1 ok 2. b^logb(x) = x x = log.5 y .5^x = .5^log.5(y) = y so y = .5^x 3. x = 2^y - 3 x+3 = 2^y log2 (x+3) = y 4. x = 6 + log y log base e or 10 ??? I will assume 10 x-6 = log y 10^(x-6) = y y = 10^x/10^6
It takes him 35 + 4 = 39 minutes to get there and 39 minutes to get back so 78 minutes for a round trip 5 hours = 300 minutes 300 + 12 = 312 minutes traveling 312/78 = 4 round trips
I do not know
Do we have any idea what f(x) is? I suppose in general you can integrate by parts int u dv = u v - int[ v du] here maybe u = f(x) and dv = x dx du = f'(x) dx and v = x^2/2 then we would get (x^2/2)f(x) - int [ (x^2/2) f'(x) dx] but that seems rather pointless.
Physics help please
Tension force in the direction of motion = 125 cos 30 = 108 N Work done by rope tension = Tension force in the direction of motion times distance moved = 108*5 = 541 Joules Since the velocity is constant, the horizontal tension force, 108 N is equal and opposite to the frictio...
F = m a G m M/r^2 = m V^2/r m, the mass of the satellite, is irrelevant, cancels. G M = V^2 r distance around circumference = 2 pi r so t = 2 pi r / V or V = 2 pi r/t so G M = (2 pi)^2 (r^3/t^2) or t^2 = (2 pi)^2 r^3/ (G M) here G = 6.67*10^=11 M = 5.98 *10^24 Kg r = 6.38*10^6...
forget the last few lines there, you do not need it
V = i R + L di/dt i will be of form i = I (1-e^kt) where I is the current after a long time determined only by the battery and the resistance because the inductor has no resistance once steady stage is reached. at t = infinity, i =V/R so I = V/R and at t = 0 , V = L di/dt and ...
Anything bigger than 1 that ends in a 0 is divisible by five and ten because ten is two times five anything bigger than 5 that ends in a five is divisible by five so I know that both top and bottom are divisible by 5 (5 * 11) / (5 * 16) or 11/16
This is a very, very, ideal problem. Nothing will happen unless the electron strays from its course perfectly down the B lines :) (unstable equilibrium)
y = -2 x^4 + 9 x^3 + 2 x^2 - 39 x + 18 make a table of x and y x _____ y -2 _____ 0 oh wow -1 _____ 48 0 ______ 18 +1 ______ -12 so a zero between 0 and 1 well, at least we had a zero at x = -2 so (x+2) is a factor so divide this mess by (x+2) so I have (x+2)(-2x^3+13x^2-24x+9...