sorry, you guys are right. for some reason there was a glitch, even if i selected converge, i got 50% of the problem correct(which means i got the first question right). but if i selected diverge, i got 100% of the problem(two problems = 100%) correct.
Consider the series
\sum_{n=1}^\infty \frac{n}{\sqrt{5n^2+5}}
Value ______
a_1 = .316227766, a_2 = 2/5, a_3 = .4242640687, a_4 = .4338609156
there doesn't seem to be any common ratio, so that means that this isn't a geometric series right?
well i think i can simplify the equation...
Determine the sum of the following series
\sum_{n=1}^\infty \frac{2^n+6^n}{9^n} or can be written as...
\sum_{n=1}^\infty \frac{8^n}{9^n}
A_1 = 8/9, A_2 = 64/81, A_3 = 512/729
common ration (r)= 8/9
first term (a)= 8/9
so plugging everything i know into the geometric series...
An = \frac{5n}{12n+5}
For both of the following answer blanks, decide whether the given sequence or series is convergent or divergent. If convergent, enter the limit (for a sequence) or the sum (for a series). If divergent, enter INF if it diverges to infinity, MINF if it diverges to minus...
Find the area of the region which is inside the polar curve r =5*cos(\theta)
and outside the curve r = 3-1*cos(\theta)
when i plugged those two functions into my calculator and found the bounds from 1.0471976 to 5.2359878.
my integral:
\int_{1.0471976}^{5.2359878}...
Find the area of the region bounded by: r^2 = 128*cos(2\theta)
\int 1/2*(\sqrt{128*cos(2\theta)})^2
that's my integral, but i don't know what the bounds are. i tried typing it into my calculator, but i could only find one bound(.78539816) and I am not even sure if this is even correct. can...
Find the area of the region bounded by: r= 6-2sin(\theta)
here's what i did:
6-2sin(\theta) = 0
sin(\theta) = 1/3
so the bounds are from arcsin(-1/3) to arcsin(1/3) right?
my integral:
\int_{-.339}^{.339} 1/2*(6-2sin(\theta))^2
i get a answer of 0.6851040673*10^11, and it's...
Find the length of parametrized curve given by
x(t) = 0t^3 +12t^2 - 24t,
y(t) = -4t^3 +12t^2+0t,
where t goes from zero to one.
Hint: The speed is a quadratic polynomial with integer coefficients.
it's an arclength question right?
x' = 24*t-24
y' = -12*t^2+24*t
\int_{0}^{1}...