Page 49


Problem 1


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Problem 2


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Problem 3


Perhaps another card that is 10 inches long an 6 inches wide? Or one that is 5 miles long and 3 miles wide?



Problem 4


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Problem 5


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Problem 6


The altitudes of one are the perpendicular bisectors of the other

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Problem 7


When I see 9 and 15 in a problem, I am thinking of the triple 9, 12, 15.

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Problem 8


We have done this problem. Here is MY proof...

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Problem 9


Page 49 #9
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Problem 10


6cm and 9cm




Page 50


Problem 1 and 2


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Try this argument when AB is r times the length of AM, and when AC is r times the length of AN.



Problem 3


The cheetah can run a total of (105)(7)=735 feet before stopping to rest. The path of the antelope will take it on a path the is closer that 735 feet to the origin between -16 and 12 seconds. Ignoring the negative times because the cheetah is assumed to begin running at t=0 (or greater), the cheetah could catch the antelope at 12 seconds, which means it could wait for 5 seconds before running.

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Problem 4


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Problem 5


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Problem 6


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Problem 7


They are describing a 30-60-90 right triangle. The smallest angle is 30 degrees. The other side is whatever it in on this special triangle.



Problem 8


THIS IS A GREAT PROBLEM, LYDIA!



Problem 9


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Problem 10


Remember to keep point O inside the square.

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Problem 11


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The problem asks for a direction. I take this to mean BEARING.

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