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truly flattered namu! hope all is well

nice hand

example question: a can will only just fall over if the tennis ball has an energy greater than 0.1 J. find the kinetic energy of the can if the tennis ball has an incoming energy of: a) 0.01 J b) 0.5 J answer: a) 0.01 J is not enough to knock over the can, so the can remains stationary...

true for n = 3 assume true for n = k: Sk: 1/k! < 1/2^(k-1) (k > 3) now consider n = k+1: Sk+1: 1/(k+1)! < 1/2^k (1/k!) . [1/(k+1)] < [1/2^(k-1)] . (1/2) using the assumption that Sk is true, then Sk+1 is true under the condition that: [1/(k+1)] < (1/2) k > 1 since we're only...

the problem with this proof is that it assumes root ((((1 + root (((1 + root ((1 + root (1 + ...)))) converges to a finite number phi. for example, how do we know it doesnt shoot off to infinity instead? allow me to illustrate my point by considering a similar "repetition": ((( (...) - 1)^2 -...

subtract (n^2) + 1 from both sides

Re: Urgent help needed, question inside. delta y = uyt + (1/2)gt^2 -0.5 = -(1/2)gt^2 t = sqrt(1/g) (time of flight) delta x = uxt ux = (delta x) / sqrt(1/g) y component of the final velocity: vy = uy + gt vy = g . sqrt(1/g) (down) final velocity using vector addition...

yay friends

you're a party!

tycho tycho tycho

= 1~2 (x^-2 + 1) dx = (-x^-1 + x) |2,1 = (-2^-1 + 2) - (-1^-1 + 1) = 3/2 = 1~4 (1 - 3x^-3) dx = [x + (3/2)x^-2] |4,1 = [4 + (3/2)*4^-2] - [1 + (3/2)*1^-2] = 51/32

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integrate the function first: (2/3)x^3 - (7/2)x^2 + 5x then sub x=1, and subtract from when x=-3: [(2/3) - (7/2) + 5] - [(2/3)*(-3)^3 - (7/2)*(-3)^2 + 5*(-3)] = 200/3 integrate first: (1/2)x^2 + (1/3)x^3 + (1/4)x^4 then sub x=3, and subtract from when x=0: [(1/2)*3^2 + (1/3)*3^3...

sketch the graph of 2x + 1 (straight line with gradient 2, cuts the y-axis at 1). area would be the area under the line, between x=3 and x=6 and the x-axis (will look like the area of a trapezium).

ihihihih

janet janet janet

xD

hi diana doing awesome *high 5* hows yourself? xD

a) first find the total length of the 6 wavelengths using the equation v = d/t v = d/t 42.3 = d/3.6 d = 152.28 dividing by 6 will give us the length of 1 wavelength: 152.28/6 = 25.38 dividing by 2 will give us the length from trough to crest: 25.38/2 = 12.69m b) percentage error =...

∫ (1 - x^2)^(1/2) dx using the substitution x = sin u what happens to the absolute signs on cos u?