T/y = x/w
Multiply each side by 'x':
Tw/y = x
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First, set up the integral W= /int_{0}^{0.12}k*x. Solve the integral on the right side. It should lead to W= (9/1250)*k. Substitute W=1 on the left side and solve for k. k should equal (1250/9) Substitute k=(1250/9) and x=0.12 in the equation F=kx and solve for F. F should equal (50/3).
r = distance she rowedw = distance she walkedt = total time to reach villaget = r/2 + w/5r^2 = 2^2 + (6 - w)^2r = (4 + 36 - 12w + w^2)^(1/2)r = (w^2 - 12w + 40)^(1/2)t = (w^2 - 12w + 40)^(1/2)/2 + w/5t' = (1/2)(1/2)(w^2 - 12w + 40)^(-1/2)(2w - 12) + 1/5 = 0(1/2)(1/2)(w^2 - 12w + 40)^(-1/2)(2w - 12) = -1/5(1/2)(w - 6) = -1/5(w^2 - 12w + 40)^(1/2)5(w - 6) = -2(w^2 - 12w + 40)^(1/2)25(w^2 - 12w + 36) = 4(w^2 - 12w + 40)(25 - 4)w^2 + (-25*12 + 4*12)w + 25*36 - 4*40 = 021w^2 - 252w + 740 = 0w = 6.87, 5.13 midiscard 6.87 ( > 6)r = (w^2 - 12w + 40)^(1/2)r = 2.18 miShe should land her boat 5.13 miles away from village
Suppose the length and width of the rectangle are L and W metres respectively.Then the perimeter, P = 20 m implies that2(L + W ) = 20 => L + W = 10 or W = 10 - L.Then Area = L * W = L * (10 - L) sq metres.
Type your answer here... C.H(w) > 0
The work done is transformed into potential energy (in the object lifted) so W = m x g x dz where W = Work m = mass g = acceleration due to gravity dz = change in height => W = 1kg x 1m x 9.81m/s/s => W = 9.81J 1J = 0.239 calories => It takes 2.345 calories to lift 1kg by 1m NOTE These are "real" calories. The calories usually talked about in relation to dieting etc are kilo-calories so to equate this value to the calories in a chocolate bar, divide by 1000