A P2 costs $9.00 per user/month.
p2+2pq+q2=1
MRS = |-p1/p2|
Even
p2 - 8p = 0 The square would be (8/2)2 = 42 = 16 So, p2 - 8p + 16 = 16 then (p - 4)2 = 42 So that (p - 4) = +/- 4 then p = 4 +/- 4 ie p = 8 or p = 0 With this particular question, factorisation is so much easier: p2 - 8p = 0 p(p - 8) = 0 So p = 0 or p = 8 Done!
3 and p2 are both factors of each part of the binomial, so divide it from each part and see what you have left. (3p4 / 3p2) - (12p2 / 3p2) (3p2)(p2-4)◄
* 0.4A * (voltage of device V) = Power in watts P1 * P1/1000 = Power in kW P2 * P2 * (electric rate in dollars per kWhr) = Cost dollars/hr C * C * (days in this month) * 24 hr/day = Cost in dollars for this month
P2-5,000/teeth.
P2-5,000/teeth.
Let p1 and p2 be the two prime numbers. Because they are prime, their divisors are div(p1) = {1,p1} and div(p2) = {1,p2}. So GCD(p1,p2) = Greatest Common Divisor of p1 and p2 = p1 if p1 equals p2 1 if p1 is different from p2
The bond order of P2 is 2.
No. Let p1 be a prime number. Let p2 be a multiple of p1 such that p2 = p1 * k. Then the factors of p2 are: 1, p1, k and p2. ==> p2 is not a prime number. Hence, a multiple of a prime number cannot be a prime number.
p2+10d+7
p2 + 3p = p (p + 3)
p2 - 25 = (p - 5)(p + 5)
The only possible geometry of a diatomic molecule such as P2 is linear.
#include <reg51.h> sbit p10=P1^0; sbit p11=P1^1; sbit p12=P1^2; void delay()//Delay process { unsigned i,j,k; for(i=0;i<500;i++) {} } main() { while(1) { if(p10==0) { P2=0X01;//Inversion of control //0011 delay(); P2=0X02;//0110 delay(); P2=0X04;//1100 delay(); P2=0X08;//1001 delay(); } if(p11==0) { P2=0X08;//Control is transferred delay(); P2=0X04; delay(); P2=0X02; delay(); P2=0X01; delay(); } if(p12==0) { P2=0X00; delay(); } } }
The molecule P2 is obtained by heating P4 over 800 oC.