A) in a group of 2,000 people, must at least 5 have the same birthday? why? b. show that if seven integers are selected from the first 10 positive integers there must be at least two pairs of these integers with the sum 11.

A) The answer is yes. 
Given 365 people, two at least must have the same birthday (obvious).
Now multiply the above number by 5 we get : 1825.
Since  then 5 at least will have the same birthday. 
b)

Think of the number 11 and how many manner we can sum to 11:
1+10=11
2+9=11
3+8=11
4+7=11
5+6=11
So there is five ways. Suppose we draw one number. There is no way to draw 7 numbers without obtaining a couple forming 11. 

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12