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To everyone who joined these forums at some point, and got discouraged by the negativity and left after a while (or even got literally scared off): I'm sorry.

I wasn't good enough at encouraging people to be kinder, and removing people who refuse to be kind. Encouraging people is hard, and removing people creates conflict, and I hate conflict... so that's why I wasn't better at it.

I was a very, very sensitive teen. The atmosphere of this forum as it is now, if it had existed in 1996, would probably have upset me far more than it would have helped.

I can handle quite a lot of negativity and even abuse now, but that isn't the point. I want to help people. I want to help the people who need it the most, and I want to help people like the 1996 version of me.

I'm still figuring out the best way to do that, but as it is now, these forums are doing more harm than good, and I can't keep running them.

Thank you to the few people who have tried to understand my point of view so far. I really, really appreciate you guys. You are beautiful people.

Everyone else: If after everything I've said so far, you still don't understand my motivations, I think it's unlikely that you will. We're just too different. Maybe someday in the future it might make sense, but until then, there's no point in arguing about it. I don't have the time or the energy for arguing anymore. I will focus my time and energy on people who support me, and those who need help.


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Get your math homework solved here.
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Michio-kun Offline

Posts: 1,975
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Given 28 thank(s) in 24 post(s)
Post: #31
Re: I'm doing everyone's math homework. (old)

To be honest, for some reason I always have trouble with these sorts of things. Even in calculus 3, you go over this again (yes, I'm serious), except you use vectors but idk, I get raped with this crap.

Anyway, here's my attempt...pretty sure they're right:


The key to solving these is to use the point-slope formula: [Image: a211b4370086efa422855a2d30c453a3.png]

And here is the standard form of a linear equation:
[Image: 7e6859919d495035e99c333d04e3e0b9.png], where A, B, C are any real numbers. A and B are coefficients of x and y.


Quote:Write the standard form of the equation of the line that passes though A(5,9) and is parallel to the graph of y=5x-9

If we want a parallel line to something, in this case, a line parallel to y=5x-9, we take the slope which is m=5 and plug it into the point-slope formula. But we still need more information so that we can get a unique line, in this case, it would be the point (5,9). For the point (5,9), x=5 and y=9 both of which we plug into the point slope equation to get:

y - 9 = 5(x - 5) which simplifies to
y - 9 = 5x - 25
y = 5x - 16 which in standard form equals
-5x + y = -16

Notice we plugged in the values from (5,9) into " y1 " and " x1 ". Just remember if you're going to be substituting values into the point-slope equation that you need substitute the variables y1 and x1 and not the "x" and "y" so that you can actually get a linear equation that makes sense.

Quote:Write the standard form of the equation of the line that passes though B(-10, -5) and is perpendicular to the graph of 6x-5y=24

This is essentially the same problem except the initial conditions ask for a line that is perpendicular and not parallel. To get a perpendicular line, we follow the same procedure as above except in this case, we will have a different value for m. (Note that the variable m is usually going to mean slope.)

Okay, so to get a perpendicular line, we need to find the "opposite reciprocal" of the slope of 6x-5y=24.

First, get the equation into slope-intercept form ( [Image: d24ebc87176b242c935535a363c5fc10.png] )
6x - 5y = 24
-5y = -6x + 24 , (divide both sides by -5 )
y = (6/5)x - (24/5)

The slope of this equation is 6/5, and the opposite reciprocal is -(5/6). All you do is negate, then invert the numerator and denominator.

So our m = -(5/6)

You can do all the work if you want. You do the same thing as above. Here's the answer:

-(5/6)x + y = -(40/3)

I can do the rest a little later. It's 5:45 AM which means time for a run.


To a mind that is still, the whole universe surrenders. - Chuang-tzu
The quieter you become, the more you can hear. - Baba Ram Dass
The whole moon and the entire sky are reflected in one dewdrop on the grass. - Dogen
Great Faith. Great Doubt. Great Effort. - The three qualities necessary for training. - Zen saying
Possessing much knowledge is like having a thousand foot fishing line with a hook, but the fish is always an inch beyond the hook. - Zen saying
09-22-2008 08:43 PM
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Re: I'm doing everyone's math homework. (old) - Michio-kun - 09-22-2008 08:43 PM

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