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.
If you dcrease the temperature you will decrease the pressure proportionately. So, T1 over T2 will equal P1 over P2. We can derive the formula P1 x T2 = P2 x T1. Substitue the values and we get 50.0 mm x 200K = P2 x 540K 10,000 mmK = P2 x 540K 10,000mmK / 540K = P2 P2 = 18.52 mm of Mercury in a constant volume
#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(); } } }