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Which gives me 3.33... and 5 which adds to 8.333.... Thank you Sent from my LGLS755 using Tapatalk

For sum_(n=0)^5(.7^n) i got 2.94117 and for sum_n=0^5(.8^n) I got 6.63045 which adds to 6.63045 And the same for the original front 0 to 5. I guess I should try to find a ratio for the 2 sums? Sent from my LGLS755 using Tapatalk

All I tried was looking for a ratio and just adding up the first 13 terms. I didn't know i could split it Sent from my LGLS755 using Tapatalk

Sum_(n=0)^inf(.7^n + .8^n) = 25/3 I had to look up the answer but I can't find how to get it myself. The book only gives me a/(1-r) but there isnt a consistent ratio between the terms. Sent from my LGLS755 using Tapatalk

Thank you Sent from my LGLS755 using Tapatalk

1 polynomials are easy to derive 2 I'm on a chapter about l'hopital's rule Take a nap Sent from my LGLS755 using Tapatalk

(8/(x^2-4))-(x/(x-2)) (8(x-2)-x(x^2-4))/(x^2-4)(x-2) (8x-16-x^3+4x)/(x^3-6x+8) 8x+4x=12x Lay hospital's rule: (12-3x^2)/(3x^2-6) Again: (-6x/6x) -1 Sent from my LGLS755 using Tapatalk

I evaluated the limit by multiplying common denominators then distributed terms so I have polynomials on top and bottom. Then derived each separately and got -6x/6x=-1 but the limit is actually -3/2. What's up with that? . . . . .\displaystyle \lim_{x \rightarrow 2^+}\, \left(\dfrac{8}{x^2\...

Ok so I evaluated the indefinite integral and got the right answer but when I plugged in the bounds, I got the wrong answer. Calculate the definite integral: . . . . .\displaystyle \int_1^3\, f(x)\, dx\, =\, F(x)\, =\, \int_1^3\, \dfrac{x\, +\, 1}{x\, (x^2\, +\, 1)}\, dx Computed by Maxima...

1 and 0 are constants too Sent from my LGLS755 using Tapatalk

What's the big deal with e^(ipi)+1=0? Like it's the 5 most common constants in math, shouldn't we expect the 5 most common ones to show up together? The fact that it has addition, multiplication, and exponentiation is impressive though. Sent from my LGLS755 using Tapatalk

Ok so from y=e^(-x^2) to x=b is being revolved around the y axis and the resulting solid is 3/2 cubic units. I set up the shell method: 2pi int_0^b (x(e^(-x^2)))dx = 3/2 u=-x^2 du=-2xdx u at b = -b^3 u at 0 = 0 -pi int_0^(-b^2) (e^u) du = 3/2 -pi [e^(-b^2) - 1] = 3/2 e^(-b^2) - 1 = (-3/2pi)...

I don't remember how i did this: "Find fluid force on vertical side of tank where..." Find the fluid force on the vertical side of the tank, where the dimensions are given in feet. Assume that the tank is full of water. (The weight-density of water is 62.4 pounds per cubic foot.) So F = (w)...

Its a parabola bounded by x=0 and y=8. Sent from my LGLS755 using Tapatalk

Found the problem Sent from my LGLS755 using Tapatalk

Sorry here's the whole thing: Y= 25-x^2 y-25=x^2 sqrt(25-y)=x Int_0^25 (25-y)^(1/2)dy U=25-y du=-dy -int_0^25(u^(1/2))du -[(2/3)u^(3/2)]_0^25 -[(2/3)(25-y)^(3/2)]_0^25 -[-(2/3)(25)^(3/2)] [(2/3)(125)] (250/3) Int_0^b ((25-y)^(1/2)dy=(250/3) U=25-y Du=-dy -int_0^b(u^(1/2))du=(250/3)...

^wrong btw Sent from my LGLS755 using Tapatalk

Y= 25-x^2 y-25=x^2 sqrt(25-y)=x Int_0^25 (25-y)^(1/2)dy U=25-y du=-dy -int_0^25(u^(1/2))du -[(2/3)u^(3/2)]_0^25 -[(2/3)(25-y)^(3/2)]_0^25 -[-(2/3)(25)^(3/2)] [(2/3)(125)] (250/3) Also how do I know when to use x and when to use y? Sent from my LGLS755 using Tapatalk

Thank you Sent from my LGLS755 using Tapatalk

1. LarCalcET6 7.1.069. Find b such that the line y = b divides the region bounded by the graphs of y = 25 - x2 and y = 0 into two regions of equal area. I got: . . . . .\displaystyle \int_0^5\, (25\, -\, x^2)\, dx\, =\, \dfrac{250}{3} Then tried to find: . . . . .\displaystyle \int_0^b\...