x f(x) 0 1 1 e 2 e^4 So, if there are 2 rectangles of width 1, then the area, using left-sides is 1*1 + 1*e = e+1 = 3.718 using right-sides, it's 1*e + 1*e^4 = 57.316 Using the trapezoidal rule, we have 1(1+e)/2 + 1(e+e^4)/2 = 30.517 Kind of a coarse approximation.
Math work problem
If there are now n sickos, then the next week there will be n+2n=3n. After x weeks there will be n*3^x sick. So, we need 2*3^x >= 30 3^x >= 28 x >= 3
(40*12*30*12*20*12)/(200000*(4*3*4)) = 4.32 so, yes, the warehouse is plenty big enough. Or, considering that 1 ft^3 holds 36 boxes, it requires only 5555.5 ft^3 for the whole batch, and the room is 24,000 ft^3, there's plenty of room.
#1. check out sin(π - π/3) #2. because 1/0 is also undefined. #3. just factor the function (sec+6)(sec+2) convert to cosines, then factor to get (6cos+1)(2cos+1)=0 #4. 2π+π/6,2π-π/6 and all multiples of 2π more or less.
(a) vertex of the parabola is at t = -b/2a (b) plug in that value to get h (c) solve for t when h=0
if the squares are of size x, then the volume is v = x(20-2x)^2 = 4x(10-x)^2 Without benefit of calculus, you still know the general shape of the curve. Since we want integer values, just start checking. x v 1 324 2 512 3 588 4 576 Now v is decreasing, so keeping all values in...
I guess 10/1
ax^2+bx+c=y 0a+0b+c=1.39 324a+18b+c=1.5 2025a+45b+c=0 a=-.00137 b=0.0308 c=1.39 y = -.00137x^2 + 0.031x + 1.39 max height on the parabola is at x = -b/2a, so at x = .031/.00274 = 11.31 y = 1.565
Trig-Weird Geometry Problem
You will find a nice discussion at http://answers.tutorvista.com/1101359/trigonometry-application-rule.html showing that if A+B+C=180, tan(A-B)+tan(B-C)+tan(C-A) = tan(A-B)*tan(B-C)*tan(C-A) Given that, one of the factors must be zero. So, A=B or B=C or A=C. That is, the trian...
first year chem
Well, since P weighs 31g and the isotopes of chlorine weigh 35 and 37 So, one mole of PCl3 is either 31+3*37=142 or 31+3*35 = 136