Recent content by ThuanSUX

An alternative solution to what Watatank said would be: 1. Combine the equations 2. Factorize y = x+2 y = x^2-5x+11 x^2-5x+11 = x+2 x^2-6x+9 = 0 (x-3)(x-3) = 0 (x-3)^2 = 0 thus, x=3 ONLY Resubstituting it into the first equation to obtain a y-coordinate and you get the point (3,5)...

Here's an alternative solution: rt[3 + rt(8)] = rt [1 + 2rt(2) + 2] = rt[(1 + rt(2))^2] <--- by completing the square = 1 + rt(2)

I'll elaborate on what followme said: Area required = Area of rectangle - Area of curve and y-axis = 2ln8 - int(x)dy (lim: 0->ln8) = 2ln8 - int(1/4e^y)dy (lim: 0->ln8) = 2ln8 - 1/4[8-1] = 2ln8 - 7/4 *I got 1/4e^y by rearranging y=ln(4x) in terms of x **0 and ln8 are...

There is no difference between the multiple subinterval and the simple formula. The complete Simpson's rule is given by: f'(x) = h/3[(y0+yn)+4(odd)+2(even)] Whereas the simplified version is designed for a function with ONLY 3 subintervals. Taking this into account, you only have y0, y1 and...

Btw, when you have boundaries for an integral, that's a definite integral. If you have an indefinite integral you should +C.

Simply put: The pollution is increasing, but at a decreasing rate. dP/dT > 0 and d2P/dT2 < 0 Thus, it's a maximum.

That's what I meant, but I couldn't figure out how to do the greater/equal and lesser/equal symbol and cbf looking. I believe what I said was correct. For f(x) > 0, you simply use the > sign instead of >, and likewise for lesser than. :)

I agree with b35ty about drawing the parabola. From the factorized form you can mark your roots on the x-axis. From here draw it out (remember to watch out for the (2-x) case because the parabola is heading downwards). If you want f(x) < 0, then it's the part of the parabola under the x-axis...

If the focus is F(-3,3) and directrix is y=5, then your vertex is (-3,4). Same x-co-ord as F and average of y-co-ord. Thus your focal length is 1 (y distance from either F or directrix to vertex). (x+3)^2=-4(1)(y-4)