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ok so

if we simplify the numerator of the fraction, we get \(((3!)!)!=(6!)!=720!\)

so our new fraction is \(\dfrac{720!}{3!}\) which is equal to \(\dfrac{720 \cdot 719!}{6}\)

we can then cancel out the 720 and 6, and we get \(120\cdot719!\)

so, \(719+120=\boxed{839}\)

I need help here too!

It looks like your sequence should go like this:

1,  1,  2,  4,  8,  16,  32, 64, 128, 256, 512......and so on.  Each term is obtained by 2 raised to the term number -1. Examples: 5th term =2^(5-1) =2^4 =16. 10th term =2^(10-1) =2^9 =512.

Therefore, the 100th term =2^(100 - 1) =2^99 =633,825,300,114,114,700,748,351,602,688

I guess it's because marshlands will temporarily store water and then slowly release it so that there won't be too much water at once in a nearby town. Also, this isn't the kind of place to ask science questions.

lol yeah

They do cars, all those.

Just in case you ever need to command a fleet.

It was a math question not a battleship strat (but thank you)

You find the gcf of 120, 96, 168

Search up gcf of 120, 96 and 168

And it's not the best idea to split your fleet into small groups.

You should split them into groups with a few capital ships and a bunch of support/destroyers

i fixed it help me now pls

Thank you for trying, appreciate the effort guest but your answer is wrong. can someone help me out? Thank you guest!!!

By similar triangles, BD = 8.2.

You must consider the probability that drawing BB, BWB, WWB, or WWBB.  This gives you a probability of 2/9 + 5/81 + 12/81 + 6/81 = 41/81.

 (10! + 11! + 12!)/10! = x^2

144 = x^2

Take the square root of both sides:

x = 12

it could also be (1,1,12)

for example the are 1 and x for the first one we get \(\cfrac{\sqrt3}{4}\)for the first area and for the height of the other triangle we have \(\dfrac{2}{\sqrt3}\) for the base and 1 for the height so the ratio is \(\dfrac{\sqrt3}{4}\) \(:\) \(\dfrac{2}{\sqrt3}\)

so we get that z=\(-\sqrt2\) since when we square it we get 2 and when we cube it we get \(-2\sqrt2\) so we get... 

but there are no values of even number roots of negative numbers so there is no value to this solution

He already solved it

Where are "For which of the following values of n does Brenda have a winning strategy? " !

Yay! I made an account!

Hi! I'm not sure about the answer, but I tried to do your problem.  Here's my solution:

First, take the square root of both sides to get +- x^2 = 2i. Then, you will get two cases:

Case #1; x^2 = 2i: We take the square root of both sides of the equation x^2 = 2i to get +- x = 1 + i. Thus, x can be either 1 + i or -1 - i in this case.

Case #2; -x^2 = 2i: We move the terms in the equation around to get x^2 = -2i. Then, taking the square root of both sides, we get

+-x = 1 - i. Therefore, this case yields the solutions x = 1 - i and x = -1 + i.

In the end, all of the complex numbers z such that z^4 = -4 are 1 + i, -1 -i, 1 - i, and -1 + i.

Maybe this video will help:

https://artofproblemsolving.com/videos/counting/chapter2/192

Thanks!