If p, q, r, ... are the roots of the equations, then (x-p), (x-q), (x-r), etc are the factors (and conversely).
Yes it is, where r is the radius
p=I*I*R ,P=V*V/R;where I is the current passing through the resistor, and V is the voltage across resistor, and R is the Resistance of the resistor,
FV of growing annuity = P * ((1+r)^n - (1+g)^n) / (r-g) P=initial payment r=discount rate or interest rate g=growth rate n=number of periods ^=raised to the power of NB: This formula breaks when r=g due to division by 0. When r=g, use P * n * (1+r)^(n-1)
d = r t t = d / r
p = r - c r - c = p r - c - r = p - r -(-c) = -(p) c = -p
Converse: If p r then p q and q rContrapositive: If not p r then not (p q and q r) = If not p r then not p q or not q r Inverse: If not p q and q r then not p r = If not p q or not q r then not p r
p+c=r.
square root of x/pi
pq-r, if p is 3, q is 4, and r is -6 is equal to 3 x 4 - (-6), which is equal to 12 + 6, which is equal to 18.
u = p r t r = u / p t
If P varies jointly as q, r and s - assume this is in direct proportion, then P ∝ qrs so P = kqrs where k is a constant.70 = k x 7 x 5 x 4 = 140k : k = 140/70 = 0.5When q = 2, r = 15 and s = 7 then,P = 0.5 x 2 x 15 x 7 = 105
PQR P=2 Q=4 R=5 2 x 4 x 5 = 40
P=s r t , so, s= P/(st)
Nothing. If I is current, V is voltage, and R is resistance, then V=I*R and V*I=P, where P is power.
You do it the obvious way. Take p, q in Q, the rationals By definition, we can write p = m/n and q = r/s where m, r are integers, n and s are natural. we define pq (p times q) = (mr)/(ns) p/q = pq^-1 where q^-1 denotes q's multiplicative inverse s/r Remark: you cannot divide by 0 here because 1) 0 have no multiplicative inverse 2) if r = 0. s/r is undefined.
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