collect x's on one side, numbers on the other: 5x = 16
Amazing. 3x-2(3x-2) -3x+6x-4 -3x+4 The answer is correct, but the middle line is just backwards. Expanding the 2nd term, I get 3x-6x+4 I hope the logic carries through on other types of problems. How ever did you arrive at that intermediate step?
assuming the diagonals are ac and bd, then they bisect each other. The midpoints of the ac abd bd are (5,2) and (5,2). Hmmmph. whaddaya know!
LCM(12,16) = 48 so, 48/12=4 and 48/16=3 laps, respectively.
The graph did not come out but it is a square with b from >(1) -> ((2)) and >(1) -> (3) and L from (4) -> (3) and (3) -> >(1) and a from ((2)) -> (4)
Prove by induction on all positive integer k that if m is any ordinary nfa with k states, and m has fewer than k - 1 transitions, then there exists a state of m that is not reachable. Let N be the λ-NFA: "L" for "λ" b >(1) -----> ((2)) | ^. ...
what's the problem? As usual, take the drivative. y' is the velocity. y' = -e^-t sin2t + 2e^-t cos2t = e^-t (sin2t + 2cos2t) so, velocity=0 when sin2t + 2cos2t = 0 sin2t = -2cos2t tan2t = -2 so, find arctan(-2) and add multiples of pi/2 to t
that's a strange function. Since x^2 >=0 and y^2 >=0, you need x=y=0. Just one point. Now, for x^2 + √(x+y) + y^2 = k for some other k>0, 2x + (1+y')/(2√(x+y)) + 2yy' = 0 y' = -(2x + 1/(2√(x+y))/(2y+1/(2√(x+y))) = -(4x√(x+...
since lim x->2 f(x) = 1, we see f is not continuous, since f(2) = -1
well, if cos^2(2x) = 1/2, cos(2x) = ±1/√2, so 2x = π/4 or 3π/4 or 5π/4 or 7π/4 That means that x is π/8,3π/8,...
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