The clue here is that we want to find extrema of f(x), which involves finding the derivative of f. Since f(x) is defined as an integral, we don't really have to do the integration. We just apply the rules for differentiating under the integral sign. (See wikipedia, and scr...
current revenue: 80 scoops @ $0.90 = $72.00 proposed revenue: 98 scoops @ $0.80 = $78.40 clearly the profit goes up by $6.40
Your substitution for u is correct, so you wind up with ∫6u^3 du Since u=x^2/3 + 14, x=1 ==> u=15 x=8 ==> u=18 and you have 3u^4/2 [15,18] = 157464 - 75937.5 = 81526.5 Not sure where you got 98304
15000seeds/oz * 1pack/300seeds * $1.50/pack = $75/oz
if the signs of x^2 and y^2 terms are the same it is a circle or ellipse If the coefficents are different, it's an ellipse Since you have 11x - x^2 - 4y^2 = 2y - 16 x^2 + 4y^2 - 11x + 2y = 16 you have an ellipse. You will wind up with something of the form (x-h)^2/4 + (y-k...
well, y = (x-2)^2 (x-3)(x+3) so, 4th degree polynomial, roots at 2,3,-3 tangent to x-axis from below at x=2 opens upward on both sides
circle, ellipse and hyperbola all need x^2 and y^2 terms. You have x = ay^2 + by + c which is a parabola
My bad. I misread it. Even after looking at it several times, I still saw 21/√2 rather than 21√2.
Hmmm. I don't buy it. You can't have x = 1/√2 , y = 41/√2 , and k = 21√2 and x^2 + (y-k)^2 = 1^2 Maybe I'll visit the figure.
since a = pi r^2 da/dt = 2pi r dr/dt Now, if da/dt = a constant k, 2pi r dr/dt = k dr/dt = k/(2pi r) since c = 2pi r, dc/dt = 2pi dr/dt = k/r So clearly dc/dt changes as r changes
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