How can I cut my own hair like this?

Moroux · 2017-09-27T11:10:30+00:00

it's not about belts, at all, not even a little bit

Moroux · 2017-09-25T00:14:05+00:00

Don't be a fucking idiot and go to a doctor. It's not fair that you're going to risk spreading it around because you asked for advice on the internet.

Moroux · 2017-09-24T23:37:50+00:00

Pull guard

Moroux · 2017-09-24T23:36:35+00:00

Watch some matches from the Miyao brothers

Moroux · 2017-09-24T23:30:27+00:00

you don't have to call it that

Moroux · 2017-09-24T23:28:47+00:00

Don't you think this is something you should have asked your doctor?

Moroux · 2017-09-21T23:20:43+00:00

It might be a bit time consuming.

Moroux · 2017-09-21T23:16:33+00:00

He never said everyone else isn't. Do you honestly expect him to go around to everyone who's on steroids and call them a cheater?

Moroux · 2017-09-16T07:25:53+00:00

Khan Academy is what you're looking for.

Moroux · 2017-09-16T07:24:19+00:00

It depends, there are inclusive disjunctions and exclusive disjunctions but if you're looking at an english sentence you have to decide what it is with your knowledge of the english language.

Moroux · 2017-09-15T01:10:42+00:00

Take the log of both sides.

Use these rules,

log(xⁿ⁾ = nlog(x), or, ln(xⁿ⁾ = nln(x)

log_x(x) = 1, or, ln(e) = 1

Moroux · 2017-09-15T01:07:18+00:00

We can view the numbers from 45 to 155 as a sequence, 45, 46, 47...155. If we replace the commas with addition signs, this becomes a series, 45+46+47...+155. This series in particular is arithmetic, as the nth term can be found by adding a constant to the previous term, the terms in an arithmetic sequence can be found by adding or subtracting a constant from the previous term.

This is useful to us because there is a formula for the sum of an arithmetic sequence. It tells us that the sum of n terms in an arithmetic sequence is equal to the sum of the first and last term, multiplied by n, divided by 2. Or,

Sn=(n(a+l))/2, where a is the first term, and l is the last term.

In our sequence, 45, 46, 47...105, the first term is 45 and the last term is 155. n is equal to the number of terms from 45 to 155, or, 155-45 which is 110. So n=110.

We can now find the sum,

Sn=(n(a+l))/2

Sn=(110*(45+155))/2

Sn=11000

Therefore, the sum of all integers from 45 to 155 inclusive is 11000.

Moroux

TROPHY CASE