Algebra 2
circle, ellipse and hyperbola all need x^2 and y^2 terms. You have x = ay^2 + by + c which is a parabola
hellllllllllllllp maaaath
My bad. I misread it. Even after looking at it several times, I still saw 21/√2 rather than 21√2.
hellllllllllllllp maaaath
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.
businessmathamatics
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
Extrema-Calc
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...
math
y=2x y=x+2 These are probably the simplest rules. Any way you can get from 2 to 4 will be acceptable.
algebra
5-12-13 is a common Pythagorean triple
Algebra
just plug in a=1000: N = 1500 + 200 ln1000 = 2882
math
correct
Reiny or Steve could you check my Math?
#3: -3h-5 The others look ok
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