The period (T) of a circle is represented by the equation: T=1/F, where F is the frequency.
F=ma a=∆v/∆t F=m(∆v/∆t)
They are inverse to one another. Since f * T = 1 f - frequency and T- time period
Suppose A is a vector with real components. A can be written as <f(t), g(t), h(t)>. Since the magnitude of A is constant we have f(t)*f(t) + g(t)*g(t) + h(t)*h(t) = c, where c is a non-negative real number. Take derivative of both sides of equation we get 2*f(t)*df(t)/dt + 2*g(t)*dg(t)/dt + 2*h(t)*dh(t)/dt = 0. Divide both sides by 2, we get f(t)*df(t)/dt + g(t)*dg(t)/dt + h(t)*dh(t)/dt = 0. Thus the dot product of A and its derivative is 0. This implies the angle between A and its derivative is Pi/2. Hence they are perpendicular.
Time period T = 1 / frequency f. Frequency f = 1 / time period T. T = 1 / f = 1 / 200 = 0.005 seconds = 5 milliseconds.
OTTFF stands for the first five numbers.OneTwoThreeFourFive
5 T on the F means 5 Toes on the Foot
tit festy tit
88 is two fat ladies in bingo
what freaking time
Do you mean f(t) = -16t^2 - 48t + 160? This is a function, which can't be "solved" as it is. - you could factor it: -16(t + 5)(t - 2) - you solve for f(t) = 0: t = 2, -5
Here is its truth-table: A B A and B F F F F T F T F F T T T
twenty fingers and toes
p > q~qTherefore, ~p| p | q | p > q | ~q | ~p || t | t | t | f | f || t | f | t | t | f || f | t | t | f | t || f | f | t | t | t |
R mean reastate the question. A mean answer it. F mean for example. F mean for example. T mean this show that. RAFFT that what it mean in Ela
P | T T F F Q | T F T F Q' | F T F T P + Q' | F T F F The layout is the best I could do with this software. Hope it is OK.
P Q (/P or /Q) T T F T F T F T T F F T