Expert: Sherry Wallin - 3/18/2009

Question
can you help me with my assignment...

pls help me
this is my assignment
i only answer question number 1,2.
pls help me to show some solutions..
thnx..


1. Evaluate:

[sqrt(-2009)][sqrt(-2009)]

2. The perimeter of a right triangle is 15 + 20sqrt(22). The sum of the squares of all sides is 396. Find the area of the triangle.

3. A certain chord of a circle is 12 inches and is the perpendicular bisector of a radius of the circle. Determine the area of the circle.

4. Find the 2009th leftmost digit starting from the decimal point in the expansion of

1/59

5. In a local factory, 10 men work 9 hours a day, all at the same rate to produce C articles. If 4 men are released, how many hours per day will the remaining men have to work to produce 2C articles?

6. Find the minimum value of

[9(x^2)(sinx)^2 + 4]/(xsinx)

for 0 < x < pi.

7. The numbers 1447, 1005 and 1231 have something in common. Each is a four-digit number beginning with 1 that has exactly two identical digits. How many such numbers are there?

8. Find the value of:

10cot[arccot(3) + arccot(7) + arccot(13) + arccot(21)]

9. The increasing sequence

1 , 3 , 4 , 9 , 10 , 12 , 13 , ...

consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the 2009th term of this sequence.

10. Evaluate the definite integral:

integral [(ln x)^2009]dx

lower limit : 0
upper limit : 1


Answer
Charise~
    Hi I only have time to answer a couple of these before I go to work.
3) since the chord bisects the radius this means that you have a right triangle with one leg [1/2]r and the other 6, use pythagoreans theorem to find r: [(1/2)r]^2 + 36 = r^2 --> (1/4)r^2 + 36 = r^2 -->
[3/4]r^2 = 36 so r^2 = 48 so  of course then the area is pi*r^2 = pi*48 = 48*pi square units.

7) since the first digit is always 1 there is only 1 way to choose the 1st digit. The 2nd digit could be 1 and if it is then the other two digits will be different, or if the 2nd digit is not 1, then either the 3rd or 4th would match one of the first, second, or 3rd or 4th. This problem can be solved by examining all the cases.

Case I:  Suppose digit 2 is 1 then there are 9 choices for the 3rd and 8 choices for the 4th(they can not be 1 and the 3rd cannot equal the 4th)so there are 1*1*9*8 = 72 numbers with 2 digits a 1 and no other digits repeated. You can generalize this to any one of the other 3 digits being a 1 and the other two not repeats.

Case II: 1 is not the duplicated digit: 1 way to choose the first digit being a 1 and there are 9 ways to choose the 2nd (it can't be a 1) and if we allow digit 2 and 3 to be the same then there is only one way to choose digit 3 and 8 ways to choose the 4th giving us the same number of 4 digits numbers beginning with a 1 and having no more than 2 digits repeated as in case I. 1*9*1*8 = 72 and again you can generalize this to digit 2,3, or 4.

Have to go to work now.

Math Prof