When (p = 3),2 p2 = 2 (3)2 = 2 (9) = 18 .(2p)2 = 4 p2 = 4 (3)2 = 4 (9) = 36
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for prime p>= 5 p = (6n+/-1) p^2 + 2 = 36n^2+1 +/- 12n +2 = 36n^2+3 +/- 12n divisible by 3 and p^2+2 >= 27 so 2nd factor >=9 so not a prime or composite
I'll try to answer the question, "If the 5th term of a geometric progression is 2, then the product of its FIRST 9 terms is --?" Given the first term is A and the ratio is r, then the progression starts out... A, Ar, Ar^2, Ar^3, Ar^4, ... So the 5th term is Ar^4, which equals 2. The series continues... Ar^5, Ar^6, Ar^7, Ar^8, ... Ar^8 is the 9th term. The product P of all 9 terms is therefore: P = A * Ar * Ar^2 *...*Ar^8 Collect all the A's P = (A^9)*(1 * r * r^2 ...* r^8) P = A^9 * r^(0+1+2+...+8) There's a formula for the sum of the first n integers (n/2)(n+1), or if you don't know just add it up. 1+2+...+8 = 36 Therefore P = A^9 * r^36 Since 36 is a multiple of 9, you can simplify: P = (Ar^4)^9 Still with me? Remember that Ar^4=2 (a given fact). So finally P = 2^9 = 512. Cute problem.
9=15+2p-15 -15 and continue to get ur answer :)p=-2
The number of trials: n = 9, the number of success: r = 4. The probability of success: P(H) = 1/2, the probability of failure: P(T) = 1/2. P(exactly 4 heads in 9 tosses) = 9C4 x (1/2)^5 x (1/2)^4 = 126(1/2)^9 = 0.246 P(at most 2 heads) = P(none heads) + P( 1 head) + P(2 heads) = 9C0[(1/2)^9][(1/2)^0] + 9C1[(1/2)^8][(1/2)^1] + 9C2[(1/2)^7][(1/2)^2] = 1(1/2)^9 + 9(1/2)^9 + 36(1/2)^9 = (1 + 9 + 36)(1/2)^9 = 46(1/2)^9 = 0.09
P= 2(l + w) P = 2( 9 + 12) P = 2(21) P = 42 cm.
answer is p/5. problem: {[(p^2)-3p]/[(p^2)-6p+9]}/{20/(4p-12)}
Parco P-I- - 2005 The Out of Towner 2-9 was released on: USA: 28 September 2006
1/2(9+p)=p-3
P = 2(2 ft 3 in + 9 in) = 2(2 ft 12 in) = 2(3 ft) = 6 ft Or, P = 2(2ft 3 in + 9 in) = 2(27 in + 9 in) = 2(36 in) = 72 in (which also are 3 ft)
When (p = 3),2 p2 = 2 (3)2 = 2 (9) = 18 .(2p)2 = 4 p2 = 4 (3)2 = 4 (9) = 36
24 - 2001 Day 2 400 p-m--500 p-m- 2-9 is rated/received certificates of: Netherlands:12 Portugal:M/12 (DVD rating)
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Police P-O-V- - 2011 2-9 was released on: USA: 24 December 2011
2.11111.... ( Recurring to infinity Let P = 2.1111.... & 10P = 21.1111.... Subtract 9P = 19 ( NB The repeating decimals subtract to zero) P = 19/9 P = 2 1/9 Done!!!!!