All Answers 359230 Answers

Hi Spongebob,

Deleted,  my apologies.

Your answer is very interesting.

r = ~0.86756

 ∑[ n*0.86756^(3n -1), n, 1, ∞] = ~6.25 

Hi Guest,

No there is no reason, you can use any log you want. 

Log 10 is an equally good alternative.  

Evaluate      ∏[n / (n+3), n, 1, 50] =1 /23,426

Setting up 2 equations:

\(y=x+21\)

\(y=4x\)

\(4x=x+21\)

\(3x=21\)

\(x=7\)

\(y=28\)

x=time and y=distance

The answer: it takes 7 seconds and the adults go 28 meters

A pair of kids and a pair of adults decided to compete in a three-legged race. The kids got to start 21 meters ahead of the adults, since they had shorter legs. When they were told to start, the kids hobbled forward at a rate of 1 meter per second, and the adults hobbled after them at a rate of 4 meters per second. Soon they were side-by-side. How long did that take? How far did the adults go?

Sorry, but that answer is incorrect, I thought it was 17 too.

S = Sum[1/n^3; n = 0; n = n + 2; n <= inf] = 1/4*zeta(5) and T = Sum[1/n^3: n = 1; n = n + 2; n <= inf] = 3/4*zeta(5), so S/T = 1/3.

Call this series S. We compare S to 1/3 S:

Subtracting the second equation from the first, we see that

Therefore,

∑[(3n - 2) / 3^n, n, 1, ∞] = 5 /4

The height (in meters) of a shot cannonball follows a trajectory given by $h(t) = -4.9t^2 + 14t - 0.4$ at time $t$ (in seconds). As an improper fraction, for how long is the cannonball above a height of $6$ meters?

Ooh, sorry I can't help you. I haven't learned this yet...

How.

So we know that 8 girls brought lunchboxes. And 4 didn't. And we also know that 2 boys brought a lunchbox, but 6 didn't.

This means that there are 10 kids in total who brought a lunchbox. 10/20 = 1/2. So if you choose a random child from the class, you have a 1/2 chance of getting a kid that has a lunchbox.

Please reply if I'm wrong.

It doesn't end, it's a infinite geometric sequence.

How do we know where it ends?

The answer is 142

The foci are at (-5*sqrt(2)) and (5*sqrt(2)), and the hyperbola represents the set of points such that |PF_1 - PF_2| = 10.  So if the distance from P to one focus is 7, then the other distance is 17.

Evaluate ∑[n(n +1)^2, n, 1, 19] =41,230

∏[n^2 / (n^2 - 1), n, 2, 2006] = 1.999003488