kids implies that P=2K, P+2(15) = K+5(15) =>
2K + 30 = K + 75 => K = 45 => P = 90.  
That means the parents would be 45 each and the kids would be 7,8,9,10, and 11, so the parents waited until they were 34 to have the first kid.

6 kids implies that P=2K, P+2(15) = K+6(15) =>
2K + 30 = K + 90 => K = 60 => average age of kids is 10.
However, the sum of the parents age would be 120, which is kind of silly.


3) The bearing of each leg will be 60 degrees farther to one side than the last.  I don't see where it says which way this hexagon goes.  If it is going clockwise, the bearing of CD would be S 35 E.
If it is going counterclockwise, the bearing of CD would be S 85 W.


4) If the rectangle's corners are at 1/3 from each of one corner and 1/3 from the opposite corner, then the lengths of the sides can be found by the triangle rule a² + b² = c².  We'll label the sides c1 and c2.  In one case that is (1/3)² + (1/3)² = 2/9 = (c1)².
IN the other case that is (2/3)² + (2/3)² = 8/9 = (c2)².
c1²*c2² = 16/9, therefore c1*c2 = 4/3.


5) This problem is not written out to where I can understand it, but I will try my best.
The angle of a rotating edge of a square if it goes in a complete circle is πa².
If it only swings far enough to stay inside the square, the farthest it could swing is 90 degrees, which is ¼ of all of the way around..  the area of the part enclosed would be ¼ of a circle, so it would have area πa²/4.
If it was swung outside of the square, it would sweep across ¾ of the circle, so it would have area 3πa²/4.
If it were swung in an angle A, it would have are Ar²/2 where A is the angle in radians.

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