S orbital
(p + 12)(p - 7) p = 7, -12
p2 - 2p + 2 can be factored as (p - 1)(p - 1)which can be written as ( p - 1)2.
You use a combination of Ohm's law ( V = I * R ) and the power formula ( P = V * I ).Substituting you get: P = ( I * R ) * I = I**2 * RDividing both sides by R you get: P / R = I**2 * R / R = I**2Taking the square root of both sides you get: SQRT( P / R) = ISwapping sides of the equals sign you get: I = SQRT( P / R )Thus the maximum current can be found using the equation I = SQRT( P / R ).
Nothing. If I is current, V is voltage, and R is resistance, then V=I*R and V*I=P, where P is power.
2 peas in a pod
2=p in a q
22
p+2 = 2-3p p+3p = 2-2 4p = 0 p = 0
p+1/4 greater than and less than 2
a = ±2.12, depending on the inequality for Z.
p = -1
It is p <= 0.
PQR P=2 Q=4 R=5 2 x 4 x 5 = 40
p + (-2) = 1 so p - 2 = 1 Add two to both sides: p = 3
no it equals one :P
Solving by elimination: p = 3 and q = -2